problem solving to equations

Solving Equations

What is an equation.

An equation says that two things are equal. It will have an equals sign "=" like this:

That equations says:

what is on the left (x − 2)  equals  what is on the right (4)

So an equation is like a statement " this equals that "

What is a Solution?

A Solution is a value we can put in place of a variable (such as x ) that makes the equation true .

Example: x − 2 = 4

When we put 6 in place of x we get:

which is true

So x = 6 is a solution.

How about other values for x ?

• For x=5 we get "5−2=4" which is not true , so x=5 is not a solution .
• For x=9 we get "9−2=4" which is not true , so x=9 is not a solution .

In this case x = 6 is the only solution.

You might like to practice solving some animated equations .

More Than One Solution

There can be more than one solution.

Example: (x−3)(x−2) = 0

When x is 3 we get:

(3−3)(3−2) = 0 × 1 = 0

And when x is 2 we get:

(2−3)(2−2) = (−1) × 0 = 0

which is also true

So the solutions are:

x = 3 , or x = 2

When we gather all solutions together it is called a Solution Set

The above solution set is: {2, 3}

Solutions Everywhere!

Some equations are true for all allowed values and are then called Identities

Example: sin(−θ) = −sin(θ) is one of the Trigonometric Identities

Let's try θ = 30°:

sin(−30°) = −0.5 and

−sin(30°) = −0.5

So it is true for θ = 30°

Let's try θ = 90°:

sin(−90°) = −1 and

−sin(90°) = −1

So it is also true for θ = 90°

Is it true for all values of θ ? Try some values for yourself!

How to Solve an Equation

There is no "one perfect way" to solve all equations.

A Useful Goal

But we often get success when our goal is to end up with:

x = something

In other words, we want to move everything except "x" (or whatever name the variable has) over to the right hand side.

Example: Solve 3x−6 = 9

Now we have x = something ,

and a short calculation reveals that x = 5

Like a Puzzle

In fact, solving an equation is just like solving a puzzle. And like puzzles, there are things we can (and cannot) do.

Here are some things we can do:

• Add or Subtract the same value from both sides
• Clear out any fractions by Multiplying every term by the bottom parts
• Divide every term by the same nonzero value
• Combine Like Terms
• Expanding (the opposite of factoring) may also help
• Recognizing a pattern, such as the difference of squares
• Sometimes we can apply a function to both sides (e.g. square both sides)

Example: Solve √(x/2) = 3

And the more "tricks" and techniques you learn the better you will get.

Special Equations

There are special ways of solving some types of equations. Learn how to ...

• solve Quadratic Equations
• solve Radical Equations
• solve Equations with Sine, Cosine and Tangent

Check Your Solutions

You should always check that your "solution" really is a solution.

How To Check

Take the solution(s) and put them in the original equation to see if they really work.

Example: solve for x:

2x x − 3 + 3 = 6 x − 3     (x≠3)

We have said x≠3 to avoid a division by zero.

Let's multiply through by (x − 3) :

2x + 3(x−3) = 6

Bring the 6 to the left:

2x + 3(x−3) − 6 = 0

Expand and solve:

2x + 3x − 9 − 6 = 0

5x − 15 = 0

5(x − 3) = 0

Which can be solved by having x=3

Let us check x=3 using the original question:

2 × 3 3 − 3 + 3  =   6 3 − 3

Hang On: 3 − 3 = 0 That means dividing by Zero!

And anyway, we said at the top that x≠3 , so ...

x = 3 does not actually work, and so:

There is No Solution!

That was interesting ... we thought we had found a solution, but when we looked back at the question we found it wasn't allowed!

This gives us a moral lesson:

"Solving" only gives us possible solutions, they need to be checked!

• Note down where an expression is not defined (due to a division by zero, the square root of a negative number, or some other reason)
• Show all the steps , so it can be checked later (by you or someone else)

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Solving equations

Here you will learn about solving equations, including linear and quadratic algebraic equations, and how to solve them.

Students will first learn about solving equations in grade 8 as a part of expressions and equations, and again in high school as a part of reasoning with equations and inequalities.

What is solving an equation?

Solving equations is a step-by-step process to find the value of the variable. A variable is the unknown part of an equation, either on the left or right side of the equals sign. Sometimes, you need to solve multi-step equations which contain algebraic expressions.

To do this, you must use the order of operations, which is a systematic approach to equation solving. When you use the order of operations, you first solve any part of an equation located within parentheses. An equation is a mathematical expression that contains an equals sign.

For example,

\begin{aligned}y+6&=11\\\\ 3(x-3)&=12\\\\ \cfrac{2x+2}{4}&=\cfrac{x-3}{3}\\\\ 2x^{2}+3&x-2=0\end{aligned}

There are two sides to an equation, with the left side being equal to the right side. Equations will often involve algebra and contain unknowns, or variables, which you often represent with letters such as x or y.

You can solve simple equations and more complicated equations to work out the value of these unknowns. They could involve fractions, decimals or integers.

Common Core State Standards

How does this relate to 8 th grade and high school math?

• Grade 8 – Expressions and Equations (8.EE.C.7) Solve linear equations in one variable.
• High school – Reasoning with Equations and Inequalities (HSA.REI.B.3) Solve linear equations and inequalities in one variable, including equations with coefficients represented by letters.

[FREE] Solving Equations Worksheet (Grade 6 to 8)

Use this worksheet to check your grade 6 to 8 students’ understanding of solving equations. 15 questions with answers to identify areas of strength and support!

How to solve equations

In order to solve equations, you need to work out the value of the unknown variable by adding, subtracting, multiplying or dividing both sides of the equation by the same value.

• Combine like terms .
• Simplify the equation by using the opposite operation to both sides.
• Isolate the variable on one side of the equation.

Solving equations examples

Example 1: solve equations involving like terms.

Solve for x.

Combine like terms.

Combine the q terms on the left side of the equation. To do this, subtract 4q from both sides.

The goal is to simplify the equation by combining like terms. Subtracting 4q from both sides helps achieve this.

After you combine like terms, you are left with q=9-4q.

2 Simplify the equation by using the opposite operation on both sides.

Add 4q to both sides to isolate q to one side of the equation.

The objective is to have all the q terms on one side. Adding 4q to both sides accomplishes this.

After you move the variable to one side of the equation, you are left with 5q=9.

3 Isolate the variable on one side of the equation.

Divide both sides of the equation by 5 to solve for q.

Dividing by 5 allows you to isolate q to one side of the equation in order to find the solution. After dividing both sides of the equation by 5, you are left with q=1 \cfrac{4}{5} \, .

Example 2: solve equations with variables on both sides

Combine the v terms on the same side of the equation. To do this, add 8v to both sides.

7v+8v=8-8v+8v

After combining like terms, you are left with the equation 15v=8.

Simplify the equation by using the opposite operation on both sides and isolate the variable to one side.

Divide both sides of the equation by 15 to solve for v. This step will isolate v to one side of the equation and allow you to solve.

15v \div 15=8 \div 15

The final solution to the equation 7v=8-8v is \cfrac{8}{15} \, .

Example 3: solve equations with the distributive property

Combine like terms by using the distributive property.

The 3 outside the parentheses needs to be multiplied by both terms inside the parentheses. This is called the distributive property.

\begin{aligned}& 3 \times c=3 c \\\\ & 3 \times(-5)=-15 \\\\ &3 c-15-4=2\end{aligned}

Once the 3 is distributed on the left side, rewrite the equation and combine like terms. In this case, the like terms are the constants on the left, –15 and –4. Subtract –4 from –15 to get –19.

Simplify the equation by using the opposite operation on both sides.

The goal is to isolate the variable, c, on one side of the equation. By adding 19 to both sides, you move the constant term to the other side.

\begin{aligned}& 3 c-19+19=2+19 \\\\ & 3 c=21\end{aligned}

Isolate the variable to one side of the equation.

To solve for c, you want to get c by itself.

Dividing both sides by 3 accomplishes this.

On the left side, \cfrac{3c}{3} simplifies to c, and on the right, \cfrac{21}{3} simplifies to 7.

The final solution is c=7.

As an additional step, you can plug 7 back into the original equation to check your work.

Example 4: solve linear equations

Combine like terms by simplifying.

Using steps to solve, you know that the goal is to isolate x to one side of the equation. In order to do this, you must begin by subtracting from both sides of the equation.

\begin{aligned} & 2x+5=15 \\\\ & 2x+5-5=15-5 \\\\ & 2x=10 \end{aligned}

Continue to simplify the equation by using the opposite operation on both sides.

Continuing with steps to solve, you must divide both sides of the equation by 2 to isolate x to one side.

\begin{aligned} & 2x \div 2=10 \div 2 \\\\ & x= 5 \end{aligned}

Isolate the variable to one side of the equation and check your work.

Plugging in 5 for x in the original equation and making sure both sides are equal is an easy way to check your work. If the equation is not equal, you must check your steps.

\begin{aligned}& 2(5)+5=15 \\\\ & 10+5=15 \\\\ & 15=15\end{aligned}

Example 5: solve equations by factoring

Solve the following equation by factoring.

Combine like terms by factoring the equation by grouping.

Multiply the coefficient of the quadratic term by the constant term.

2 x (-20) = -40

Look for two numbers that multiply to give you –40 and add up to the coefficient of 3. In this case, the numbers are 8 and –5 because 8 x -5=–40, and 8+–5=3.

Split the middle term using those two numbers, 8 and –5. Rewrite the middle term using the numbers 8 and –5.

2x^2+8x-5x-20=0

Group the terms in pairs and factor out the common factors.

2x^2+8x-5x-20=2x(x + 4)-5(x+4)=0

Now, you’ve factored the equation and are left with the following simpler equations 2x-5 and x+4.

This step relies on understanding the zero product property, which states that if two numbers multiply to give zero, then at least one of those numbers must equal zero.

Let’s relate this back to the factored equation (2x-5)(x+4)=0

Because of this property, either (2x-5)=0 or (x+4)=0

Isolate the variable for each equation and solve.

When solving these simpler equations, remember that you must apply each step to both sides of the equation to maintain balance.

\begin{aligned}& 2 x-5=0 \\\\ & 2 x-5+5=0+5 \\\\ & 2 x=5 \\\\ & 2 x \div 2=5 \div 2 \\\\ & x=\cfrac{5}{2} \end{aligned}

\begin{aligned}& x+4=0 \\\\ & x+4-4=0-4 \\\\ & x=-4\end{aligned}

The solution to this equation is x=\cfrac{5}{2} and x=-4.

Example 6: solve quadratic equations

Solve the following quadratic equation.

Combine like terms by factoring the quadratic equation when terms are isolated to one side.

To factorize a quadratic expression like this, you need to find two numbers that multiply to give -5 (the constant term) and add to give +2 (the coefficient of the x term).

The two numbers that satisfy this are -1 and +5.

So you can split the middle term 2x into -1x+5x: x^2-1x+5x-5-1x+5x

Now you can take out common factors x(x-1)+5(x-1).

And since you have a common factor of (x-1), you can simplify to (x+5)(x-1).

The numbers -1 and 5 allow you to split the middle term into two terms that give you common factors, allowing you to simplify into the form (x+5)(x-1).

Let’s relate this back to the factored equation (x+5)(x-1)=0.

Because of this property, either (x+5)=0 or (x-1)=0.

Now, you can solve the simple equations resulting from the zero product property.

\begin{aligned}& x+5=0 \\\\ & x+5-5=0-5 \\\\ & x=-5 \\\\\\ & x-1=0 \\\\ & x-1+1=0+1 \\\\ & x=1\end{aligned}

The solutions to this quadratic equation are x=1 and x=-5.

Teaching tips for solving equations

• Use physical manipulatives like balance scales as a visual aid. Show how you need to keep both sides of the equation balanced, like a scale. Add or subtract the same thing from both sides to keep it balanced when solving. Use this method to practice various types of equations.
• Emphasize the importance of undoing steps to isolate the variable. If you are solving for x and 3 is added to x, subtracting 3 undoes that step and isolates the variable x.
• Relate equations to real-world, relevant examples for students. For example, word problems about tickets for sports games, cell phone plans, pizza parties, etc. can make the concepts click better.
• Allow time for peer teaching and collaborative problem solving. Having students explain concepts to each other, work through examples on whiteboards, etc. reinforces the process and allows peers to ask clarifying questions. This type of scaffolding would be beneficial for all students, especially English-Language Learners. Provide supervision and feedback during the peer interactions.

Easy mistakes to make

• Forgetting to distribute or combine like terms One common mistake is neglecting to distribute a number across parentheses or combine like terms before isolating the variable. This error can lead to an incorrect simplified form of the equation.
• Misapplying the distributive property Incorrectly distributing a number across terms inside parentheses can result in errors. Students may forget to multiply each term within the parentheses by the distributing number, leading to an inaccurate equation.
• Failing to perform the same operation on both sides It’s crucial to perform the same operation on both sides of the equation to maintain balance. Forgetting this can result in an imbalanced equation and incorrect solutions.
• Making calculation errors Simple arithmetic mistakes, such as addition, subtraction, multiplication, or division errors, can occur during the solution process. Checking calculations is essential to avoid errors that may propagate through the steps.
• Ignoring fractions or misapplying operations When fractions are involved, students may forget to multiply or divide by the common denominator to eliminate them. Misapplying operations on fractions can lead to incorrect solutions or complications in the final answer.

Related math equations lessons

• Math equations
• Rearranging equations
• How to find the equation of a line
• Solve equations with fractions
• Linear equations
• Writing linear equations
• Substitution
• Identity math
• One step equation

Practice solving equations questions

1. Solve 4x-2=14.

Add 2 to both sides.

Divide both sides by 4.

2. Solve 3x-8=x+6.

Add 8 to both sides.

Subtract x from both sides.

Divide both sides by 2.

3. Solve 3(x+3)=2(x-2).

Expanding the parentheses.

Subtract 9 from both sides.

Subtract 2x from both sides.

4. Solve \cfrac{2 x+2}{3}=\cfrac{x-3}{2}.

Multiply by 6 (the lowest common denominator) and simplify.

Expand the parentheses.

Subtract 4 from both sides.

Subtract 3x from both sides.

5. Solve \cfrac{3 x^{2}}{2}=24.

Multiply both sides by 2.

Divide both sides by 3.

Square root both sides.

6. Solve by factoring:

Use factoring to find simpler equations.

Set each set of parentheses equal to zero and solve.

x=3 or x=10

Solving equations FAQs

The first step in solving a simple linear equation is to simplify both sides by combining like terms. This involves adding or subtracting terms to isolate the variable on one side of the equation.

Performing the same operation on both sides of the equation maintains the equality. This ensures that any change made to one side is also made to the other, keeping the equation balanced and preserving the solutions.

To handle variables on both sides of the equation, start by combining like terms on each side. Then, move all terms involving the variable to one side by adding or subtracting, and simplify to isolate the variable. Finally, perform any necessary operations to solve for the variable.

To deal with fractions in an equation, aim to eliminate them by multiplying both sides of the equation by the least common denominator. This helps simplify the equation and make it easier to isolate the variable. Afterward, proceed with the regular steps of solving the equation.

The next lessons are

• Inequalities
• Types of graph
• Math formulas
• Coordinate plane

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Chapter 1: Solving Equations and Inequalities

Problem solving, learning objectives.

• Translate words into algebraic expressions and equations
• Define a process for solving word problems
• Apply the steps for solving word problems to distance, rate, and time problems
• Apply the steps for solving word problems to interest rate problems
• Evaluate a formula using substitution
• Rearrange formulas to isolate specific variables
• Identify an unknown given a formula
• Apply the steps for solving word problems to geometry problems
• Use the formula for converting between Fahrenheit and Celsius

Define a Process for Problem Solving

Word problems can be tricky. Often it takes a bit of practice to convert an English sentence into a mathematical sentence, which is one of the first steps to solving word problems. In the table below, words or phrases commonly associated with mathematical operators are categorized. Word problems often contain these or similar words, so it’s good to see what mathematical operators are associated with them.

