7.4 Solve Rational Equations
Learning objectives.
By the end of this section, you will be able to:
- Solve rational equations
- Use rational functions
- Solve a rational equation for a specific variable
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Be Prepared 7.10
Before you get started, take this readiness quiz.
Solve: 1 6 x + 1 2 = 1 3 . 1 6 x + 1 2 = 1 3 . If you missed this problem, review Example 2.9 .
Be Prepared 7.11
Solve: n 2 − 5 n − 36 = 0 . n 2 − 5 n − 36 = 0 . If you missed this problem, review Example 6.45 .
Be Prepared 7.12
Solve the formula 5 x + 2 y = 10 5 x + 2 y = 10 for y . y . If you missed this problem, review Example 2.31 .
After defining the terms ‘expression’ and ‘equation’ earlier, we have used them throughout this book. We have simplified many kinds of expressions and solved many kinds of equations . We have simplified many rational expressions so far in this chapter. Now we will solve a rational equation .
Rational Equation
A rational equation is an equation that contains a rational expression.
You must make sure to know the difference between rational expressions and rational equations. The equation contains an equal sign.
Solve Rational Equations
We have already solved linear equations that contained fractions. We found the LCD of all the fractions in the equation and then multiplied both sides of the equation by the LCD to “clear” the fractions.
We will use the same strategy to solve rational equations. We will multiply both sides of the equation by the LCD. Then, we will have an equation that does not contain rational expressions and thus is much easier for us to solve. But because the original equation may have a variable in a denominator, we must be careful that we don’t end up with a solution that would make a denominator equal to zero.
So before we begin solving a rational equation, we examine it first to find the values that would make any denominators zero. That way, when we solve a rational equation we will know if there are any algebraic solutions we must discard.
An algebraic solution to a rational equation that would cause any of the rational expressions to be undefined is called an extraneous solution to a rational equation .
Extraneous Solution to a Rational Equation
An extraneous solution to a rational equation is an algebraic solution that would cause any of the expressions in the original equation to be undefined.
We note any possible extraneous solutions, c , by writing x ≠ c x ≠ c next to the equation.
Example 7.33
How to solve a rational equation.
Solve: 1 x + 1 3 = 5 6 . 1 x + 1 3 = 5 6 .
Try It 7.65
Solve: 1 y + 2 3 = 1 5 . 1 y + 2 3 = 1 5 .
Try It 7.66
Solve: 2 3 + 1 5 = 1 x . 2 3 + 1 5 = 1 x .
The steps of this method are shown.
Solve equations with rational expressions.
- Step 1. Note any value of the variable that would make any denominator zero.
- Step 2. Find the least common denominator of all denominators in the equation.
- Step 3. Clear the fractions by multiplying both sides of the equation by the LCD.
- Step 4. Solve the resulting equation.
- If any values found in Step 1 are algebraic solutions, discard them.
- Check any remaining solutions in the original equation.
We always start by noting the values that would cause any denominators to be zero.
Example 7.34
How to solve a rational equation using the zero product property.
Solve: 1 − 5 y = − 6 y 2 . 1 − 5 y = − 6 y 2 .
Note any value of the variable that would make any denominator zero. | |
Find the least common denominator of all denominators in the equation. The LCD is . | |
Clear the fractions by multiplying both sides of the equation by the LCD. | |
Distribute. | |
Multiply. | |
Solve the resulting equation. First write the quadratic equation in standard form. | |
Factor. | |
Use the Zero Product Property. | |
Solve. | |
Check. We did not get 0 as an algebraic solution. | |
The solution is |
Try It 7.67
Solve: 1 − 2 x = 15 x 2 . 1 − 2 x = 15 x 2 .
Try It 7.68
Solve: 1 − 4 y = 12 y 2 . 1 − 4 y = 12 y 2 .
In the next example, the last denominators is a difference of squares. Remember to factor it first to find the LCD.