Some examples follow:

• [latex]x\text{ is }5[/latex]  becomes [latex]x=5[/latex]
• Three more than a number becomes [latex]x+3[/latex]
• Four less than a number becomes [latex]x-4[/latex]
• Double the cost becomes [latex]2\cdot\text{ cost }[/latex]
• Groceries and gas together for the week cost $250 means [latex]\text{ groceries }+\text{ gas }=250[/latex]
• The difference of 9 and a number becomes [latex]9-x[/latex]. Notice how 9 is first in the sentence and the expression

Let’s practice translating a few more English phrases into algebraic expressions.

Translate the table into algebraic expressions:

In this example video, we show how to translate more words into mathematical expressions.

The power of algebra is how it can help you model real situations in order to answer questions about them.

Here are some steps to translate problem situations into algebraic equations you can solve. Not every word problem fits perfectly into these steps, but they will help you get started.

• Read and understand the problem.
• Determine the constants and variables in the problem.
• Translate words into algebraic expressions and equations.
• Write an equation to represent the problem.
• Solve the equation.
• Check and interpret your answer. Sometimes writing a sentence helps.

Twenty-eight less than five times a certain number is 232. What is the number?

Following the steps provided:

• Read and understand: we are looking for a number.
• Constants and variables: 28 and 232 are constants, “a certain number” is our variable because we don’t know its value, and we are asked to find it. We will call it x.
• Translate:  five times a certain number translates to [latex]5x[/latex] Twenty-eight less than five times a certain number translates to [latex]5x-28[/latex] because subtraction is built backward. is 232 translates to [latex]=232[/latex] because “is” is associated with equals.
• Write an equation:  [latex]5x-28=232[/latex]

[latex]\begin{array}{r}5x-28=232\\5x=260\\x=52\,\,\,\end{array}[/latex]

[latex]\begin{array}{r}5\left(52\right)-28=232\\5\left(52\right)=260\\260=260\end{array}[/latex].

In the video that follows, we show another example of how to translate a sentence into a mathematical expression using a problem solving method.

Another type of number problem involves consecutive numbers. Consecutive numbers are numbers that come one after the other, such as 3, 4, 5. If we are looking for several consecutive numbers it is important to first identify what they look like with variables before we set up the equation.

For example, let’s say I want to know the next consecutive integer after 4. In mathematical terms, we would add 1 to 4 to get 5. We can generalize this idea as follows: the consecutive integer of any number, x , is [latex]x+1[/latex]. If we continue this pattern we can define any number of consecutive integers from any starting point. The following table shows how to describe four consecutive integers using algebraic notation.

We apply the idea of consecutive integers to solving a word problem in the following example.

The sum of three consecutive integers is 93. What are the integers?

• Read and understand:  We are looking for three numbers, and we know they are consecutive integers.
• Constants and Variables:  93 is a constant. The first integer we will call x . Second: [latex]x+1[/latex] Third: [latex]x+2[/latex]
• Translate:  The sum of three consecutive integers translates to [latex]x+\left(x+1\right)+\left(x+2\right)[/latex], based on how we defined the first, second, and third integers. Notice how we placed parentheses around the second and third integers. This is just to make each integer more distinct. is 93 translates to [latex]=93[/latex] because is is associated with equals.
• Write an equation:  [latex]x+\left(x+1\right)+\left(x+2\right)=93[/latex]

[latex]x+x+1+x+2=93[/latex]

Combine like terms, simplify, and solve.

[latex]\begin{array}{r}x+x+1+x+2=93\\3x+3 = 93\\\underline{-3\,\,\,\,\,-3}\\3x=90\\\frac{3x}{3}=\frac{90}{3}\\x=30\end{array}[/latex]

• Check and Interpret: Okay, we have found a value for x . We were asked to find the value of three consecutive integers, so we need to do a couple more steps. Remember how we defined our variables: The first integer we will call [latex]x[/latex], [latex]x=30[/latex] Second: [latex]x+1[/latex] so [latex]30+1=31[/latex] Third: [latex]x+2[/latex] so [latex]30+2=32[/latex] The three consecutive integers whose sum is [latex]93[/latex] are [latex]30\text{, }31\text{, and }32[/latex]

There is often a well-known formula or relationship that applies to a word problem. For example, if you were to plan a road trip, you would want to know how long it would take you to reach your destination. [latex]d=rt[/latex] is a well-known relationship that associates distance traveled, the rate at which you travel, and how long the travel takes.

Distance, Rate, and Time

If you know two of the quantities in the relationship [latex]d=rt[/latex], you can easily find the third using methods for solving linear equations. For example, if you know that you will be traveling on a road with a speed limit of [latex]30\frac{\text{ miles }}{\text{ hour }}[/latex] for 2 hours, you can find the distance you would travel by multiplying rate times time or [latex]\left(30\frac{\text{ miles }}{\text{ hour }}\right)\left(2\text{ hours }\right)=60\text{ miles }[/latex].

We can generalize this idea depending on what information we are given and what we are looking for. For example, if we need to find time, we could solve the [latex]d=rt[/latex] equation for t using division:

[latex]d=rt\\\frac{d}{r}=t[/latex]

Likewise, if we want to find rate, we can isolate r using division:

[latex]d=rt\\\frac{d}{t}=r[/latex]

In the following examples you will see how this formula is applied to answer questions about ultra marathon running.

Ultra marathon running (defined as anything longer than 26.2 miles) is becoming very popular among women even though it remains a male-dominated niche sport. Ann Trason has broken twenty world records in her career. One such record was the American River 50-mile Endurance Run which begins in Sacramento, California, and ends in Auburn, California. [1] In 1993 Trason finished the run with a time of 6:09:08.  The men’s record for the same course was set in 1994 by Tom Johnson who finished the course with a time of 5:33:21. [2]

In the next examples we will use the [latex]d=rt[/latex] formula to answer the following questions about the two runners.

What was each runner’s rate for their record-setting runs?

By the time Johnson had finished, how many more miles did Trason have to run?

How much further could Johnson have run if he had run as long as Trason?

What was each runner’s time for running one mile?

To make answering the questions easier, we will round the two runners’ times to 6 hours and 5.5 hours.

Read and Understand:  We are looking for rate and we know distance and time, so we can use the idea: [latex]d=rt\\\frac{d}{t}=r[/latex]

Define and Translate: Because there are two runners, making a table to organize this information helps. Note how we keep units to help us keep track of what how all the terms are related to each other.

Write and Solve:

Trason’s rate:

[latex]\begin{array}{c}d=rt\\\\50\text{ miles }=\text{r}\left(6\text{ hours }\right)\\\frac{50\text{ miles }}{6\text{ hours }}=\frac{8.33\text{ miles }}{\text{ hour }}\end{array}[/latex].

(rounded to two decimal places)

Johnson’s rate:

[latex]\begin{array}{c}d=rt\\\\,\,\,\,\,\,\,50\text{ miles }=\text{r}\left(5.5\text{ hours }\right)\\\frac{50\text{ miles }}{6\text{ hours }}=\frac{9.1\text{ miles }}{\text{ hour }}\end{array}[/latex]

Check and Interpret:

We can fill in our table with this information.

Now that we know each runner’s rate we can answer the second question.

Here is the table we created for reference:

Read and Understand:  We are looking for how many miles Trason still had on the trail when Johnson had finished after 5.5 hours. This is a distance, and we know rate and time.

Define and Translate:  We can use the formula [latex]d=rt[/latex] again. This time the unknown is d , and the time Trason had run is 5.5 hours.

[latex]\begin{array}{l}d=rt\\\\d=8.33\frac{\text{ miles }}{\text{ hour }}\left(5.5\text{ hours }\right)\\\\d=45.82\text{ miles }\end{array}[/latex].

Have we answered the question? We were asked to find how many more miles she had to run after 5.5 hours.  What we have found is how long she had run after 5.5 hours. We need to subtract [latex]d=45.82\text{ miles }[/latex] from the total distance of the course.

[latex]50\text{ miles }-45.82\text{ miles }=1.48\text{ miles }[/latex]

The third question is similar to the second. Now that we know each runner’s rate, we can answer questions about individual distances or times.

Read and Understand:  The word further implies we are looking for a distance.

Define and Translate:  We can use the formula [latex]d=rt[/latex] again. This time the unknown is d , the time is 6 hours, and Johnson’s rate is [latex]9.1\frac{\text{ miles }}{\text{ hour }}[/latex]

[latex]\begin{array}{l}d=rt\\\\d=9.1\frac{\text{ miles }}{\text{ hour }}\left(6\text{ hours }\right)\\\\d=54.6\text{ miles }\end{array}[/latex].

Have we answered the question? We were asked to find how many more miles Johnson would have run if he had run at his rate of [latex]9.1\frac{\text{ miles }}{\text{ hour }}[/latex] for 6 hours.

Johnson would have run 54.6 miles, so that’s 4.6 more miles than than he ran for the race.

Now we will tackle the last question where we are asked to find a time for each runner.

Read and Understand:  we are looking for time, and this time our distance has changed from 50 miles to 1 mile, so we can use

Define and Translate: we can use the formula [latex]d=rt[/latex] again. This time the unknown is t , the distance is 1 mile, and we know each runner’s rate. It may help to create a new table:

We will need to divide to isolate time.

[latex]\begin{array}{c}d=rt\\\\1\text{ mile }=8.33\frac{\text{ miles }}{\text{ hour }}\left(t\text{ hours }\right)\\\\\frac{1\text{ mile }}{\frac{8.33\text{ miles }}{\text{ hour }}}=t\text{ hours }\\\\0.12\text{ hours }=t\end{array}[/latex].

0.12 hours is about 7.2 minutes, so Trason’s time for running one mile was about 7.2 minutes. WOW! She did that for 6 hours!

[latex]\begin{array}{c}d=rt\\\\1\text{ mile }=9.1\frac{\text{ miles }}{\text{ hour }}\left(t\text{ hours }\right)\\\\\frac{1\text{ mile }}{\frac{9.1\text{ miles }}{\text{ hour }}}=t\text{ hours }\\\\0.11\text{ hours }=t\end{array}[/latex].

0.11 hours is about 6.6 minutes, so Johnson’s time for running one mile was about 6.6 minutes. WOW! He did that for 5.5 hours!

Have we answered the question? We were asked to find how long it took each runner to run one mile given the rate at which they ran the whole 50-mile course.  Yes, we answered our question.

Trason’s mile time was [latex]7.2\frac{\text{minutes}}{\text{mile}}[/latex] and Johnsons’ mile time was [latex]6.6\frac{\text{minutes}}{\text{mile}}[/latex]

In the following video, we show another example of answering many rate questions given distance and time.

Simple Interest

In order to entice customers to invest their money, many banks will offer interest-bearing accounts. The accounts work like this: a customer deposits a certain amount of money (called the Principal, or P ), which then grows slowly according to the interest rate ( R , measured in percent) and the length of time ( T , usually measured in months) that the money stays in the account. The amount earned over time is called the interest ( I ), which is then given to the customer.

The simplest way to calculate interest earned on an account is through the formula [latex]\displaystyle I=P\,\cdot \,R\,\cdot \,T[/latex].

If we know any of the three amounts related to this equation, we can find the fourth. For example, if we want to find the time it will take to accrue a specific amount of interest, we can solve for T using division:

[latex]\displaystyle\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,I=P\,\cdot \,R\,\cdot \,T\\\\ \frac{I}{{P}\,\cdot \,R}=\frac{P\cdot\,R\,\cdot \,T}{\,P\,\cdot \,R}\\\\\,\,\,\,\,\,\,\,\,\,\,{T}=\frac{I}{\,R\,\cdot \,T}\end{array}[/latex]

Below is a table showing the result of solving for each individual variable in the formula.

In the next examples, we will show how to substitute given values into the simple interest formula, and decipher which variable to solve for.

If a customer deposits a principal of $2000 at a monthly rate of 0.7%, what is the total amount that she has after 24 months?

Substitute in the values given for the Principal, Rate, and Time.

[latex]\displaystyle\begin{array}{l}I=P\,\cdot \,R\,\cdot \,T\\I=2000\cdot 0.7\%\cdot 24\end{array}[/latex]

Rewrite 0.7% as the decimal 0.007, then multiply.

[latex]\begin{array}{l}I=2000\cdot 0.007\cdot 24\\I=336\end{array}[/latex]

Add the interest and the original principal amount to get the total amount in her account.

[latex] \displaystyle 2000+336=2336[/latex]

She has $2336 after 24 months.

The following video shows another example of finding an account balance after a given amount of time, principal invested, and a rate.

In the following example you will see why it is important to make sure the units of the interest rate match the units of time when using the simple interest formula.

Alex invests $600 at 3.25% monthly interest for 3 years. What amount of interest has Alex earned?

Read and Understand: The question asks for an amount, so we can substitute what we are given into the simple interest formula [latex]I=P\,\cdot \,R\,\cdot \,T[/latex]

Define and Translate:  we know P, R, and T so we can use substitution. R  = 0.0325, P = $600, and T = 3 years. We have to be careful! R is in months, and T is in years.  We need to change T into months because we can’t change the rate—it is set by the bank.

[latex]{T}=3\text{ years }\cdot12\frac{\text{ months }}{ year }=36\text{ months }[/latex]

Substitute the given values into the formula.

[latex]\begin{array}{l} I=P\,\cdot \,R\,\cdot \,T\\\\I=600\,\cdot \,0.035\,\cdot \,36\\\\{I}=756\end{array}[/latex]

We were asked what amount Alex earned, which is the amount provided by the formula. In the previous example we were asked the total amount in the account, which included the principal and interest earned.

Alex has earned $756.

After 10 years, Jodi’s account balance has earned $1080 in interest. The rate on the account is 0.09% monthly. What was the original amount she invested in the account?

Read and Understand: The question asks for the original amount invested, the principal. We are given a length of time in years, and an interest rate in months, and the amount of interest earned.

Define and Translate:  we know I = $1080, R = 0.009, and T = 10 years so we can use [latex]{P}=\frac{I}{{R}\,\cdot \,T}[/latex]

We also need to make sure the units on the interest rate and the length of time match, and they do not. We need to change time into months again.

[latex]{T}=10\text{ years }\cdot12\frac{\text{ months }}{ year }=120\text{ months }[/latex]

Substitute the given values into the formula

[latex]\begin{array}{l}{P}=\frac{I}{{R}\,\cdot \,T}\\\\{P}=\frac{1080}{{0.009}\,\cdot \,120}\\\\{P}=\frac{1080}{1.08}=1000\end{array}[/latex]

We were asked to find the principal given the amount of interest earned on an account.  If we substitute P = $1000 into the formula [latex]I=P\,\cdot \,R\,\cdot \,T[/latex] we get

[latex]I=1000\,\cdot \,0.009\,\cdot \,120\\I=1080[/latex]

Our solution checks out. Jodi invested $1000.

Further Applications of Linear Equations

Formulas come up in many different areas of life. We have seen the formula that relates distance, rate, and time and the formula for simple interest on an investment. In this section we will look further at formulas and see examples of formulas for dimensions of geometric shapes as well as the formula for converting temperature between Fahrenheit and Celsius.

There are many geometric shapes that have been well studied over the years. We know quite a bit about circles, rectangles, and triangles. Mathematicians have proven many formulas that describe the dimensions of geometric shapes including area, perimeter, surface area, and volume.

Perimeter is the distance around an object. For example, consider a rectangle with a length of 8 and a width of 3. There are two lengths and two widths in a rectangle (opposite sides), so we add [latex]8+8+3+3=22[/latex]. Since there are two lengths and two widths in a rectangle, you may find the perimeter of a rectangle using the formula [latex]{P}=2\left({L}\right)+2\left({W}\right)[/latex] where

In the following example, we will use the problem-solving method we developed to find an unknown width using the formula for the perimeter of a rectangle. By substituting the dimensions we know into the formula, we will be able to isolate the unknown width and find our solution.

You want to make another garden box the same size as the one you already have. You write down the dimensions of the box and go to the lumber store to buy some boards. When you get there you realize you didn’t write down the width dimension—only the perimeter and length. You want the exact dimensions so you can have the store cut the lumber for you.

Here is what you have written down:

Perimeter = 16.4 feet Length = 4.7 feet

Can you find the dimensions you need to have your boards cut at the lumber store? If so, how many boards do you need and what lengths should they be?

Read and Understand:  We know perimeter = 16.4 feet and length = 4.7 feet, and we want to find width.

Define and Translate:

Define the known and unknown dimensions:

First we will substitute the dimensions we know into the formula for perimeter:

[latex]\begin{array}{l}\,\,\,\,\,P=2{W}+2{L}\\\\16.4=2\left(w\right)+2\left(4.7\right)\end{array}[/latex]

Then we will isolate w to find the unknown width.

[latex]\begin{array}{l}16.4=2\left(w\right)+2\left(4.7\right)\\16.4=2{w}+9.4\\\underline{-9.4\,\,\,\,\,\,\,\,\,\,\,\,\,-9.4}\\\,\,\,\,\,\,\,7=2\left(w\right)\\\,\,\,\,\,\,\,\frac{7}{2}=\frac{2\left(w\right)}{2}\\\,\,\,\,3.5=w\end{array}[/latex]

Write the width as a decimal to make cutting the boards easier and replace the units on the measurement, or you won’t get the right size of board!

If we replace the width we found, [latex]w=3.5\text{ feet }[/latex] into the formula for perimeter with the dimensions we wrote down, we can check our work:

[latex]\begin{array}{l}\,\,\,\,\,{P}=2\left({L}\right)+2\left({W}\right)\\\\{16.4}=2\left({4.7}\right)+2\left({3.5}\right)\\\\{16.4}=9.4+7\\\\{16.4}=16.4\end{array}[/latex]

Our calculation for width checks out. We need to ask for 2 boards cut to 3.5 feet and 2 boards cut to 4.7 feet so we can make the new garden box.

This video shows a similar garden box problem.

We could have isolated the w in the formula for perimeter before we solved the equation, and if we were going to use the formula many times, it could save a lot of time. The next example shows how to isolate a variable in a formula before substituting known dimensions or values into the formula.

Isolate the term containing the variable, w, from the formula for the perimeter of a rectangle :  

[latex]{P}=2\left({L}\right)+2\left({W}\right)[/latex].

First, isolate the term with  w by subtracting 2 l from both sides of the equation.

[latex] \displaystyle \begin{array}{l}\,\,\,\,\,\,\,\,\,\,p\,=\,\,\,\,2l+2w\\\underline{\,\,\,\,\,-2l\,\,\,\,\,-2l\,\,\,\,\,\,\,\,\,\,\,}\\p-2l=\,\,\,\,\,\,\,\,\,\,\,\,\,2w\end{array}[/latex]

Next, clear the coefficient of w by dividing both sides of the equation by 2.

[latex]\displaystyle \begin{array}{l}\underline{p-2l}=\underline{2w}\\\,\,\,\,\,\,2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\\ \,\,\,\frac{p-2l}{2}\,\,=\,\,w\\\,\,\,\,\,\,\,\,\,\,\,w=\frac{p-2l}{2}\end{array}[/latex]

You can rewrite the equation so the isolated variable is on the left side.

[latex]w=\frac{p-2l}{2}[/latex]

The area of a triangle is given by [latex] A=\frac{1}{2}bh[/latex] where

A = area b = the length of the base h = the height of the triangle

Remember that when two variables or a number and a variable are sitting next to each other without a mathematical operator between them, you can assume they are being multiplied. This can seem frustrating, but you can think of it like mathematical slang. Over the years, people who use math frequently have just made that shortcut enough that it has been adopted as convention.

In the next example we will use the formula for area of a triangle to find a missing dimension, as well as use substitution to solve for the base of a triangle given the area and height.

Find the base ( b) of a triangle with an area ( A ) of 20 square feet and a height ( h) of 8 feet.

Use the formula for the area of a triangle, [latex] {A}=\frac{{1}}{{2}}{bh}[/latex] .

Substitute the given lengths into the formula and solve for  b.

[latex]\displaystyle \begin{array}{l}\,\,A=\frac{1}{2}bh\\\\20=\frac{1}{2}b\cdot 8\\\\20=\frac{8}{2}b\\\\20=4b\\\\\frac{20}{4}=\frac{4b}{4}\\\\ \,\,\,5=b\end{array}[/latex]

The base of the triangle measures 5 feet.

We can rewrite the formula in terms of b or h as we did with perimeter previously. This probably seems abstract, but it can help you develop your equation-solving skills, as well as help you get more comfortable with working with all kinds of variables, not just x .

Use the multiplication and division properties of equality to isolate the variable b .

[latex]\begin{array}{l}\,\,\,\,\,\,\,\,A=\frac{1}{2}bh\\\\\left(2\right)A=\left(2\right)\frac{1}{2}bh\\\\\,\,\,\,\,\,2A=bh\\\\\,\,\,\,\,\,\,\frac{2A}{h}=\frac{bh}{h}\\\\\,\,\,\,\,\,\,\,\frac{2A}{h}=\frac{b\cancel{h}}{\cancel{h}}\end{array}[/latex]

Write the equation with the desired variable on the left-hand side as a matter of convention:

Use the multiplication and division properties of equality to isolate the variable h .

[latex]\begin{array}{l}\,\,\,\,\,\,\,\,A=\frac{1}{2}bh\\\\\left(2\right)A=\left(2\right)\frac{1}{2}bh\\\\\,\,\,\,\,\,2A=bh\\\\\,\,\,\,\,\,\,\frac{2A}{b}=\frac{bh}{b}\\\\\,\,\,\,\,\,\,\,\frac{2A}{b}=\frac{h\cancel{b}}{\cancel{b}}\end{array}[/latex]

Temperature

Let’s look at another formula that includes parentheses and fractions, the formula for converting from the Fahrenheit temperature scale to the Celsius scale.

[latex]C=\left(F-32\right)\cdot \frac{5}{9}[/latex]

Given a temperature of [latex]12^{\circ}{C}[/latex], find the equivalent in [latex]{}^{\circ}{F}[/latex].

Substitute the given temperature in[latex]{}^{\circ}{C}[/latex] into the conversion formula:

[latex]12=\left(F-32\right)\cdot \frac{5}{9}[/latex]

Isolate the variable F to obtain the equivalent temperature.

[latex]\begin{array}{r}12=\left(F-32\right)\cdot \frac{5}{9}\\\\\left(\frac{9}{5}\right)12=F-32\,\,\,\,\,\,\,\,\,\,\,\,\,\\\\\left(\frac{108}{5}\right)12=F-32\,\,\,\,\,\,\,\,\,\,\,\,\,\\\\21.6=F-32\,\,\,\,\,\,\,\,\,\,\,\,\,\\\underline{+32\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+32}\,\,\,\,\,\,\,\,\,\,\,\,\\\\53.6={}^{\circ}{F}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\end{array}[/latex]

As with the other formulas we have worked with, we could have isolated the variable F first, then substituted in the given temperature in Celsius.

Solve the formula shown below for converting from the Fahrenheit scale to the Celsius scale for F.

To isolate the variable F, it would be best to clear the fraction involving F first. Multiply both sides of the equation by [latex] \displaystyle \frac{9}{5}[/latex].

[latex]\begin{array}{l}\\\,\,\,\,\left(\frac{9}{5}\right)C=\left(F-32\right)\left(\frac{5}{9}\right)\left(\frac{9}{5}\right)\\\\\,\,\,\,\,\,\,\,\,\,\,\,\frac{9}{5}C=F-32\end{array}[/latex]

Add 32 to both sides.

[latex]\begin{array}{l}\frac{9}{5}\,C+32=F-32+32\\\\\frac{9}{5}\,C+32=F\end{array}[/latex]

[latex]F=\frac{9}{5}C+32[/latex]

Think About It

Express the formula for the surface area of a cylinder, [latex]s=2\pi rh+2\pi r^{2}[/latex], in terms of the height, h .

In this example, the variable h is buried pretty deeply in the formula for surface area of a cylinder. Using the order of operations, it can be isolated. Before you look at the solution, use the box below to write down what you think is the best first step to take to isolate h .

[latex]\begin{array}{r}S\,\,=2\pi rh+2\pi r^{2} \\ \underline{-2\pi r^{2}\,\,\,\,\,\,\,\,\,\,\,\,\,-2\pi r^{2}}\\S-2\pi r^{2}\,\,\,\,=\,\,\,\,2\pi rh\,\,\,\,\,\,\,\,\,\,\,\,\,\,\end{array}[/latex]

Next, isolate the variable h by dividing both sides of the equation by [latex]2\pi r[/latex].

[latex]\begin{array}{r}\frac{S-2\pi r^{2}}{2\pi r}=\frac{2\pi rh}{2\pi r} \\\\ \frac{S-2\pi r^{2}}{2\pi r}=h\,\,\,\,\,\,\,\,\,\,\end{array}[/latex]

• "Ann Trason." Wikipedia. Accessed May 05, 2016. https://en.wikipedia.org/wiki/Ann_Trason . ↵
•  "American River 50 Mile Endurance Run." Wikipedia. Accessed May 05, 2016. https://en.wikipedia.org/wiki/American_River_50_Mile_Endurance_Run . ↵
• Writing Algebraic Expressions. Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/uD_V5t-6Kzs . License : CC BY: Attribution
• Write and Solve a Linear Equations to Solve a Number Problem (1). Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/izIIqOztUyI . License : CC BY: Attribution
• Write and Solve a Linear Equations to Solve a Number Problem (Consecutive Integers). Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/S5HZy3jKodg . License : CC BY: Attribution
• Screenshot: Ann Trason Trail Running. Authored by : Lumen Learning. License : CC BY: Attribution
• Problem Solving Using Distance, Rate, Time (Running). Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/3WLp5mY1FhU . License : CC BY: Attribution
• Simple Interest - Determine Account Balance (Monthly Interest). Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/XkGgEEMR_00 . License : CC BY: Attribution
• Simple Interest - Determine Interest Balance (Monthly Interest). Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/mRV5ljj32Rg . License : CC BY: Attribution
• Simple Interest - Determine Principal Balance (Monthly Interest). Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/vbMqN6lVoOM . License : CC BY: Attribution
• Find the Width of a Rectangle Given the Perimeter / Literal Equation. Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/jlxPgKQfhQs . License : CC BY: Attribution
• Find the Base of a Triangle Given Area / Literal Equation. Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/VQZQvJ3rXYg . License : CC BY: Attribution
• Convert Celsius to Fahrenheit / Literal Equation. Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/DRydX8V-JwY . License : CC BY: Attribution
• Ann Trason. Provided by : Wikipedia. Located at : https://en.wikipedia.org/wiki/Ann_Trason . License : CC BY-SA: Attribution-ShareAlike
• American River 50 Mile Endurance Run. Provided by : Wikipedia. Located at : https://en.wikipedia.org/wiki/American_River_50_Mile_Endurance_Run . License : CC BY-SA: Attribution-ShareAlike

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4.9: Strategies for Solving Applications and Equations

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Learning Objectives

By the end of this section, you will be able to:

• Use a problem solving strategy for word problems
• Solve number word problems
• Solve percent applications
• Solve simple interest applications

Before you get started, take this readiness quiz.

• Translate “six less than twice x ” into an algebraic expression. If you missed this problem, review [link] .
• Convert 4.5% to a decimal. If you missed this problem, review [link] .
• Convert 0.6 to a percent. If you missed this problem, review [link] .

Have you ever had any negative experiences in the past with word problems? When we feel we have no control, and continue repeating negative thoughts, we set up barriers to success. Realize that your negative experiences with word problems are in your past. To move forward you need to calm your fears and change your negative feelings.

Start with a fresh slate and begin to think positive thoughts. Repeating some of the following statements may be helpful to turn your thoughts positive. Thinking positive thoughts is a first step towards success.

• I think I can! I think I can!
• While word problems were hard in the past, I think I can try them now.
• I am better prepared now—I think I will begin to understand word problems.
• I am able to solve equations because I practiced many problems and I got help when I needed it—I can try that with word problems.
• It may take time, but I can begin to solve word problems.
• You are now well prepared and you are ready to succeed. If you take control and believe you can be successful, you will be able to master word problems.

Use a Problem Solving Strategy for Word Problems

Now that we can solve equations, we are ready to apply our new skills to word problems. We will develop a strategy we can use to solve any word problem successfully.

EXAMPLE \(\PageIndex{1}\)

Normal yearly snowfall at the local ski resort is 12 inches more than twice the amount it received last season. The normal yearly snowfall is 62 inches. What was the snowfall last season at the ski resort?

EXAMPLE \(\PageIndex{2}\)

Guillermo bought textbooks and notebooks at the bookstore. The number of textbooks was three more than twice the number of notebooks. He bought seven textbooks. How many notebooks did he buy?

He bought two notebooks

EXAMPLE \(\PageIndex{3}\)

Gerry worked Sudoku puzzles and crossword puzzles this week. The number of Sudoku puzzles he completed is eight more than twice the number of crossword puzzles. He completed 22 Sudoku puzzles. How many crossword puzzles did he do?

He did seven crosswords puzzles

We summarize an effective strategy for problem solving.

PROBLEM SOLVING STRATEGY FOR WORD PROBLEMS

• Read the problem. Make sure all the words and ideas are understood.
• Identify what you are looking for.
• Name what you are looking for. Choose a variable to represent that quantity.
• Translate into an equation. It may be helpful to restate the problem in one sentence with all the important information. Then, translate the English sentence into an algebra equation.
• Solve the equation using proper algebra techniques.
• Check the answer in the problem to make sure it makes sense.
• Answer the question with a complete sentence.

Solve Number Word Problems

We will now apply the problem solving strategy to “number word problems.” Number word problems give some clues about one or more numbers and we use these clues to write an equation. Number word problems provide good practice for using the Problem Solving Strategy.

EXAMPLE \(\PageIndex{4}\)

The sum of seven times a number and eight is thirty-six. Find the number.

Did you notice that we left out some of the steps as we solved this equation? If you’re not yet ready to leave out these steps, write down as many as you need.

EXAMPLE \(\PageIndex{5}\)

The sum of four times a number and two is fourteen. Find the number.

EXAMPLE \(\PageIndex{6}\)

The sum of three times a number and seven is twenty-five. Find the number.

Some number word problems ask us to find two or more numbers. It may be tempting to name them all with different variables, but so far, we have only solved equations with one variable. In order to avoid using more than one variable, we will define the numbers in terms of the same variable. Be sure to read the problem carefully to discover how all the numbers relate to each other.

EXAMPLE \(\PageIndex{7}\)

The sum of two numbers is negative fifteen. One number is nine less than the other. Find the numbers.

EXAMPLE \(\PageIndex{8}\)

The sum of two numbers is negative twenty-three. One number is seven less than the other. Find the numbers.

\(−15,−8\)

EXAMPLE \(\PageIndex{9}\)

The sum of two numbers is negative eighteen. One number is forty more than the other. Find the numbers.

\(−29,11\)

Consecutive Integers (optional)

Some number problems involve consecutive integers . Consecutive integers are integers that immediately follow each other. Examples of consecutive integers are:

\[\begin{array}{rrrr} 1, & 2, & 3, & 4 \\ −10, & −9, & −8, & −7\\ 150, & 151, & 152, & 153 \end{array}\]

Notice that each number is one more than the number preceding it. Therefore, if we define the first integer as n , the next consecutive integer is \(n+1\). The one after that is one more than \(n+1\), so it is \(n+1+1\), which is \(n+2\).

\[\begin{array}{ll} n & 1^{\text{st}} \text{integer} \\ n+1 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; & 2^{\text{nd}}\text{consecutive integer} \\ n+2 & 3^{\text{rd}}\text{consecutive integer} \;\;\;\;\;\;\;\; \text{etc.} \end{array}\]

We will use this notation to represent consecutive integers in the next example.