Example 7.35
Solve: 2 x + 2 + 4 x − 2 = x − 1 x 2 − 4 . 2 x + 2 + 4 x − 2 = x − 1 x 2 − 4 .
Note any value of the variable that would make any denominator zero. | |
Find the least common denominator of all denominators in the equation. The LCD is | |
Clear the fractions by multiplying both sides of the equation by the LCD. | |
Distribute. | |
Remove common factors. | |
Simplify. | |
Distribute. | |
Solve. | |
Check: We did not get 2 or −2 as algebraic solutions. | |
The solution is |
Try It 7.69
Solve: 2 x + 1 + 1 x − 1 = 1 x 2 − 1 . 2 x + 1 + 1 x − 1 = 1 x 2 − 1 .
Try It 7.70
Solve: 5 y + 3 + 2 y − 3 = 5 y 2 − 9 . 5 y + 3 + 2 y − 3 = 5 y 2 − 9 .
In the next example, the first denominator is a trinomial . Remember to factor it first to find the LCD.
Example 7.36
Solve: m + 11 m 2 − 5 m + 4 = 5 m − 4 − 3 m − 1 . m + 11 m 2 − 5 m + 4 = 5 m − 4 − 3 m − 1 .
Note any value of the variable that would make any denominator zero. Use the factored form of the quadratic denominator. | |
Find the least common denominator of all denominators in the equation. The LCD is | |
Clear the fractions by multiplying both sides of the equation by the LCD. | |
Distribute. | |
Remove common factors. | |
Simplify. | |
Solve the resulting equation. | |
Check. The only algebraic solution was 4, but we said that 4 would make a denominator equal to zero. The algebraic solution is an extraneous solution. | |
There is no solution to this equation. |
Try It 7.71
Solve: x + 13 x 2 − 7 x + 10 = 6 x − 5 − 4 x − 2 . x + 13 x 2 − 7 x + 10 = 6 x − 5 − 4 x − 2 .
Try It 7.72
Solve: y − 6 y 2 + 3 y − 4 = 2 y + 4 + 7 y − 1 . y − 6 y 2 + 3 y − 4 = 2 y + 4 + 7 y − 1 .
The equation we solved in the previous example had only one algebraic solution, but it was an extraneous solution. That left us with no solution to the equation. In the next example we get two algebraic solutions. Here one or both could be extraneous solutions.
Example 7.37
Solve: y y + 6 = 72 y 2 − 36 + 4 . y y + 6 = 72 y 2 − 36 + 4 .
Factor all the denominators, so we can note any value of the variable that would make any denominator zero. | |
Find the least common denominator. The LCD is | |
Clear the fractions. | |
Simplify. | |
Simplify. | |
Solve the resulting equation. | |
Check. | |
The solution is |
Try It 7.73
Solve: x x + 4 = 32 x 2 − 16 + 5 . x x + 4 = 32 x 2 − 16 + 5 .
Try It 7.74
Solve: y y + 8 = 128 y 2 − 64 + 9 . y y + 8 = 128 y 2 − 64 + 9 .
In some cases, all the algebraic solutions are extraneous.
Example 7.38
Solve: x 2 x − 2 − 2 3 x + 3 = 5 x 2 − 2 x + 9 12 x 2 − 12 . x 2 x − 2 − 2 3 x + 3 = 5 x 2 − 2 x + 9 12 x 2 − 12 .
We will start by factoring all denominators, to make it easier to identify extraneous solutions and the LCD. | |
Note any value of the variable that would make any denominator zero. | |
Find the least common denominator. The LCD is | |
Clear the fractions. | |
Simplify. | |
Simplify. | |
Solve the resulting equation. | |
Check. and are extraneous solutions. | |
The equation has no solution. |
Try It 7.75
Solve: y 5 y − 10 − 5 3 y + 6 = 2 y 2 − 19 y + 54 15 y 2 − 60 . y 5 y − 10 − 5 3 y + 6 = 2 y 2 − 19 y + 54 15 y 2 − 60 .
Try It 7.76
Solve: z 2 z + 8 − 3 4 z − 8 = 3 z 2 − 16 z − 16 8 z 2 + 16 z − 64 . z 2 z + 8 − 3 4 z − 8 = 3 z 2 − 16 z − 16 8 z 2 + 16 z − 64 .
Example 7.39
Solve: 4 3 x 2 − 10 x + 3 + 3 3 x 2 + 2 x − 1 = 2 x 2 − 2 x − 3 . 4 3 x 2 − 10 x + 3 + 3 3 x 2 + 2 x − 1 = 2 x 2 − 2 x − 3 .
Factor all the denominators, so we can note any value of the variable that would make any denominator zero. | |
Find the least common denominator. The LCD is | |
Clear the fractions. | |
Simplify. | |
Distribute. | |
Simplify. | |
The only algebraic solution was but we said that would make a denominator equal to zero. The algebraic solution is an extraneous solution. | |
There is no solution to this equation. |
Try It 7.77
Solve: 15 x 2 + x − 6 − 3 x − 2 = 2 x + 3 . 15 x 2 + x − 6 − 3 x − 2 = 2 x + 3 .
Try It 7.78
Solve: 5 x 2 + 2 x − 3 − 3 x 2 + x − 2 = 1 x 2 + 5 x + 6 . 5 x 2 + 2 x − 3 − 3 x 2 + x − 2 = 1 x 2 + 5 x + 6 .
Use Rational Functions
Working with functions that are defined by rational expressions often lead to rational equations. Again, we use the same techniques to solve them.
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Example 7.40
For rational function, f ( x ) = 2 x − 6 x 2 − 8 x + 15 , f ( x ) = 2 x − 6 x 2 − 8 x + 15 , ⓐ find the domain of the function, ⓑ solve f ( x ) = 1 , f ( x ) = 1 , and ⓒ find the points on the graph at this function value.
ⓐ The domain of a rational function is all real numbers except those that make the rational expression undefined. So to find them, we will set the denominator equal to zero and solve.
Factor the trinomial. | |
Use the Zero Product Property. | |
Solve. | |
The domain is all real numbers except |
Substitute in the rational expression. | |
Factor the denominator. | |
Multiply both sides by the LCD, | |
Simplify. | |
Solve. | |
Factor. | |
Use the Zero Product Property. | |
Solve. |
However, x = 3 x = 3 is outside the domain of this function, so we discard that root as extraneous.
ⓒ The value of the function is 1 when x = 7 . x = 7 . So the points on the graph of this function when f ( x ) = 1 f ( x ) = 1 is ( 7 , 1 ) ) ( 7 , 1 ) )
Try It 7.79
For rational function, f ( x ) = 8 − x x 2 − 7 x + 12 , f ( x ) = 8 − x x 2 − 7 x + 12 , ⓐ find the domain of the function ⓑ solve f ( x ) = 3 f ( x ) = 3 ⓒ find the points on the graph at this function value.
Try It 7.80
For rational function, f ( x ) = x − 1 x 2 − 6 x + 5 , f ( x ) = x − 1 x 2 − 6 x + 5 , ⓐ find the domain of the function ⓑ solve f ( x ) = 4 f ( x ) = 4 ⓒ find the points on the graph at this function value.
Solve a Rational Equation for a Specific Variable
When we solved linear equations, we learned how to solve a formula for a specific variable. Many formulas used in business, science, economics, and other fields use rational equations to model the relation between two or more variables. We will now see how to solve a rational equation for a specific variable.