EXAMPLE \(\PageIndex{10}\)

Find three consecutive integers whose sum is \(−54\).

EXAMPLE \(\PageIndex{11}\)

Find three consecutive integers whose sum is \(−96\).

\(−33,−32,−31\)

EXAMPLE \(\PageIndex{12}\)

Find three consecutive integers whose sum is \(−36\).

\(−13,−12,−11\)

Now that we have worked with consecutive integers, we will expand our work to include consecutive even integers and consecutive odd integers . Consecutive even integers are even integers that immediately follow one another. Examples of consecutive even integers are:

\[24, 26, 28\]

\[−12,−10,−8\]

Notice each integer is two more than the number preceding it. If we call the first one n , then the next one is \(n+2\). The one after that would be \(n+2+2\) or \(n+4\).

Consecutive odd integers are odd integers that immediately follow one another. Consider the consecutive odd integers 63, 65, and 67.

\[63, 65, 67\]

\[n,n+2,n+4\]

Does it seem strange to have to add two (an even number) to get the next odd number? Do we get an odd number or an even number when we add 2 to 3? to 11? to 47?

Whether the problem asks for consecutive even numbers or odd numbers, you do not have to do anything different. The pattern is still the same—to get to the next odd or the next even integer, add two.

EXAMPLE \(\PageIndex{13}\)

Find three consecutive even integers whose sum is \(120\).

EXAMPLE \(\PageIndex{14}\)

Find three consecutive even integers whose sum is 102.

\(32, 34, 36\)

EXAMPLE \(\PageIndex{15}\)

Find three consecutive even integers whose sum is \(−24\).

\(−10,−8,−6\)

When a number problem is in a real life context, we still use the same strategies that we used for the previous examples.

EXAMPLE \(\PageIndex{16}\)

A married couple together earns $110,000 a year. The wife earns $16,000 less than twice what her husband earns. What does the husband earn?

According to the National Automobile Dealers Association, the average cost of a car in 2014 was $28,400. This was $1,600 less than six times the cost in 1975. What was the average cost of a car in 1975?

The average cost was $5,000.

EXAMPLE \(\PageIndex{18}\)

US Census data shows that the median price of new home in the U.S. in November 2014 was $280,900. This was $10,700 more than 14 times the price in November 1964. What was the median price of a new home in November 1964?

The median price was $19,300.

Solve Percent Applications

There are several methods to solve percent equations. In algebra, it is easiest if we just translate English sentences into algebraic equations and then solve the equations. Be sure to change the given percent to a decimal before you use it in the equation.

EXAMPLE \(\PageIndex{19}\)

Translate and solve:

• What number is 45% of 84?
• 8.5% of what amount is $4.76?
• 168 is what percent of 112?
• What number is 45% of 80?
• 7.5% of what amount is $1.95?
• 110 is what percent of 88?

ⓐ 36 ⓑ $26 ⓒ \(125 \% \)

EXAMPLE \(\PageIndex{21}\)

• What number is 55% of 60?
• 8.5% of what amount is $3.06?
• 126 is what percent of 72?

ⓐ 33 ⓑ $36 ⓐ \(175 \% \)

Now that we have a problem solving strategy to refer to, and have practiced solving basic percent equations, we are ready to solve percent applications. Be sure to ask yourself if your final answer makes sense—since many of the applications we will solve involve everyday situations, you can rely on your own experience.

EXAMPLE \(\PageIndex{22}\)

The label on Audrey’s yogurt said that one serving provided 12 grams of protein, which is 24% of the recommended daily amount. What is the total recommended daily amount of protein?

EXAMPLE \(\PageIndex{23}\)

One serving of wheat square cereal has 7 grams of fiber, which is 28% of the recommended daily amount. What is the total recommended daily amount of fiber?

EXAMPLE \(\PageIndex{24}\)

One serving of rice cereal has 190 mg of sodium, which is 8% of the recommended daily amount. What is the total recommended daily amount of sodium?

Remember to put the answer in the form requested. In the next example we are looking for the percent.

EXAMPLE \(\PageIndex{25}\)

Veronica is planning to make muffins from a mix. The package says each muffin will be 240 calories and 60 calories will be from fat. What percent of the total calories is from fat?

EXAMPLE \(\PageIndex{26}\)

Mitzi received some gourmet brownies as a gift. The wrapper said each 28% brownie was 480 calories, and had 240 calories of fat. What percent of the total calories in each brownie comes from fat? Round the answer to the nearest whole percent.

EXAMPLE \(\PageIndex{27}\)

The mix Ricardo plans to use to make brownies says that each brownie will be 190 calories, and 76 calories are from fat. What percent of the total calories are from fat? Round the answer to the nearest whole percent.

It is often important in many fields—business, sciences, pop culture—to talk about how much an amount has increased or decreased over a certain period of time. This increase or decrease is generally expressed as a percent and called the percent change .

To find the percent change, first we find the amount of change, by finding the difference of the new amount and the original amount. Then we find what percent the amount of change is of the original amount.

FIND PERCENT CHANGE

\[\text{change}= \text{new amount}−\text{original amount}\]

change is what percent of the original amount?

EXAMPLE \(\PageIndex{28}\)

Recently, the California governor proposed raising community college fees from $36 a unit to $46 a unit. Find the percent change. (Round to the nearest tenth of a percent.)

EXAMPLE \(\PageIndex{29}\)

Find the percent change. (Round to the nearest tenth of a percent.) In 2011, the IRS increased the deductible mileage cost to 55.5 cents from 51 cents.

\(8.8 \% \)

EXAMPLE \(\PageIndex{30}\)

Find the percent change. (Round to the nearest tenth of a percent.) In 1995, the standard bus fare in Chicago was $1.50. In 2008, the standard bus fare was 2.25.

Applications of discount and mark-up are very common in retail settings.

When you buy an item on sale, the original price has been discounted by some dollar amount. The discount rate , usually given as a percent, is used to determine the amount of the discount . To determine the amount of discount, we multiply the discount rate by the original price.

The price a retailer pays for an item is called the original cost . The retailer then adds a mark-up to the original cost to get the list price , the price he sells the item for. The mark-up is usually calculated as a percent of the original cost. To determine the amount of mark-up, multiply the mark-up rate by the original cost.

\[ \begin{align} \text{amount of discount} &= \text{discount rate}· \text{original price} \\ \text{sale price} &= \text{original amount}– \text{discount price} \end{align}\]

The sale price should always be less than the original price.

\[\begin{align} \text{amount of mark-up} &= \text{mark-up rate}·\text{original price} \\ \text{list price} &= \text{original cost}–\text{mark-up} \end{align}\]

The list price should always be more than the original cost.

EXAMPLE \(\PageIndex{31}\)

Liam’s art gallery bought a painting at an original cost of $750. Liam marked the price up 40%. Find

• the amount of mark-up and
• the list price of the painting.

EXAMPLE \(\PageIndex{32}\)

Find ⓐ the amount of mark-up and ⓑ the list price: Jim’s music store bought a guitar at original cost $1,200. Jim marked the price up 50%.

ⓐ $600 ⓑ $1,800

EXAMPLE \(\PageIndex{33}\)

Find ⓐ the amount of mark-up and ⓑ the list price: The Auto Resale Store bought Pablo’s Toyota for $8,500. They marked the price up 35%.

ⓐ $2,975 ⓑ $11,475

Solve Simple Interest Applications

Interest is a part of our daily lives. From the interest earned on our savings to the interest we pay on a car loan or credit card debt, we all have some experience with interest in our lives.

The amount of money you initially deposit into a bank is called the principal , P , and the bank pays you interest, I. When you take out a loan, you pay interest on the amount you borrow, also called the principal.

In either case, the interest is computed as a certain percent of the principal, called the rate of interest , r . The rate of interest is usually expressed as a percent per year, and is calculated by using the decimal equivalent of the percent. The variable t , (for time) represents the number of years the money is saved or borrowed.

Interest is calculated as simple interest or compound interest. Here we will use simple interest.

SIMPLE INTEREST

If an amount of money, P , called the principal, is invested or borrowed for a period of t years at an annual interest rate r , the amount of interest, I , earned or paid is

\[ \begin{array}{ll} I=Prt \; \; \; \; \; \; \; \; \; \; \; \; \text{where} & { \begin{align} I &= \text{interest} \\ P &= \text{principal} \\ r &= \text{rate} \\ t &= \text{time} \end{align}} \end{array}\]

Interest earned or paid according to this formula is called simple interest .

The formula we use to calculate interest is \(I=Prt\). To use the formula we substitute in the values for variables that are given, and then solve for the unknown variable. It may be helpful to organize the information in a chart.

EXAMPLE \(\PageIndex{34}\)

Areli invested a principal of $950 in her bank account that earned simple interest at an interest rate of 3%. How much interest did she earn in five years?

\( \begin{aligned} I & = \; ? \\ P & = \; \$ 950 \\ r & = \; 3 \% \\ t & = \; 5 \text{ years} \end{aligned}\)

\(\begin{array}{ll} \text{Identify what you are asked to find, and choose a} & \text{What is the simple interest?} \\ \text{variable to represent it.} & \text{Let } I= \text{interest.} \\ \text{Write the formula.} & I=Prt \\ \text{Substitute in the given information.} & I=(950)(0.03)(5) \\ \text{Simplify.} & I=142.5 \\ \text{Check.} \\ \text{Is } \$142.50 \text{ a reasonable amount of interest on } \$ \text{ 950?} \; \;\;\;\;\; \;\;\;\;\;\; \\ \text{Yes.} \\ \text{Write a complete sentence.} & \text{The interest is } \$ \text{142.50.} \end{array}\)

EXAMPLE \(\PageIndex{35}\)

Nathaly deposited $12,500 in her bank account where it will earn 4% simple interest. How much interest will Nathaly earn in five years?

He will earn $2,500.

EXAMPLE \(\PageIndex{36}\)

Susana invested a principal of $36,000 in her bank account that earned simple interest at an interest rate of 6.5%.6.5%. How much interest did she earn in three years?

She earned $7,020.

There may be times when we know the amount of interest earned on a given principal over a certain length of time, but we do not know the rate.

EXAMPLE \(\PageIndex{37}\)

Hang borrowed $7,500 from her parents to pay her tuition. In five years, she paid them $1,500 interest in addition to the $7,500 she borrowed. What was the rate of simple interest?

\( \begin{aligned} I & = \; \$ 1500 \\ P & = \; \$ 7500 \\ r & = \; ? \\ t & = \; 5 \text{ years} \end{aligned}\)

Identify what you are asked to find, and choose What is the rate of simple interest? a variable to represent it. Write the formula. Substitute in the given information. Multiply. Divide. Change to percent form. Let r = rate of interest. I = P r t 1,500 = ( 7,500 ) r ( 5 ) 1,500 = 37,500 r 0.04 = r 4 % = r Check. I = P r t 1,500 = ? ( 7,500 ) ( 0.04 ) ( 5 ) 1,500 = 1,500 ✓ Write a complete sentence. The rate of interest was 4%. Identify what you are asked to find, and choose What is the rate of simple interest? a variable to represent it. Write the formula. Substitute in the given information. Multiply. Divide. Change to percent form. Let r = rate of interest. I = P r t 1 ,500 = ( 7,500 ) r ( 5 ) 1,500 = 37,500 r 0.04 = r 4 % = r Check. I = P r t 1 ,500 = ? ( 7,500 ) ( 0.04 ) ( 5 ) 1,500 = 1, 500 ✓ Write a complete sentence. The rate of interest was 4%.

EXAMPLE \(\PageIndex{38}\)

Jim lent his sister $5,000 to help her buy a house. In three years, she paid him the $5,000, plus $900 interest. What was the rate of simple interest?

The rate of simple interest was 6%.

EXAMPLE \(\PageIndex{39}\)

Loren lent his brother $3,000 to help him buy a car. In four years, his brother paid him back the $3,000 plus $660 in interest. What was the rate of simple interest?

The rate of simple interest was 5.5%.

In the next example, we are asked to find the principal—the amount borrowed.

EXAMPLE \(\PageIndex{40}\)

Sean’s new car loan statement said he would pay $4,866,25 in interest from a simple interest rate of 8.5% over five years. How much did he borrow to buy his new car?

\( \begin{aligned} I & = \; 4,866.25 \\ P & = \; ? \\ r & = \; 8.5 \% \\ t & = \; 5 \text{ years} \end{aligned}\)

Identify what you are asked to find, What is the amount borrowed (the principal)? and choose a variable to represent it. Write the formula. Substitute in the given information. Multiply. Divide. Let P = principal borrowed. I = P r t 4,866.25 = P ( 0.085 ) ( 5 ) 4,866.25 = 0.425 P 11,450 = P Check. I = P r t 4,866.25 = ? ( 11,450 ) ( 0.085 ) ( 5 ) 4,866.25 = 4,866.25 ✓ Write a complete sentence. The principal was $11,450. Identify what you are asked to find, What is the amount borrowed (the principal)? and choose a variable to represent it. Write the formula. Substitute in the given information. Multiply. Divide. Let P = principal borrowed. I = P r t 4 ,866.25 = P ( 0.085 ) ( 5 ) 4,866.25 = 0.425 P 11,450 = P Check. I = P r t 4 ,866.25 = ? ( 11,450 ) ( 0.085 ) ( 5 ) 4,866.25 = 4,866.25 ✓ Write a complete sentence. The principal was $11,450.

EXAMPLE \(\PageIndex{41}\)

Eduardo noticed that his new car loan papers stated that with a 7.5% simple interest rate, he would pay $6,596.25 in interest over five years. How much did he borrow to pay for his car?

He paid $17,590.

EXAMPLE \(\PageIndex{42}\)

In five years, Gloria’s bank account earned $2,400 interest at 5% simple interest. How much had she deposited in the account?

She deposited $9,600.

Access this online resource for additional instruction and practice with using a problem solving strategy.

• Begining Arithmetic Problems

Key Concepts

\(\text{change}=\text{new amount}−\text{original amount}\)

\(\text{change is what percent of the original amount?}\)

• \( \begin{align} \text{amount of discount} &= \text{discount rate}· \text{original price} \\ \text{sale price} &= \text{original amount}– \text{discount price} \end{align}\)
• \(\begin{align} \text{amount of mark-up} &= \text{mark-up rate}·\text{original price} \\ \text{list price} &= \text{original cost}–\text{mark-up} \end{align}\)
• If an amount of money, P , called the principal, is invested or borrowed for a period of t years at an annual interest rate r , the amount of interest, I , earned or paid is: \[\begin{aligned} &{} &{} &{I=interest} \nonumber\\ &{I=Prt} &{\text{where} \space} &{P=principal} \nonumber\\ &{} &{\space} &{r=rate} \nonumber\\ &{} &{\space} &{t=time} \nonumber \end{aligned}\]

Practice Makes Perfect

1. List five positive thoughts you can say to yourself that will help you approach word problems with a positive attitude. You may want to copy them on a sheet of paper and put it in the front of your notebook, where you can read them often.

Answers will vary.

2. List five negative thoughts that you have said to yourself in the past that will hinder your progress on word problems. You may want to write each one on a small piece of paper and rip it up to symbolically destroy the negative thoughts.

In the following exercises, solve using the problem solving strategy for word problems. Remember to write a complete sentence to answer each question.

3. There are \(16\) girls in a school club. The number of girls is four more than twice the number of boys. Find the number of boys.

4. There are \(18\) Cub Scouts in Troop 645. The number of scouts is three more than five times the number of adult leaders. Find the number of adult leaders.

5. Huong is organizing paperback and hardback books for her club’s used book sale. The number of paperbacks is \(12\) less than three times the number of hardbacks. Huong had \(162\) paperbacks. How many hardback books were there?

58 hardback books

6. Jeff is lining up children’s and adult bicycles at the bike shop where he works. The number of children’s bicycles is nine less than three times the number of adult bicycles. There are \(42\) adult bicycles. How many children’s bicycles are there?