When we developed the point-slope formula from our slope formula, we cleared the fractions by multiplying by the LCD.
m = y − y 1 x − x 1 Multiply both sides of the equation by x − x 1 . m ( x − x 1 ) = ( y − y 1 x − x 1 ) ( x − x 1 ) Simplify. m ( x − x 1 ) = y − y 1 Rewrite the equation with the y terms on the left. y − y 1 = m ( x − x 1 ) m = y − y 1 x − x 1 Multiply both sides of the equation by x − x 1 . m ( x − x 1 ) = ( y − y 1 x − x 1 ) ( x − x 1 ) Simplify. m ( x − x 1 ) = y − y 1 Rewrite the equation with the y terms on the left. y − y 1 = m ( x − x 1 )
In the next example, we will use the same technique with the formula for slope that we used to get the point-slope form of an equation of a line through the point ( 2 , 3 ) . ( 2 , 3 ) . We will add one more step to solve for y .
Example 7.41
Solve: m = y − 2 x − 3 m = y − 2 x − 3 for y . y .
Note any value of the variable that would make any denominator zero. | |
Clear the fractions by multiplying both sides of the equation by the LCD, | |
Simplify. | |
Isolate the term with . |
Try It 7.81
Solve: m = y − 5 x − 4 m = y − 5 x − 4 for y . y .
Try It 7.82
Solve: m = y − 1 x + 5 m = y − 1 x + 5 for y . y .
Remember to multiply both sides by the LCD in the next example.
Example 7.42
Solve: 1 c + 1 m = 1 1 c + 1 m = 1 for c .
Note any value of the variable that would make any denominator zero. | |
Clear the fractions by multiplying both sides of the equations by the LCD, . | |
Distribute. | |
Simplify. | |
Collect the terms with to the right. | |
Factor the expression on the right. | |
To isolate , divide both sides by | |
Simplify by removing common factors. | |
Notice that even though we excluded from the original equation, we must also now state that |
Try It 7.83
Solve: 1 a + 1 b = c 1 a + 1 b = c for a .
Try It 7.84
Solve: 2 x + 1 3 = 1 y 2 x + 1 3 = 1 y for y .
Access this online resource for additional instruction and practice with equations with rational expressions.
- Equations with Rational Expressions
Section 7.4 Exercises
Practice makes perfect.
In the following exercises, solve each rational equation.
1 a + 2 5 = 1 2 1 a + 2 5 = 1 2
6 3 − 2 d = 4 9 6 3 − 2 d = 4 9
4 5 + 1 4 = 2 v 4 5 + 1 4 = 2 v
3 8 + 2 y = 1 4 3 8 + 2 y = 1 4
1 − 2 m = 8 m 2 1 − 2 m = 8 m 2
1 + 4 n = 21 n 2 1 + 4 n = 21 n 2
1 + 9 p = −20 p 2 1 + 9 p = −20 p 2
1 − 7 q = −6 q 2 1 − 7 q = −6 q 2
5 3 v − 2 = 7 4 v 5 3 v − 2 = 7 4 v
8 2 w + 1 = 3 w 8 2 w + 1 = 3 w
3 x + 4 + 7 x − 4 = 8 x 2 − 16 3 x + 4 + 7 x − 4 = 8 x 2 − 16