In the following exercises, solve each number word problem.

7. The difference of a number and \(12\) is three. Find the number.

8. The difference of a number and eight is four. Find the number.

9. The sum of three times a number and eight is \(23\). Find the number.

10. The sum of twice a number and six is \(14\). Find the number.

11 . The difference of twice a number and seven is \(17\). Find the number.

12. The difference of four times a number and seven is \(21\). Find the number.

13. Three times the sum of a number and nine is \(12\). Find the number.

14. Six times the sum of a number and eight is \(30\). Find the number.

15. One number is six more than the other. Their sum is \(42\). Find the numbers.

\(18, \;24\)

16. One number is five more than the other. Their sum is \(33\). Find the numbers.

17. The sum of two numbers is \(20\). One number is four less than the other. Find the numbers.

\(8, \;12\)

18 . The sum of two numbers is \(27\). One number is seven less than the other. Find the numbers.

19. One number is \(14\) less than another. If their sum is increased by seven, the result is \(85\). Find the numbers.

\(32,\; 46\)

20 . One number is \(11\) less than another. If their sum is increased by eight, the result is \(71\). Find the numbers.

21. The sum of two numbers is \(14\). One number is two less than three times the other. Find the numbers.

\(4,\; 10\)

22. The sum of two numbers is zero. One number is nine less than twice the other. Find the numbers.

23. The sum of two consecutive integers is \(77\). Find the integers.

\(38,\; 39\)

24. The sum of two consecutive integers is \(89\). Find the integers.

25. The sum of three consecutive integers is \(78\). Find the integers.

\(25,\; 26,\; 27\)

26. The sum of three consecutive integers is \(60\). Find the integers.

27. Find three consecutive integers whose sum is \(−36\).

\(−11,\;−12,\;−13\)

28. Find three consecutive integers whose sum is \(−3\).

29. Find three consecutive even integers whose sum is \(258\).

\(84,\; 86,\; 88\)

30. Find three consecutive even integers whose sum is \(222\).

31. Find three consecutive odd integers whose sum is \(−213\).

\(−69,\;−71,\;−73\)

32. Find three consecutive odd integers whose sum is \(−267\).

33. Philip pays \($1,620\) in rent every month. This amount is \($120\) more than twice what his brother Paul pays for rent. How much does Paul pay for rent?

34. Marc just bought an SUV for \($54,000\). This is \($7,400\) less than twice what his wife paid for her car last year. How much did his wife pay for her car?

35. Laurie has \($46,000\) invested in stocks and bonds. The amount invested in stocks is \($8,000\) less than three times the amount invested in bonds. How much does Laurie have invested in bonds?

\($13,500\)

36. Erica earned a total of \($50,450\) last year from her two jobs. The amount she earned from her job at the store was \($1,250\) more than three times the amount she earned from her job at the college. How much did she earn from her job at the college?

In the following exercises, translate and solve.

37. a. What number is 45% of 120? b. 81 is 75% of what number? c. What percent of 260 is 78?

a. 54 b. 108 c. 30%

38. a. What number is 65% of 100? b. 93 is 75% of what number? c. What percent of 215 is 86?

39. a. 250% of 65 is what number? b. 8.2% of what amount is $2.87? c. 30 is what percent of 20?

a. 162.5 b. $35 c. 150%

40. a. 150% of 90 is what number? b. 6.4% of what amount is $2.88? c. 50 is what percent of 40?

In the following exercises, solve.

41. Geneva treated her parents to dinner at their favorite restaurant. The bill was $74.25. Geneva wants to leave 16% of the total bill as a tip. How much should the tip be?

42. When Hiro and his co-workers had lunch at a restaurant near their work, the bill was $90.50. They want to leave 18% of the total bill as a tip. How much should the tip be?

43. One serving of oatmeal has 8 grams of fiber, which is 33% of the recommended daily amount. What is the total recommended daily amount of fiber?

44. One serving of trail mix has 67 grams of carbohydrates, which is 22% of the recommended daily amount. What is the total recommended daily amount of carbohydrates?

45. A bacon cheeseburger at a popular fast food restaurant contains 2070 milligrams (mg) of sodium, which is 86% of the recommended daily amount. What is the total recommended daily amount of sodium?

46. A grilled chicken salad at a popular fast food restaurant contains 650 milligrams (mg) of sodium, which is 27% of the recommended daily amount. What is the total recommended daily amount of sodium?

47. The nutrition fact sheet at a fast food restaurant says the fish sandwich has 380 calories, and 171 calories are from fat. What percent of the total calories is from fat?

48. The nutrition fact sheet at a fast food restaurant says a small portion of chicken nuggets has 190 calories, and 114 calories are from fat. What percent of the total calories is from fat?

49. Emma gets paid $3,000 per month. She pays $750 a month for rent. What percent of her monthly pay goes to rent?

50. Dimple gets paid $3,200 per month. She pays $960 a month for rent. What percent of her monthly pay goes to rent?

51. Tamanika received a raise in her hourly pay, from $15.50 to $17.36. Find the percent change.

52. Ayodele received a raise in her hourly pay, from $24.50 to $25.48. Find the percent change.

53. Annual student fees at the University of California rose from about $4,000 in 2000 to about $12,000 in 2010. Find the percent change.

54. The price of a share of one stock rose from $12.50 to $50. Find the percent change.

55. A grocery store reduced the price of a loaf of bread from $2.80 to $2.73. Find the percent change.

−2.5%

56. The price of a share of one stock fell from $8.75 to $8.54. Find the percent change.

57. Hernando’s salary was $49,500 last year. This year his salary was cut to $44,055. Find the percent change.

58. In ten years, the population of Detroit fell from 950,000 to about 712,500. Find the percent change.

In the following exercises, find a. the amount of discount and b. the sale price.

59. Janelle bought a beach chair on sale at 60% off. The original price was $44.95.

a. $26.97 b. $17.98

60. Errol bought a skateboard helmet on sale at 40% off. The original price was $49.95.

In the following exercises, find a. the amount of discount and b. the discount rate (Round to the nearest tenth of a percent if needed.)

61. Larry and Donna bought a sofa at the sale price of $1,344. The original price of the sofa was $1,920.

a. $576 b. 30%

62. Hiroshi bought a lawnmower at the sale price of $240. The original price of the lawnmower is $300.

In the following exercises, find a. the amount of the mark-up and b. the list price.

63. Daria bought a bracelet at original cost $16 to sell in her handicraft store. She marked the price up 45%. What was the list price of the bracelet?

a. $7.20 b. $23.20

64. Regina bought a handmade quilt at original cost $120 to sell in her quilt store. She marked the price up 55%. What was the list price of the quilt?

65. Tom paid $0.60 a pound for tomatoes to sell at his produce store. He added a 33% mark-up. What price did he charge his customers for the tomatoes?

a. $0.20 b. $0.80

66. Flora paid her supplier $0.74 a stem for roses to sell at her flower shop. She added an 85% mark-up. What price did she charge her customers for the roses?

67. Casey deposited $1,450 in a bank account that earned simple interest at an interest rate of 4%. How much interest was earned in two years?

68 . Terrence deposited $5,720 in a bank account that earned simple interest at an interest rate of 6%. How much interest was earned in four years?

69. Robin deposited $31,000 in a bank account that earned simple interest at an interest rate of 5.2%. How much interest was earned in three years?

70. Carleen deposited $16,400 in a bank account that earned simple interest at an interest rate of 3.9% How much interest was earned in eight years?

71. Hilaria borrowed $8,000 from her grandfather to pay for college. Five years later, she paid him back the $8,000, plus $1,200 interest. What was the rate of simple interest?

72. Kenneth lent his niece $1,200 to buy a computer. Two years later, she paid him back the $1,200, plus $96 interest. What was the rate of simple interest?

73. Lebron lent his daughter $20,000 to help her buy a condominium. When she sold the condominium four years later, she paid him the $20,000, plus $3,000 interest. What was the rate of simple interest?

74. Pablo borrowed $50,000 to start a business. Three years later, he repaid the $50,000, plus $9,375 interest. What was the rate of simple interest?

75. In 10 years, a bank account that paid 5.25% simple interest earned $18,375 interest. What was the principal of the account?

76. In 25 years, a bond that paid 4.75% simple interest earned $2,375 interest. What was the principal of the bond?

77. Joshua’s computer loan statement said he would pay $1,244.34 in simple interest for a three-year loan at 12.4%. How much did Joshua borrow to buy the computer?

78. Margaret’s car loan statement said she would pay $7,683.20 in simple interest for a five-year loan at 9.8%. How much did Margaret borrow to buy the car?

Everyday Math

79 . Tipping At the campus coffee cart, a medium coffee costs $1.65. MaryAnne brings $2.00 with her when she buys a cup of coffee and leaves the change as a tip. What percent tip does she leave?

80 . Tipping Four friends went out to lunch and the bill came to $53.75 They decided to add enough tip to make a total of $64, so that they could easily split the bill evenly among themselves. What percent tip did they leave?

Writing Exercises

81. What has been your past experience solving word problems? Where do you see yourself moving forward?

82. Without solving the problem “44 is 80% of what number” think about what the solution might be. Should it be a number that is greater than 44 or less than 44? Explain your reasoning.

83. After returning from vacation, Alex said he should have packed 50% fewer shorts and 200% more shirts. Explain what Alex meant.

84. Because of road construction in one city, commuters were advised to plan that their Monday morning commute would take 150% of their usual commuting time. Explain what this means.

a. After completing the exercises, use this checklist to evaluate your mastery of the objective of this section.

b. After reviewing this checklist, what will you do to become confident for all objectives?

Equation Solver

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• Solve for Variable
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• Solving Simple Equations

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Problem solving with equations

Solving a task usually consists of finding the value of some quantity by logical reasoning and calculations. For example, find the speed, time, distance, mass of an object, or amount of something.

Such a task can be solved by using an equation. To do this, the desired value is denoted by a variable, then by logical reasoning the equation is composed and solved. After solving the equation, you check to see if the solution to the equation satisfies the conditions of the task.

Writing expressions containing the unknown

The solution of the task is accompanied by the composition of the equation to this task. At the initial stage of studying tasks it is desirable to learn to make up letter expressions describing this or that situation in life. This stage is not complicated and can be studied in the process of solving the task itself.

Consider a few situations that can be written down using a mathematical expression.

Task 1. The father is x years old. Mom is two years younger. The son is three times younger than the father. Write the age of each using expressions.

Task 2. Father is x years old, mother is 2 years younger than father. Son is 3 times younger than father, daughter is 3 times younger than mother. Write the age of each using expressions.

Task 3. Father is x years old, mother is 3 years younger than father. The son is 3 times younger than the father, the daughter is 3 times younger than the mother. How old are each of them if the total age of the father, mother, son, and daughter is 92?

In this task, in addition to writing down the expressions, we need to calculate the age of each family member.

First, write down the age of each family member using expressions. We will take the age of the father as the variable x, and then use that variable to make the rest of the expressions:

Now let's determine the age of each family member. To do this, we need to make and solve an equation. We have all the components of the equation ready. The only thing left to do is to put them together.

The total age of 92 was obtained by adding the ages of the father, mother, son and daughter:

father + mother + son + daughter = 92

For each age, we made up a mathematical expression. These expressions will be the components of our equation. Let's assemble our equation according to this diagram and the table above. That is, replace the words father, mother, son, daughter with their corresponding expression in the table:

The expression corresponding to mom's age x - 3 was put in brackets for clarity.

Now let's solve the resulting equation. To begin with, you can open the brackets where you can:

To free the equation from fractions, multiply both parts by 3

Solve the resulting equation using the known identity transformations:

We found the value of variable x. This variable was responsible for the age of the father. So the age of the father is 36 years old.

Knowing the age of the father, we can calculate the ages of the other family members. To do this, substitute the value of variable x in those expressions that correspond to the age of a particular family member.

The task said that my mother is 3 years younger than my father. We denoted her age by the expression x - 3. The value of the variable x is now known, and to calculate mom's age, substitute the found value 36 instead of x in the expression x - 3

x - 3 = 36 - 3 = 33 years old mom.

The age of the other family members is determined similarly:

Task 4. A kilogram of caviar costs x dollars. Write an expression that calculates how many kilograms of caviar you can buy for $300.

Task 5. x dollars was used to buy 5 kilograms of strawberries. Write an expression that calculates how many dollars one kilogram of strawberries costs.

Task 6. Tom, John, and Leo went to the cafeteria during recess and bought a large pizza and a glass of fresh juice. The large pizza cost x dollars and the coffee cost $15. Determine the cost of the pizza if you know that $120 was paid for everything?

Of course, this task is simple and can be solved without resorting to an equation. To do this, subtract the cost of three glasses of juice from $120 (15 × 3), and divide the result by 3

But our goal is to make an equation to the task and solve this equation. So, the cost of a large pizza is x dollars. Only three of them were bought. So if we increase the cost by three times, we get an expression describing how many dollars were paid for the three large pizzas.

3x — the cost of three large pizzas

And the cost of three glasses of juice can be written as 15 × 3. 15 is the cost of one glass of juice, and 3 is the multiplier (Tom, John, and Leo) that triples that cost.

Under the terms of the task, $120 is paid for everything. We already have an approximate scheme of what needs to be done:

Cost of three large pizzas + cost of three glasses of fresh juice = $120

The expressions describing the cost of three pizzas and three glasses of juice are ready. These are expressions 3x and 15 × 3. Use the scheme to make an equation and solve it:

So, the cost of one large pizza is $25.

The task is solved correctly only if the equation to it is written correctly. Unlike ordinary equations by which we learn to find the roots, problem-solving equations have their own specific application. Each component of such an equation can be described in verbal form. When making an equation, it is imperative that we understand why we are including one component or another and why it is needed.

You must also remember that after solving the equation, the left-hand side will have to equal the right-hand side. The equation composed must not contradict this idea.

Let's imagine that the equation is a scale with two bowls and a screen showing the state of the scale.

At this point, the screen shows an equal sign. It is clear why the left bowl is equal to the right bowl - there is nothing on the scales. The state of the scales and the absence of anything on the bowls will be recorded with the following equality:

Put a watermelon on the left side of the scale:

The left bowl outweighed the right bowl and the screen sounded an alarm, showing a not equal sign ( ≠ ). This sign indicates that the left bowl is not equal to the right bowl.

Now let's try to solve the problem. Let's find out how much the watermelon on the left bowl weighs. But how can we find this out? Our scales are only for checking if the left bowl is equal to the right one.

Equations come to the rescue. Recall that an equation contains a variable whose value must be found. The scale in this case is the equation itself, and the mass of the watermelon is the variable, the value of which needs to be found. Our goal is to get this equation right. That is, to align the scales so that we can calculate the mass of the watermelon.

To level the scales, we can put some heavy object on the right-hand bowl. For example, let's put a weight of 7 kilograms there.

Now the right bowl outweighs the left bowl. The screen still shows that the bowls are not equal.

Let's try to put a weight of 4 kg on the left bowl

Now the scales are aligned. In the picture you can see that the left bowl is at the level of the right bowl. And the screen shows an equal sign. This sign says that the left bowl is equal to the right bowl.

So we got an equation - an equality that contains the unknown. The left bowl is the left side of the equation, consisting of the components 4 and the variable x (watermelon mass), and the right bowl is the right side of the equation, consisting of the component 7.

Well, it's not hard to guess that the root of the equation 4 + x = 7 is 3. So the mass of the watermelon is 3 kg.

The same is true for the other tasks. To find some unknown value, various elements are added to the left or right side of the equation: terms, multipliers, expressions. In school tasks, these elements may already be given. All that remains is to structure them correctly and construct the equation. In this example, we were trying weights of different weights to calculate the mass of the watermelon.

Naturally, the data given in the task must first be reduced to a form in which they can be included in the equation.

Consider the following task. The father's age is equal to the age of the son and daughter together. The son is twice as old as the daughter and twenty years younger than the father. How old are each?