5 y − 9 + 1 y + 9 = 18 y 2 − 81 5 y − 9 + 1 y + 9 = 18 y 2 − 81
8 z − 10 − 7 z + 10 = 5 z 2 − 100 8 z − 10 − 7 z + 10 = 5 z 2 − 100
9 a + 11 − 6 a − 11 = 6 a 2 − 121 9 a + 11 − 6 a − 11 = 6 a 2 − 121
−10 q − 2 − 7 q + 4 = 1 −10 q − 2 − 7 q + 4 = 1
2 s + 7 − 3 s − 3 = 1 2 s + 7 − 3 s − 3 = 1
v − 10 v 2 − 5 v + 4 = 3 v − 1 − 6 v − 4 v − 10 v 2 − 5 v + 4 = 3 v − 1 − 6 v − 4
w + 8 w 2 − 11 w + 28 = 5 w − 7 + 2 w − 4 w + 8 w 2 − 11 w + 28 = 5 w − 7 + 2 w − 4
x − 10 x 2 + 8 x + 12 = 3 x + 2 + 4 x + 6 x − 10 x 2 + 8 x + 12 = 3 x + 2 + 4 x + 6
y − 5 y 2 − 4 y − 5 = 1 y + 1 + 1 y − 5 y − 5 y 2 − 4 y − 5 = 1 y + 1 + 1 y − 5
b + 3 3 b + b 24 = 1 b b + 3 3 b + b 24 = 1 b
c + 3 12 c + c 36 = 1 4 c c + 3 12 c + c 36 = 1 4 c
d d + 3 = 18 d 2 − 9 + 4 d d + 3 = 18 d 2 − 9 + 4
m m + 5 = 50 m 2 − 25 + 6 m m + 5 = 50 m 2 − 25 + 6
n n + 2 − 3 = 8 n 2 − 4 n n + 2 − 3 = 8 n 2 − 4
p p + 7 − 8 = 98 p 2 − 49 p p + 7 − 8 = 98 p 2 − 49
q 3 q − 9 − 3 4 q + 12 = 7 q 2 + 6 q + 63 24 q 2 − 216 q 3 q − 9 − 3 4 q + 12 = 7 q 2 + 6 q + 63 24 q 2 − 216
r 3 r − 15 − 1 4 r + 20 = 3 r 2 + 17 r + 40 12 r 2 − 300 r 3 r − 15 − 1 4 r + 20 = 3 r 2 + 17 r + 40 12 r 2 − 300
s 2 s + 6 − 2 5 s + 5 = 5 s 2 − 3 s − 7 10 s 2 + 40 s + 30 s 2 s + 6 − 2 5 s + 5 = 5 s 2 − 3 s − 7 10 s 2 + 40 s + 30
t 6 t − 12 − 5 2 t + 10 = t 2 − 23 t + 70 12 t 2 + 36 t − 120 t 6 t − 12 − 5 2 t + 10 = t 2 − 23 t + 70 12 t 2 + 36 t − 120
2 x 2 + 2 x − 8 − 1 x 2 + 9 x + 20 = 4 x 2 + 3 x − 10 2 x 2 + 2 x − 8 − 1 x 2 + 9 x + 20 = 4 x 2 + 3 x − 10
5 x 2 + 4 x + 3 + 2 x 2 + x − 6 = 3 x 2 − x − 2 5 x 2 + 4 x + 3 + 2 x 2 + x − 6 = 3 x 2 − x − 2
3 x 2 − 5 x − 6 + 3 x 2 − 7 x + 6 = 6 x 2 − 1 3 x 2 − 5 x − 6 + 3 x 2 − 7 x + 6 = 6 x 2 − 1
2 x 2 + 2 x − 3 + 3 x 2 + 4 x + 3 = 6 x 2 − 1 2 x 2 + 2 x − 3 + 3 x 2 + 4 x + 3 = 6 x 2 − 1
Solve Rational Equations that Involve Functions
For rational function, f ( x ) = x − 2 x 2 + 6 x + 8 , f ( x ) = x − 2 x 2 + 6 x + 8 , ⓐ find the domain of the function ⓑ solve f ( x ) = 5 f ( x ) = 5 ⓒ find the points on the graph at this function value.
For rational function, f ( x ) = x + 1 x 2 − 2 x − 3 , f ( x ) = x + 1 x 2 − 2 x − 3 , ⓐ find the domain of the function ⓑ solve f ( x ) = 1 f ( x ) = 1 ⓒ find the points on the graph at this function value.
For rational function, f ( x ) = 2 − x x 2 − 7 x + 10 , f ( x ) = 2 − x x 2 − 7 x + 10 , ⓐ find the domain of the function ⓑ solve f ( x ) = 2 f ( x ) = 2 ⓒ find the points on the graph at this function value.