The daughter's age can be denoted by x. If the son is twice as old as the daughter, his age will be denoted by 2x. The condition of the problem says that together the age of the daughter and the son are equal to the age of the father. So the father's age will be denoted by the sum x + 2x

In the expression x + 2x you can give like terms. Then the age of the father will be denoted as 3x

Now let's make an equation. We need to get an equation where we can find the unknown x. Let's use weights. On the left hand side put the age of the father (3x), and on the right hand side put the age of the son (2x)

It is clear why the left bowl outweighs the right and why the screen shows the sign ( ≠ ). After all, it is logical that the age of the father is greater than the age of the son.

But we need to equalize the scales to be able to calculate the unknown x. To do this we need to add some number to the right-hand scale. What exact number is given in the task. The condition said that the son is 20 years younger than his father. So 20 years is the number to put on the scale.

The scales will even out if we add these 20 years to the right side of the scale. In other words, raise the son to the age of the father.

Now the scales are aligned. We got the equation 3x = 2x + 20, which is easy to solve:

At the beginning of this task, we used the variable x to represent our daughter's age. Now we found the value of this variable. The daughter is 20 years old.

Next, it was said that the son is two years older than the daughter, so the son is (20 × 2), that is, 40 years old.

Finally, let's calculate the age of the father. It was said in the problem that he is equal to the sum of the ages of the son and daughter, that is, (20 + 40) years old.

Let's return to the middle of the task and note one point. When we put the age of the father and the age of the son on the scale, the left cup outweighed the right

But we solved this problem by adding another 20 years to the right-hand scale. As a result, the scales aligned and we got the equality 3x = 2x + 20

But we could not add those 20 years to the right cup, but subtract them from the left. We would have obtained equality and in that case

This time you get the equation 3x -20 = 2x. The root of the equation is still 20

That is, the equations 3x = 2x + 20 and 3x -20 = 2x are equipotent. And we remember that equal equations have the same roots. If you look closely at these two equations, you can see that the second equation is obtained by moving the number 20 from the right side to the left side with the opposite sign. And this action, as stated in the previous lesson, does not change the roots of the equation.

Also note that at the beginning of the task, the ages of each family member could be given by other expressions.

Let's solve this equation

In other words, the task can be solved by different methods. So you should not be discouraged if you cannot solve a particular problem. But you should keep in mind that there are the easiest ways to solve the problem. It is possible to get to the city center by different routes, but there is always the most convenient, fastest and safest route.

Examples of problem solving

Task 1. There are a total of 30 notebooks in two packs. If two notebooks were moved from the first stack to the second stack, the first stack would contain twice as many notebooks as the second stack. How many notebooks were in each pack?

Denote by x the number of notebooks that were in the first stack. If the total number of notebooks was 30, and the variable x is the number of notebooks in the first stack, then the number of notebooks in the second stack will be denoted by the expression 30 - x. That is, from the total number of notebooks we subtract the number of notebooks in the first stack, and thus we obtain the number of notebooks in the second stack.

Further it is said that if you move two notebooks from the first pack to the second pack, there will be twice as many notebooks in the first pack. So let's remove two notebooks from the first stack

and add these two notebooks to the second pack

The expressions from which we will compose the equation now take the following form:

Let's try to make an equation out of the available expressions. Put both stacks of notebooks on the scales

The left bowl is heavier than the right one. This is because the problem statement says that after taking two notebooks from the first stack and putting them into the second stack, the number of notebooks in the first stack became twice as many as in the second stack.

To equalize the scales and get the equation, let's double the right-hand side. To do this, multiply it by 2

We obtain the equation x-2 = 2(30 - x +2) . Solve this equation:

We denoted the first packet by the variable x. Now we have found its value. The variable x is 22. So there were 22 notebooks in the first stack.

And we denoted the second tutu by the expression 30 - x, and since the value of the variable x is now known, we can calculate the number of notebooks in the second tutu. It is equal to 30 - 22, i.e. 8 pieces.

Task 2. Two people were peeling potatoes. One peeled two potatoes per minute and the other peeled three potatoes. Together they peeled 400 pieces. How long did each work, if the second worked 25 minutes longer than the first?

Denote by x the time the first person worked. Since the second person worked 25 minutes longer than the first person, his time will be denoted by x + 25

The first worker peeled 2 potatoes per minute, and since he worked x minutes, he peeled a total of 2x potatoes.

The second person peeled three potatoes per minute, and since he worked x + 25 minutes, he peeled a total of 3(x + 25) potatoes.

Together they peeled 400 potatoes

From the available components let us make and solve the equation. The left side of the equation will be the potatoes peeled by each person, and the right side will be their sum:

At the beginning of this task, we used the variable x to denote the working time of the first person. Now we found the value of this variable. The first person worked for 65 minutes.

And the second person worked x + 25 minutes, and since the value of the variable x is now known, we can calculate the working time of the second person - it is 65 + 25, that is, 90 minutes.

The task from the Russian algebra textbook . (Their currency is rubles and kopecks (like dollars and cents))

Of the varieties of tea is a mixture of 32 kg. A kilogram of the first grade costs 8 rubles, and the second grade 6 rubles. 50 kop. How many kilos of both varieties, if a kilo of mixture costs (no profit and loss) 7 rubles. 10 kop.

Denote by x the mass of first grade tea. Then the mass of second grade tea will be denoted by the expression 32 - x

A kilogram of first grade tea costs 8 rubles. If you multiply these eight rubles by the number of kilograms of first grade tea, you can find out how many rubles it cost x kilograms of first grade tea.

A kg of second class tea costs 6 roubles. 50 kopecks. If the price is 6 rubles. 50 kopecks times 32 - x, we will find out how much it cost 32 - x kg of tea of the second sort.

The condition says that a kilogram of mixture costs 7 rubles. 10 kopecks. All in all 32 kg of the mixture were produced. Multiply 7 rubles. 10 kopecks by 32 we can find out how much 32 kg of the mixture costs.

The expressions from which we will make the equation now take the following form:

Let's try to make an equation out of the available expressions. Let's put the cost of mixtures of first and second grade teas on the left scale, and on the right scale let's put the cost of 32 kg of mixture, i.e. the total cost of the mixture, which includes both varieties of tea:

We get the equation 8x + 6.50(32 - x) = 7.10 * 32 . Let's solve it:

At the beginning of this task, we used the variable x to denote the mass of the first grade tea. Now we have found the value of this variable. The variable x is 12.8. This means that 12.8 kg of first grade tea was used to make the mixture.

We have used the expression 32 - x to represent the mass of second-grade tea, and since the value of the variable x is now known, we can calculate the mass of second-grade tea. It is 32 - 12.8, i.e. 19.2. This means that 19.2 kg of second grade tea were used to prepare the mixture.

Some tasks may involve topics that a person may not have studied. This task is one of those tasks. It touches on the concepts of distance, speed, and time. Accordingly, to solve such a problem, it is necessary to have an idea of the things mentioned in the problem. In our case, it is necessary to know what represents distance, speed and time.

In the task we need to find the distances of the two roads. We have to make an equation that will allow us to calculate these distances.

Let's remember how distance, speed, and time are interrelated. Each of these quantities can be described using a letter equation:

We will use the right side of one of these equations to make our own equation. To find out which one, go back to the text of the problem and pay attention to the following point:

Now let's make an equation out of the available expressions

Through the variable S we denoted the distance of the first road. Now we have found the value of this variable. The variable S is 15. So the distance of the first road is 15 km.

And the distance of the second road is S + 3. Since the value of the variable S is now known, we can calculate the distance of the second road. This distance is equal to the sum of 15 + 3, that is 18 km.

Task 4. Two cars walk along the highway at the same speed. If the first increases the speed by 10 km/h, and the second decreases the speed by 10 km/h, then the first car will pass the same distance in 2 h as the second car did in 3 h. At what speed are the cars traveling?

Denote by v the speed of each car. Further in the problem there are hints: increase the speed of the first car by 10 km/h, and decrease the speed of the second car by 10 km/h. Let us use this hint

Further it is said that at such speeds (increased and decreased by 10 km/h) the first car will cover the same distance in 2 hours as the second car did in 3 hours. The phrase "as much" can be understood as "the distance traveled by the first car will be equal to the distance traveled by the second car" .

The distance, as we remember, is determined by the formula S = vt. We are interested in the right part of this letter equation - it will allow us to make an equation containing the variable v.

So, at speed v + 10 km/h, the first car will travel 2(v+10) km and the second car will travel 3(v - 10) km. Under this condition, the cars will travel the same distance, so it is sufficient to connect the two expressions with an equal sign to obtain the equation. Then we obtain the equation 2(v+10) = 3(v - 10). Solve it:

In the task condition it was said that the cars go at the same speed. We denoted this speed by the variable v. Now we have found the value of this variable. The variable v is 50. So the speed of both cars was 50 km/h.

Task 5. In 9 hours along the river the ship travels the same distance as in 11 hours against the stream. Find the boat's own speed if the speed of the river flow is 2 km/h.

Denote by v the boat's own speed. The speed of the river flow is 2 km/h. The speed of the ship along the river will be v + 2 km/h, and against the current - (v - 2) km/h.

The statement of the task says that the boat takes the same distance in 9 hours upstream as it took 11 hours upstream. The phrase "the same distance" can be understood as "the distance traveled by the ship upstream in 9 hours is equal to the distance traveled by the ship against the river in 11 hours". That is, the distances will be the same.

The distance is determined by the formula S = vt. Let's use the right part of this letter equation to make our own equation.

So, the boat will travel 9(v + 2) km in 9 hours upstream, and 11(v - 2) km in 11 hours upstream. Since both expressions describe the same distance, let's equate the first expression to the second one. The resulting equation is 9(v + 2) = 11(v - 2) 

So the proper speed of the motorboat is 20 km/h.

When solving tasks, it is a useful habit to determine in advance on which set the solution is sought.

Suppose that in the task we needed to find the time in which the pedestrian travels the specified path. We denoted the time by the variable t, then made an equation containing this variable and found its value.

We know from practice that the time of motion of an object can take both integer and fractional values, for example 2 h, 1.5 h, and 0.5 h. Then we can say that the solution of this problem is searched for on the set of rational numbers Q, because each value of 2 h, 1.5 h, 0.5 h can be represented as a fraction.

So after you denote an unknown quantity by a variable, it is useful to specify which set this quantity belongs to. In our example, time t belongs to the set of rational numbers Q

You can also introduce a restriction for the variable t, stating that it can only take positive values. Indeed, if an object has spent a certain amount of time on the path, then this time cannot be negative. Therefore next to the expression t ∈ Q we specify that its value must be greater than zero:

t  ∈  R , t  > 0

If by solving the equation we get a negative value for the variable t, then we can conclude that the problem is solved incorrectly, because this solution will not satisfy the condition t ∈ Q, t > 0.

Another example. If we were solving a task that required us to find the number of people to do a particular job, we would denote this number by the variable x. In such a task the solution would be searched for on the set of natural numbers

x  ∈ N

Indeed, the number of people is a whole number, e.g., 2 people, 3 people, 5 people. But not 1.5 (one whole person and half a person) or 2.3 (two whole persons and three tenths of a person).

Here we could specify that the number of people must be greater than zero, but the numbers included in the set of natural numbers N are themselves positive and greater than zero. In this set there are no negative numbers and no number 0. Therefore the expression x > 0 can be omitted. 

Task 6. A team of 2.5 times as many painters as carpenters arrived to repair a school. Soon the foreman added four more painters to the brigade, and transferred two carpenters to another site. As a result, there were four times as many painters as carpenters in the brigade. How many painters and how many carpenters were in the brigade originally

Denote by x the carpenters who arrived initially for repairs.

The number of carpenters is an integer greater than zero. Therefore let us specify that x belongs to the set of natural numbers

x  ∈ N

There were 2.5 times more painters than carpenters. Therefore, the number of painters will be denoted as 2.5x.

It goes on to say that the foreman included four more painters in the crew, and transferred two carpenters to another site. Let's do the same for our expressions. Reduce the number of carpenters by 2

And the number of painters will increase by 4

Now the number of carpenters and painters will be denoted by the following expressions:

Let's try to make an equation out of the available expressions:

The right bowl is larger because after adding four more painters to the brigade and moving two carpenters to another site, the number of painters in the brigade is 4 times more than the number of carpenters. To equalize the scales, you need to quadruple the left bowl:

The variable x was used to represent the original number of carpenters. Now we have found the value of this variable.  The variable x is equal to 8. So there were 8 carpenters in the team originally.

The number of painters was expressed as 2.5x, and since the value of the variable x is now known, we can calculate the number of painters, which is 2.5 × 8, that is, 20.

Go back to the beginning of the task and make sure that the condition x ∈ N is satisfied. The variable x is 8, and the elements of the set of natural numbers N are all numbers starting with 1, 2, 3 and so on to infinity. The same set includes the number 8, which we found.

8  ∈ N

The same can be said of the number of painters. The number 20 belongs to the set of natural numbers:

20  ∈ N

To understand the essence of the task and to correctly compose the equation, it is not necessary to use the model of scales with bowls. You can use other models: segments, tables, diagrams. You can come up with your own model, which would describe the essence of the task well.

Task 9. 30% of the milk was poured out of the can. This left 14 liters of milk in the can. How many liters of milk was in the can originally?

We need to find the original number of liters in the can. Let's represent the number of liters as a line and sign this line as X

It is said that 30% of the milk was poured out of the beaker. Let's draw an approximate figure of 30%

A percentage is, by definition, one hundredth of something. If 30% of the milk is poured, the other 70% is left in the can. This 70% is the 14 liters in the problem. Let's draw a picture of the remaining 70%

Now you can make an equation. Recall how to find the percentage of a number. To do this, the total amount of something is divided by 100 and the result is multiplied by the number of percentages you are looking for. Note that 14 liters, which is 70 percent, can be obtained in the same way: the original number of liters X divided by 100 and the result multiplied by 70. Equate all this to the number 14

Or get a simpler equation: write 70% as 0.70, then multiply by X and equate that expression to 14

So originally there were 20 liters of milk in the can.

Task 9. Two alloys of gold and silver are taken. One has the quantity of these metals in the ratio of 1 : 9, and the other 2 : 3. How much of each alloy do you need to take to get 15 kg of a new alloy in which the gold and silver would be 1 : 4?

Let's first try to find out how much gold and silver will be contained in 15 kg of the new alloy. The task says that the content of these metals should be 1 : 4, that is, one part of the alloy should contain gold, and four parts should contain silver. Then the total number of parts in the alloy will be 1 + 4 = 5, and the mass of one part will be 15 : 5 = 3 kg.

Let's determine how much gold is contained in 15 kg of the alloy. To do this, multiply 3 kg by the number of parts of gold:

3 kg × 1 = 3 kg

Let's determine how much silver will be contained in 15 kg of alloy:

3 kg × 4 = 12 kg

So an alloy with a mass of 15 kg will contain 3 kg of gold and 12 kg of silver. Now let's go back to the original alloys. We have to use each of them. Denote by x the mass of the first alloy, and the mass of the second alloy can be denoted by 15 - x

Express in percentages all the ratios given in the problem and fill in the following table with them:

Let's transfer these data to the table. Let's enter 10% in the first line in the column "percentage of gold in the alloy" , 90% in the first line in the column "percentage of silver in the alloy" , and in the last column "mass of the alloy" we will enter the variable x, because that is how we designated the mass of the first alloy:

Let's transfer these data to the table. Let's enter 40% in the second line in the column "percentage of gold in the alloy", 60% in the second line in the column "percentage of silver in the alloy", and in the last column "mass of the alloy" let's enter the expression 15 - x, because that is how we denoted the mass of the second alloy:

Now you can use this table to make equations. Recall the tasks of concentration, alloys, and mixtures . If we separately add up the gold of both alloys and equate this sum to the mass of gold of the resulting alloy, we can find out what the value of x equals.

Further, for convenience we will express percentages in decimals.

The first alloy had 0.10x gold and the second alloy had 0.40(15 - x) gold. Then the mass of gold in the resulting alloy will be the sum of the masses of gold of the first and second alloys and this mass is 20% of the new alloy. And 20% of the new alloy is 3 kg of gold, calculated earlier. The resulting equation is 0.10x + 0.40(15 - x) = 3 . Let us solve this equation:

Initially we denoted by x the mass of the first alloy. Now we have found the value of this variable. The variable x is 10. And the mass of the second alloy we denoted by 15 - x, and since the value of the variable x is now known, we can calculate the mass of the second alloy, it is equal to 15 - 10 = 5 kg.

So in order to obtain a new alloy with a mass of 15 kg in which gold and silver would be 1 : 4, we would have to take 10 kg of the first alloy and 5 kg of the second alloy.

The equation could also be made using the second column of the resulting table. Then we would get the equation 0.90x + 0.60(15 - x) = 12 . The root of this equation is also 10.

Task 10. There is an ore from two layers with copper grades of 6% and 11%. How much poor ore must be taken to get 20 tons with a copper content of 8% when mixed with the rich ore?

Denote by x the mass of poor ore. Since 20 tons of ore are to be obtained, 20 - x will be taken from the rich ore. Since the copper content in the poor ore is 6%, x tons of ore will contain 0.06x tons of copper. The rich ore has a copper content of 11%, and 20 - x tons of rich ore will contain 0.11(20 - x) tons of copper.

The resulting 20 tons of ore should have a copper content of 8%. This means that 20 tons of ore will contain 20 × 0.08 = 1.6 tons of copper.

Add 0.06x and 0.11(20 - x) and equate the sum to 1.6. We obtain the equation 0.06x + 0.11(20 - x) = 1.6

Let's solve this equation:

So, to get 20 tons of ore with 8% copper content, you need to take 12 tons of poor ore. The rich ore will be taken 20 - 12 = 8 tons.

Task 11. By increasing the average speed from 250 to 300 meters/minute, the athlete began to run the distance 1 minute faster. What is the length of the distance?

The length of the course (or distance of the course) can be described by the following letter equation:

Let's use the right side of this equation to make our own equation. Initially, the athlete ran the distance at a speed of 250 meters per minute. At this speed, the length of the course would be described by the expression 250t

Then, the athlete increased her speed to 300 metres per minute. At this speed, the length of the course would be described by the expression 300t

Note that the length of the course is a constant. Whether the athlete increases her speed or decreases it, the length of the distance will remain the same.

This allows us to equate 250t with 300t, since both describe the same length.

250 t = 300 t

But the problem says that at a speed of 300 meters per minute, the athlete began to run the distance 1 minute faster. In other words, at 300 meters per minute, the running time will decrease by one. Therefore, in the equation 250t = 300t on the right hand side, the time should be reduced by one:

We get the simplest equation. Let's solve it:

At a speed of 250 meters per minute, the athlete runs the distance in 6 minutes. Knowing the speed and time, you can determine the length of the course:

S = 250 × 6 = 1500 m

And at 300 meters per minute, the athlete runs the distance in t - 1, that is, in 5 minutes. As stated earlier, the length of the distance does not change:

S  = 300 × 5 = 1500 m

Task 12. A rider catches up with a pedestrian 15 km ahead of him. In how many hours will the rider catch up with the pedestrian, if the first rides 10 km every hour and the second rides only 4 km?

This task is a motion task . It can be solved by determining the speed of approach and dividing the initial distance between the rider and the pedestrian by that speed.

The speed of convergence is determined by subtracting the slower speed from the faster speed:

10 km/h - 4 km/h = 6 km/h (approach speed)

With each hour, the distance of 15 kilometers will decrease by 6 kilometers. To find out when it is completely shortened (when the rider catches up with the pedestrian), divide 15 by 6

15 : 6 = 2,5 h

2.5 hours is two whole hours and a half hour. And half an hour is 30 minutes. So the rider will catch up with the pedestrian in two hours and 30 minutes.

Let's solve this problem using an equation.

Assume that the pedestrian and the rider set out from the same place. The pedestrian left before the rider and managed to cover 15 km

After that, the rider followed him on the road at a speed of 10 km/h. The speed of the pedestrian is only 4 km/h. This means that in some time the rider will catch up with the pedestrian. We need to find this time.

When the rider catches up with the pedestrian it will mean that they have traveled the same distance together. The distance traveled by the rider and the pedestrian is described by the following equation:

Let us use the right side of this equation to make our own equation.

The distance traveled by the rider will be described by the expression 10t. Since the pedestrian went ahead of the rider and managed to cover 15 km, the distance traveled by him will be described by the expression 4t + 15.

By the time the rider catches up with the pedestrian, they have both traveled the same distance. This allows us to equate the distances traveled by the rider and the pedestrian:

Tasks for independent decision

Train speeds in this problem are measured in kilometers per hour. Therefore, convert the 45 min specified in the problem to hours. 45 min is 0.75 h

Denote the time it takes for the freight train to arrive in the city by the variable t . Since the passenger train arrives in this city 0.75 h faster, its travel time will be denoted by the expression t −  0,75

The passenger train covered 48(t-0.75) km, and the freight train 36t km. Since we are talking about the same distance, let us equate the first expression with the second one. The resulting equation is 48(t - 0.75) = 36t. Solve it:

Now calculate the distance between the cities. To do this, multiply the speed of the freight train (36 km/h) by its travel time t. The value of the variable t is now known - it is equal to three hours

36 × 3 = 108 km

You can also use the speed of the passenger train to calculate the distance. But in this case the value of the variable t should be reduced by 0.75 because the passenger train spent 0.75 h less time

48 × (3 − 0,75) = 144 − 36 = 108 km

Answer: The distance between the cities is 108 km.

Let t be the time after which the cars meet. Then the first car at the time of the meeting will have traveled 65t km, and the second 60t km. Add these distances and equate them to 150. We get the equation 65t + 60t = 150

The value of the variable t is 1.2. So the cars met after 1.2 hours.

Answer: the cars met after 1.2 hours.

Let x be the number of workers in the first workshop. The second workshop had three times as many workers as the first workshop, so the number of workers in the second workshop can be denoted by the expression 3x. The third workshop had 15 fewer workers than the second workshop. Therefore, the number of workers in the third workshop can be denoted by the expression 3x - 15.

The task says that the total number of workers was 685. Therefore we can add the expressions x, 3x, 3x - 15 and equate that sum to the number 685. The result is x + 3x + (3x - 15) = 685

The variable x was used to denote the number of workers in the first workshop. Now we have found the value of this variable, it is 100. So there were 100 workers in the first workshop.

The second workshop had 3x workers, so 3 × 100 = 300. And the third workshop had 3x - 15, so 3 × 100 - 15 = 285

Answer: In the first shop there were 100 workers, in the second - 300, in the third - 285.

Let x motors be repaired by the first workshop. Then the second workshop had to repair 18 - x motors.

Since the first workshop fulfilled its plan by 120%, that means it repaired 1.2x motors . And the second repair shop fulfilled its plan by 125%, so it repaired 1.25(18 - x) motors.

The problem says that 22 motors were repaired. Therefore we can add the expressions 1.2x and 1.25(18 - x) , then equate that sum to the number 22. The resulting equation is 1.2x + 1.25(18 - x) = 22

The variable x was used to denote the number of motors to be repaired by the first workshop. Now we have found the value of this variable, it is 10. So the first workshop had to repair 10 motors.

The expression 18 - x represents the number of motors to be repaired by the second workshop. So the second workshop had to repair 18 - 10 = 8 motors.

Answer: The first workshop was to repair 10 engines and the second workshop was to repair 8 engines..

Let x dollars was the price of the product before the price increase. If the price increased by 30%, it means that it increased by $0.30x dollars. After the price increase, the product began to cost $91. Add x to 0.30x and equate that sum to 91. As a result we get the equation x + 0.30x = 91

So before the price increase, the product cost $70.

Answer: Before the price increase, the product cost $70.

Let x be the initial number. Increase it by 25%. We obtain the expression x + 0.25x. Given the like terms, we obtain x + 0.25x = 1.25x.

Find what part of the original number x is from the new number 1.25x

If the new number 1.25x is 100%, and the original number x is 80% of it, then reducing the new number by 20% you can get the original number x>

Answer: to get the original number, the new number must be reduced by 20%.

Let x be the original number. Increase it by 20%. We obtain the expression x + 0.20x. Equate this sum to the number 144, so we obtain the equation x + 0.20x = 144

Answer: the original value of the number is 120.

Let x be the original number. Decrease it by 10%. We obtain the expression x - 0.10x. Equate this difference to the number 45, so we get the equation x - 0.10x = 45

Answer: the original value of the number is 50.

Let x be the original price of the album. Decrease this price by 15%, we get x - 0.15x. Reduce the price by another $15, so we get x - 0.15x - 15. After these reductions, the album now costs $19. Equate the expression x - 0.15x - 15 to the number 19, we get the equation x - 0.15x - 15 = 19

Answer: The original album price is $40.

If 80% of the mass is lost, the remaining 20% will account for 4 tons of hay. Let x tons of grass be required to produce 4 tons of hay. If 4 tons will account for 20% of the grass, then we can make the equation:

Answer: to get 4 tons of hay, you need to cut 20 tons of grass.

Let x kg of 20% salt solution be added to 1 kg of 10% salt solution.

1 kg of 10% salt solution contains 0.1 kg of salt. And x kg of 20% salt solution contains 0.20x kg of salt.

After adding x kg of the 20% solution, the new solution will contain 0.12(1 + x) kg of salt. We add 0.1 and 0.20x, then equate that sum to 0.12(1 + x). The resulting equation is 0.1 + 0.20x = 0.12(1 + x)

Answer: to get a 12% salt solution, you need to add 0.25 kg of 20% solution to 1 kg of 10% solution.

Let x kg of the first solution be taken. Since 25 kg of solution must be prepared, the mass of the second solution can be denoted by the expression 25 - x.

The first solution will contain 0.20x kg of salt, and the second solution will contain 0.30(25 - x) kg of salt. The resulting solution contains 25 × 0.252 = 6.3 kg of salt. Add the expressions 0.20x and 0.30(25 - x), then equate that sum to 6.3. The resulting equation is

So the first solution should take 12 kg, and the second 25 - 12 = 13 kg.

Answer: the first solution should take 12 kg, and the second 13 kg.

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Do You Understand the Problem You’re Trying to Solve?

To solve tough problems at work, first ask these questions.

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Problem solving skills are invaluable in any job. But all too often, we jump to find solutions to a problem without taking time to really understand the dilemma we face, according to Thomas Wedell-Wedellsborg , an expert in innovation and the author of the book, What’s Your Problem?: To Solve Your Toughest Problems, Change the Problems You Solve .

In this episode, you’ll learn how to reframe tough problems by asking questions that reveal all the factors and assumptions that contribute to the situation. You’ll also learn why searching for just one root cause can be misleading.

Key episode topics include: leadership, decision making and problem solving, power and influence, business management.

HBR On Leadership curates the best case studies and conversations with the world’s top business and management experts, to help you unlock the best in those around you. New episodes every week.

• Listen to the original HBR IdeaCast episode: The Secret to Better Problem Solving (2016)
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• Discover 100 years of Harvard Business Review articles, case studies, podcasts, and more at HBR.org .

HANNAH BATES: Welcome to HBR on Leadership , case studies and conversations with the world’s top business and management experts, hand-selected to help you unlock the best in those around you.

Problem solving skills are invaluable in any job. But even the most experienced among us can fall into the trap of solving the wrong problem.

Thomas Wedell-Wedellsborg says that all too often, we jump to find solutions to a problem – without taking time to really understand what we’re facing.

He’s an expert in innovation, and he’s the author of the book, What’s Your Problem?: To Solve Your Toughest Problems, Change the Problems You Solve .

  In this episode, you’ll learn how to reframe tough problems, by asking questions that reveal all the factors and assumptions that contribute to the situation. You’ll also learn why searching for one root cause can be misleading. And you’ll learn how to use experimentation and rapid prototyping as problem-solving tools.

This episode originally aired on HBR IdeaCast in December 2016. Here it is.

SARAH GREEN CARMICHAEL: Welcome to the HBR IdeaCast from Harvard Business Review. I’m Sarah Green Carmichael.

Problem solving is popular. People put it on their resumes. Managers believe they excel at it. Companies count it as a key proficiency. We solve customers’ problems.

The problem is we often solve the wrong problems. Albert Einstein and Peter Drucker alike have discussed the difficulty of effective diagnosis. There are great frameworks for getting teams to attack true problems, but they’re often hard to do daily and on the fly. That’s where our guest comes in.

Thomas Wedell-Wedellsborg is a consultant who helps companies and managers reframe their problems so they can come up with an effective solution faster. He asks the question “Are You Solving The Right Problems?” in the January-February 2017 issue of Harvard Business Review. Thomas, thank you so much for coming on the HBR IdeaCast .

THOMAS WEDELL-WEDELLSBORG: Thanks for inviting me.

SARAH GREEN CARMICHAEL: So, I thought maybe we could start by talking about the problem of talking about problem reframing. What is that exactly?

THOMAS WEDELL-WEDELLSBORG: Basically, when people face a problem, they tend to jump into solution mode to rapidly, and very often that means that they don’t really understand, necessarily, the problem they’re trying to solve. And so, reframing is really a– at heart, it’s a method that helps you avoid that by taking a second to go in and ask two questions, basically saying, first of all, wait. What is the problem we’re trying to solve? And then crucially asking, is there a different way to think about what the problem actually is?

SARAH GREEN CARMICHAEL: So, I feel like so often when this comes up in meetings, you know, someone says that, and maybe they throw out the Einstein quote about you spend an hour of problem solving, you spend 55 minutes to find the problem. And then everyone else in the room kind of gets irritated. So, maybe just give us an example of maybe how this would work in practice in a way that would not, sort of, set people’s teeth on edge, like oh, here Sarah goes again, reframing the whole problem instead of just solving it.

THOMAS WEDELL-WEDELLSBORG: I mean, you’re bringing up something that’s, I think is crucial, which is to create legitimacy for the method. So, one of the reasons why I put out the article is to give people a tool to say actually, this thing is still important, and we need to do it. But I think the really critical thing in order to make this work in a meeting is actually to learn how to do it fast, because if you have the idea that you need to spend 30 minutes in a meeting delving deeply into the problem, I mean, that’s going to be uphill for most problems. So, the critical thing here is really to try to make it a practice you can implement very, very rapidly.

There’s an example that I would suggest memorizing. This is the example that I use to explain very rapidly what it is. And it’s basically, I call it the slow elevator problem. You imagine that you are the owner of an office building, and that your tenants are complaining that the elevator’s slow.

Now, if you take that problem framing for granted, you’re going to start thinking creatively around how do we make the elevator faster. Do we install a new motor? Do we have to buy a new lift somewhere?

The thing is, though, if you ask people who actually work with facilities management, well, they’re going to have a different solution for you, which is put up a mirror next to the elevator. That’s what happens is, of course, that people go oh, I’m busy. I’m busy. I’m– oh, a mirror. Oh, that’s beautiful.

And then they forget time. What’s interesting about that example is that the idea with a mirror is actually a solution to a different problem than the one you first proposed. And so, the whole idea here is once you get good at using reframing, you can quickly identify other aspects of the problem that might be much better to try to solve than the original one you found. It’s not necessarily that the first one is wrong. It’s just that there might be better problems out there to attack that we can, means we can do things much faster, cheaper, or better.

SARAH GREEN CARMICHAEL: So, in that example, I can understand how A, it’s probably expensive to make the elevator faster, so it’s much cheaper just to put up a mirror. And B, maybe the real problem people are actually feeling, even though they’re not articulating it right, is like, I hate waiting for the elevator. But if you let them sort of fix their hair or check their teeth, they’re suddenly distracted and don’t notice.