For rational function, f ( x ) = 5 − x x 2 + 5 x + 6 , f ( x ) = 5 − x x 2 + 5 x + 6 , ⓐ find the domain of the function ⓑ solve f ( x ) = 3 f ( x ) = 3 ⓒ the points on the graph at this function value.
In the following exercises, solve.
C r = 2 π C r = 2 π for r . r .
I r = P I r = P for r . r .
v + 3 w − 1 = 1 2 v + 3 w − 1 = 1 2 for w . w .
x + 5 2 − y = 4 3 x + 5 2 − y = 4 3 for y . y .
a = b + 3 c − 2 a = b + 3 c − 2 for c . c .
m = n 2 − n m = n 2 − n for n . n .
1 p + 2 q = 4 1 p + 2 q = 4 for p . p .
3 s + 1 t = 2 3 s + 1 t = 2 for s . s .
2 v + 1 5 = 3 w 2 v + 1 5 = 3 w for w . w .
6 x + 2 3 = 1 y 6 x + 2 3 = 1 y for y . y .
m + 3 n − 2 = 4 5 m + 3 n − 2 = 4 5 for n . n .
r = s 3 − t r = s 3 − t for t . t .
E c = m 2 E c = m 2 for c . c .
R T = W R T = W for T . T .
3 x − 5 y = 1 4 3 x − 5 y = 1 4 for y . y .
c = 2 a + b 5 c = 2 a + b 5 for a . a .
Writing Exercises
Your class mate is having trouble in this section. Write down the steps you would use to explain how to solve a rational equation.
Alek thinks the equation y y + 6 = 72 y 2 − 36 + 4 y y + 6 = 72 y 2 − 36 + 4 has two solutions, y = −6 y = −6 and y = 4 . y = 4 . Explain why Alek is wrong.
ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.
ⓑ On a scale of 1 − 10 , 1 − 10 , how would you rate your mastery of this section in light of your responses on the checklist? How can you improve this?
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Access for free at https://openstax.org/books/intermediate-algebra-2e/pages/1-introduction
- Authors: Lynn Marecek, Andrea Honeycutt Mathis
- Publisher/website: OpenStax
- Book title: Intermediate Algebra 2e
- Publication date: May 6, 2020
- Location: Houston, Texas
- Book URL: https://openstax.org/books/intermediate-algebra-2e/pages/1-introduction
- Section URL: https://openstax.org/books/intermediate-algebra-2e/pages/7-4-solve-rational-equations
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Critter Calculations
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Equation Grid
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Equation Decoder
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Algebra Attack
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Ocean Equations
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Equation Maze
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Shadow Solving
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Sweet Solutions
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Equation Explorers
About these 15 worksheets.
These worksheets help students practice and master the process of solving for variables in algebraic equations. These worksheets are designed with a variety of formats and problem types to engage students and reinforce their understanding of algebraic principles. The worksheets guide students through the fundamental skills needed to manipulate and simplify equations, ultimately solving for unknowns. They often incorporate creative elements, such as mazes, matching exercises, or themed designs, to keep the practice engaging and dynamic.
One common type of problem found on these worksheets is the straightforward linear equation. These problems focus on basic one-step or two-step equations, where students are required to isolate the variable by performing inverse operations. For example, students may need to add, subtract, multiply, or divide both sides of an equation to balance it and solve for the unknown. These types of problems are crucial because they form the foundation for more complex algebraic concepts. By repeatedly practicing these steps, students solidify their understanding of the equality principle and learn how to reverse operations to simplify equations.
In addition to traditional linear equations, many of these worksheets also feature problems that involve negative numbers, fractions, and decimals. This inclusion helps students develop fluency in handling a variety of number types, which is critical as they progress in their math studies. Solving equations that involve fractions requires students to multiply by the reciprocal or find common denominators, while working with decimals involves careful attention to place value. These challenges enhance a student’s precision and calculation skills, as even small errors in handling fractions or decimals can lead to incorrect solutions.