But if you have, this is sort of a pedestrian example, but say you have a roommate or a spouse who doesn’t clean up the kitchen. Facing that problem and not having your elegant solution already there to highlight the contrast between the perceived problem and the real problem, how would you take a problem like that and attack it using this method so that you can see what some of the other options might be?

THOMAS WEDELL-WEDELLSBORG: Right. So, I mean, let’s say it’s you who have that problem. I would go in and say, first of all, what would you say the problem is? Like, if you were to describe your view of the problem, what would that be?

SARAH GREEN CARMICHAEL: I hate cleaning the kitchen, and I want someone else to clean it up.

THOMAS WEDELL-WEDELLSBORG: OK. So, my first observation, you know, that somebody else might not necessarily be your spouse. So, already there, there’s an inbuilt assumption in your question around oh, it has to be my husband who does the cleaning. So, it might actually be worth, already there to say, is that really the only problem you have? That you hate cleaning the kitchen, and you want to avoid it? Or might there be something around, as well, getting a better relationship in terms of how you solve problems in general or establishing a better way to handle small problems when dealing with your spouse?

SARAH GREEN CARMICHAEL: Or maybe, now that I’m thinking that, maybe the problem is that you just can’t find the stuff in the kitchen when you need to find it.

THOMAS WEDELL-WEDELLSBORG: Right, and so that’s an example of a reframing, that actually why is it a problem that the kitchen is not clean? Is it only because you hate the act of cleaning, or does it actually mean that it just takes you a lot longer and gets a lot messier to actually use the kitchen, which is a different problem. The way you describe this problem now, is there anything that’s missing from that description?

SARAH GREEN CARMICHAEL: That is a really good question.

THOMAS WEDELL-WEDELLSBORG: Other, basically asking other factors that we are not talking about right now, and I say those because people tend to, when given a problem, they tend to delve deeper into the detail. What often is missing is actually an element outside of the initial description of the problem that might be really relevant to what’s going on. Like, why does the kitchen get messy in the first place? Is it something about the way you use it or your cooking habits? Is it because the neighbor’s kids, kind of, use it all the time?

There might, very often, there might be issues that you’re not really thinking about when you first describe the problem that actually has a big effect on it.

SARAH GREEN CARMICHAEL: I think at this point it would be helpful to maybe get another business example, and I’m wondering if you could tell us the story of the dog adoption problem.

THOMAS WEDELL-WEDELLSBORG: Yeah. This is a big problem in the US. If you work in the shelter industry, basically because dogs are so popular, more than 3 million dogs every year enter a shelter, and currently only about half of those actually find a new home and get adopted. And so, this is a problem that has persisted. It’s been, like, a structural problem for decades in this space. In the last three years, where people found new ways to address it.

So a woman called Lori Weise who runs a rescue organization in South LA, and she actually went in and challenged the very idea of what we were trying to do. She said, no, no. The problem we’re trying to solve is not about how to get more people to adopt dogs. It is about keeping the dogs with their first family so they never enter the shelter system in the first place.

In 2013, she started what’s called a Shelter Intervention Program that basically works like this. If a family comes and wants to hand over their dog, these are called owner surrenders. It’s about 30% of all dogs that come into a shelter. All they would do is go up and ask, if you could, would you like to keep your animal? And if they said yes, they would try to fix whatever helped them fix the problem, but that made them turn over this.

And sometimes that might be that they moved into a new building. The landlord required a deposit, and they simply didn’t have the money to put down a deposit. Or the dog might need a $10 rabies shot, but they didn’t know how to get access to a vet.

And so, by instigating that program, just in the first year, she took her, basically the amount of dollars they spent per animal they helped went from something like $85 down to around $60. Just an immediate impact, and her program now is being rolled out, is being supported by the ASPCA, which is one of the big animal welfare stations, and it’s being rolled out to various other places.

And I think what really struck me with that example was this was not dependent on having the internet. This was not, oh, we needed to have everybody mobile before we could come up with this. This, conceivably, we could have done 20 years ago. Only, it only happened when somebody, like in this case Lori, went in and actually rethought what the problem they were trying to solve was in the first place.

SARAH GREEN CARMICHAEL: So, what I also think is so interesting about that example is that when you talk about it, it doesn’t sound like the kind of thing that would have been thought of through other kinds of problem solving methods. There wasn’t necessarily an After Action Review or a 5 Whys exercise or a Six Sigma type intervention. I don’t want to throw those other methods under the bus, but how can you get such powerful results with such a very simple way of thinking about something?

THOMAS WEDELL-WEDELLSBORG: That was something that struck me as well. This, in a way, reframing and the idea of the problem diagnosis is important is something we’ve known for a long, long time. And we’ve actually have built some tools to help out. If you worked with us professionally, you are familiar with, like, Six Sigma, TRIZ, and so on. You mentioned 5 Whys. A root cause analysis is another one that a lot of people are familiar with.

Those are our good tools, and they’re definitely better than nothing. But what I notice when I work with the companies applying those was those tools tend to make you dig deeper into the first understanding of the problem we have. If it’s the elevator example, people start asking, well, is that the cable strength, or is the capacity of the elevator? That they kind of get caught by the details.

That, in a way, is a bad way to work on problems because it really assumes that there’s like a, you can almost hear it, a root cause. That you have to dig down and find the one true problem, and everything else was just symptoms. That’s a bad way to think about problems because problems tend to be multicausal.

There tend to be lots of causes or levers you can potentially press to address a problem. And if you think there’s only one, if that’s the right problem, that’s actually a dangerous way. And so I think that’s why, that this is a method I’ve worked with over the last five years, trying to basically refine how to make people better at this, and the key tends to be this thing about shifting out and saying, is there a totally different way of thinking about the problem versus getting too caught up in the mechanistic details of what happens.

SARAH GREEN CARMICHAEL: What about experimentation? Because that’s another method that’s become really popular with the rise of Lean Startup and lots of other innovation methodologies. Why wouldn’t it have worked to, say, experiment with many different types of fixing the dog adoption problem, and then just pick the one that works the best?

THOMAS WEDELL-WEDELLSBORG: You could say in the dog space, that’s what’s been going on. I mean, there is, in this industry and a lot of, it’s largely volunteer driven. People have experimented, and they found different ways of trying to cope. And that has definitely made the problem better. So, I wouldn’t say that experimentation is bad, quite the contrary. Rapid prototyping, quickly putting something out into the world and learning from it, that’s a fantastic way to learn more and to move forward.

My point is, though, that I feel we’ve come to rely too much on that. There’s like, if you look at the start up space, the wisdom is now just to put something quickly into the market, and then if it doesn’t work, pivot and just do more stuff. What reframing really is, I think of it as the cognitive counterpoint to prototyping. So, this is really a way of seeing very quickly, like not just working on the solution, but also working on our understanding of the problem and trying to see is there a different way to think about that.

If you only stick with experimentation, again, you tend to sometimes stay too much in the same space trying minute variations of something instead of taking a step back and saying, wait a minute. What is this telling us about what the real issue is?

SARAH GREEN CARMICHAEL: So, to go back to something that we touched on earlier, when we were talking about the completely hypothetical example of a spouse who does not clean the kitchen–

THOMAS WEDELL-WEDELLSBORG: Completely, completely hypothetical.

SARAH GREEN CARMICHAEL: Yes. For the record, my husband is a great kitchen cleaner.

You started asking me some questions that I could see immediately were helping me rethink that problem. Is that kind of the key, just having a checklist of questions to ask yourself? How do you really start to put this into practice?

THOMAS WEDELL-WEDELLSBORG: I think there are two steps in that. The first one is just to make yourself better at the method. Yes, you should kind of work with a checklist. In the article, I kind of outlined seven practices that you can use to do this.

But importantly, I would say you have to consider that as, basically, a set of training wheels. I think there’s a big, big danger in getting caught in a checklist. This is something I work with.

My co-author Paddy Miller, it’s one of his insights. That if you start giving people a checklist for things like this, they start following it. And that’s actually a problem, because what you really want them to do is start challenging their thinking.

So the way to handle this is to get some practice using it. Do use the checklist initially, but then try to step away from it and try to see if you can organically make– it’s almost a habit of mind. When you run into a colleague in the hallway and she has a problem and you have five minutes, like, delving in and just starting asking some of those questions and using your intuition to say, wait, how is she talking about this problem? And is there a question or two I can ask her about the problem that can help her rethink it?

SARAH GREEN CARMICHAEL: Well, that is also just a very different approach, because I think in that situation, most of us can’t go 30 seconds without jumping in and offering solutions.

THOMAS WEDELL-WEDELLSBORG: Very true. The drive toward solutions is very strong. And to be clear, I mean, there’s nothing wrong with that if the solutions work. So, many problems are just solved by oh, you know, oh, here’s the way to do that. Great.

But this is really a powerful method for those problems where either it’s something we’ve been banging our heads against tons of times without making progress, or when you need to come up with a really creative solution. When you’re facing a competitor with a much bigger budget, and you know, if you solve the same problem later, you’re not going to win. So, that basic idea of taking that approach to problems can often help you move forward in a different way than just like, oh, I have a solution.

I would say there’s also, there’s some interesting psychological stuff going on, right? Where you may have tried this, but if somebody tries to serve up a solution to a problem I have, I’m often resistant towards them. Kind if like, no, no, no, no, no, no. That solution is not going to work in my world. Whereas if you get them to discuss and analyze what the problem really is, you might actually dig something up.

Let’s go back to the kitchen example. One powerful question is just to say, what’s your own part in creating this problem? It’s very often, like, people, they describe problems as if it’s something that’s inflicted upon them from the external world, and they are innocent bystanders in that.

SARAH GREEN CARMICHAEL: Right, or crazy customers with unreasonable demands.

THOMAS WEDELL-WEDELLSBORG: Exactly, right. I don’t think I’ve ever met an agency or consultancy that didn’t, like, gossip about their customers. Oh, my god, they’re horrible. That, you know, classic thing, why don’t they want to take more risk? Well, risk is bad.

It’s their business that’s on the line, not the consultancy’s, right? So, absolutely, that’s one of the things when you step into a different mindset and kind of, wait. Oh yeah, maybe I actually am part of creating this problem in a sense, as well. That tends to open some new doors for you to move forward, in a way, with stuff that you may have been struggling with for years.

SARAH GREEN CARMICHAEL: So, we’ve surfaced a couple of questions that are useful. I’m curious to know, what are some of the other questions that you find yourself asking in these situations, given that you have made this sort of mental habit that you do? What are the questions that people seem to find really useful?

THOMAS WEDELL-WEDELLSBORG: One easy one is just to ask if there are any positive exceptions to the problem. So, was there day where your kitchen was actually spotlessly clean? And then asking, what was different about that day? Like, what happened there that didn’t happen the other days? That can very often point people towards a factor that they hadn’t considered previously.

SARAH GREEN CARMICHAEL: We got take-out.

THOMAS WEDELL-WEDELLSBORG: S,o that is your solution. Take-out from [INAUDIBLE]. That might have other problems.

Another good question, and this is a little bit more high level. It’s actually more making an observation about labeling how that person thinks about the problem. And what I mean with that is, we have problem categories in our head. So, if I say, let’s say that you describe a problem to me and say, well, we have a really great product and are, it’s much better than our previous product, but people aren’t buying it. I think we need to put more marketing dollars into this.

Now you can go in and say, that’s interesting. This sounds like you’re thinking of this as a communications problem. Is there a different way of thinking about that? Because you can almost tell how, when the second you say communications, there are some ideas about how do you solve a communications problem. Typically with more communication.

And what you might do is go in and suggest, well, have you considered that it might be, say, an incentive problem? Are there incentives on behalf of the purchasing manager at your clients that are obstructing you? Might there be incentive issues with your own sales force that makes them want to sell the old product instead of the new one?

So literally, just identifying what type of problem does this person think about, and is there different potential way of thinking about it? Might it be an emotional problem, a timing problem, an expectations management problem? Thinking about what label of what type of problem that person is kind of thinking as it of.

SARAH GREEN CARMICHAEL: That’s really interesting, too, because I think so many of us get requests for advice that we’re really not qualified to give. So, maybe the next time that happens, instead of muddying my way through, I will just ask some of those questions that we talked about instead.

THOMAS WEDELL-WEDELLSBORG: That sounds like a good idea.

SARAH GREEN CARMICHAEL: So, Thomas, this has really helped me reframe the way I think about a couple of problems in my own life, and I’m just wondering. I know you do this professionally, but is there a problem in your life that thinking this way has helped you solve?

THOMAS WEDELL-WEDELLSBORG: I’ve, of course, I’ve been swallowing my own medicine on this, too, and I think I have, well, maybe two different examples, and in one case somebody else did the reframing for me. But in one case, when I was younger, I often kind of struggled a little bit. I mean, this is my teenage years, kind of hanging out with my parents. I thought they were pretty annoying people. That’s not really fair, because they’re quite wonderful, but that’s what life is when you’re a teenager.

And one of the things that struck me, suddenly, and this was kind of the positive exception was, there was actually an evening where we really had a good time, and there wasn’t a conflict. And the core thing was, I wasn’t just seeing them in their old house where I grew up. It was, actually, we were at a restaurant. And it suddenly struck me that so much of the sometimes, kind of, a little bit, you love them but they’re annoying kind of dynamic, is tied to the place, is tied to the setting you are in.

And of course, if– you know, I live abroad now, if I visit my parents and I stay in my old bedroom, you know, my mother comes in and wants to wake me up in the morning. Stuff like that, right? And it just struck me so, so clearly that it’s– when I change this setting, if I go out and have dinner with them at a different place, that the dynamic, just that dynamic disappears.

SARAH GREEN CARMICHAEL: Well, Thomas, this has been really, really helpful. Thank you for talking with me today.

THOMAS WEDELL-WEDELLSBORG: Thank you, Sarah.  

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Title: machine learning based optimization workflow for tuning numerical settings of differential equation solvers for boundary value problems.

Abstract: Several numerical differential equation solvers have been employed effectively over the years as an alternative to analytical solvers to quickly and conveniently solve differential equations. One category of these is boundary value solvers, which are used to solve real-world problems formulated as differential equations with boundary conditions. These solvers require certain numerical settings to solve the differential equations that affect their solvability and performance. A systematic fine-tuning of these settings is required to obtain the desired solution and performance. Currently, these settings are either selected by trial and error or require domain expertise. In this paper, we propose a machine learning-based optimization workflow for fine-tuning the numerical settings to reduce the time and domain expertise required in the process. In the evaluation section, we discuss the scalability, stability, and reliability of the proposed workflow. We demonstrate our workflow on a numerical boundary value problem solver.

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New results for fractional ordinary differential equations in fuzzy metric space

• Li Chen ,  , 
• Suyun Wang , 
• Yongjun Li , 
• Jinying Wei
• School of Information Engineering, Lanzhou City University, Lanzhou, 730070, China
• Received: 17 January 2024 Revised: 27 March 2024 Accepted: 08 April 2024 Published: 15 April 2024

MSC : 34A08, 34B15, 35J05

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In this paper, we primarily focused on the existence and uniqueness of the initial value problem for fractional order fuzzy ordinary differential equations in a fuzzy metric space. First, definitions and relevant properties of the Gamma function and Beta function within a fuzzy metric space were provided. Second, by employing the principle of fuzzy compression mapping and Choquet integral of fuzzy numerical functions, we established the existence and uniqueness of solutions to initial value problems for fuzzy ordinary differential equations. Finally, several examples were presented to demonstrate the validity of our obtained results.

• fuzzy metric space ,
• fractional fuzzy differential equation ,
• power compression mapping principle ,
• existence and uniqueness

Citation: Li Chen, Suyun Wang, Yongjun Li, Jinying Wei. New results for fractional ordinary differential equations in fuzzy metric space[J]. AIMS Mathematics, 2024, 9(6): 13861-13873. doi: 10.3934/math.2024674

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• © 2024 the Author(s), licensee AIMS Press. This is an open access article distributed under the terms of the Creative Commons Attribution License ( http://creativecommons.org/licenses/by/4.0 )

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