Another engaging format found on these worksheets is the maze or path-solving activity. In these exercises, students must follow a sequence of correct answers to navigate through a maze. Each correct solution leads them to the next step in the maze, while incorrect answers might lead them astray. This format adds a layer of fun to equation solving while reinforcing accuracy and critical thinking. Students are motivated to double-check their answers to ensure they’re on the right path, which encourages attention to detail. These mazes are an excellent way to build both speed and precision in solving equations.
Many worksheets also incorporate matching or fill-in-the-blank exercises where students solve an equation and then match their solution with a corresponding answer. This type of exercise can be particularly useful for self-checking, as it provides immediate feedback. If a student’s solution doesn’t match any of the available options, they know to revisit their work and look for errors. This format fosters independence and allows students to take ownership of their learning process, developing problem-solving skills that extend beyond the classroom.
These worksheets are designed with a theme or a creative twist, such as solving for variables in a secret message. In these problems, each correct answer corresponds to a letter, and when all the problems are solved, the letters spell out a hidden word or phrase. This type of worksheet adds an element of surprise and fun, keeping students engaged and motivated to complete the task. The process of solving each equation remains the same, but the added challenge of uncovering a hidden message makes the practice more enjoyable and rewarding.
Another important aspect of these worksheets is that they often include a mix of equation types to ensure students are exposed to a broad range of problem-solving scenarios. For instance, students might encounter equations with variables on both sides, which require them to combine like terms and move all variable terms to one side of the equation. These problems teach students how to simplify more complex algebraic expressions and develop a deeper understanding of how to manipulate equations to find solutions.
Equations involving parentheses and the distributive property are also commonly featured. These problems require students to expand expressions by applying the distributive property before solving the equation. This introduces another layer of complexity, as students must remember to correctly distribute terms across all elements within the parentheses. These exercises reinforce the importance of following the correct order of operations and provide practice in breaking down more complex expressions into manageable steps.
The skills taught through these worksheets are foundational for success in higher-level math. By practicing with these worksheets, students develop a strong understanding of how to manipulate algebraic expressions and solve for unknowns, a skill that is critical for more advanced math topics such as systems of equations, quadratic equations, and functions. The variety of problems and formats ensures that students are not only practicing routine calculations but also developing flexibility in their problem-solving approaches.
These worksheets promote a growth mindset by encouraging students to view mistakes as learning opportunities. With immediate feedback from matching or maze activities, students can identify errors and correct their understanding, reinforcing the idea that persistence and practice lead to improvement. This mindset is crucial for building confidence in math, as it teaches students that challenges are an integral part of the learning process.
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COMMENTS
3w − 1 = the length of the rectangle. Step 4: Translate into an equation. We know the area. Write the formula for the area of a rectangle. Step 5: Solve the equation. Substitute in the values. Distribute. This is a quadratic equation; rewrite it in standard form. Solve the equation using the Quadratic Formula.
Your mastery of the lessons is an important tool to solve many real-world problems related to quadratic equations. In this module, you will learn how to solve word problems that involves quadratic equations. LEARNING COMPETENCY. The learners will be able to: solve problems involving quadratic equations. M9AL-1e-1; Math9_Quarter1_Module9_FINAL-V3-1
This video is about solving word problems involving quadratic equations and rational algebraic equations.D&E's videos are intended to help people who want to...
INTO QUADRATIC EQUATIONS In module 8, you have learned solving rational algebraic expressions reducible into quadratic equations. Mastery to this topic is a valuable tool to answer many real-life problems which you will learn from this module. Find out how much you already know about solving word problems on rational algebraic expression ...
Solving problems involving quadratic equations and rational algebraic equations (Part 2) | Teacherrie TVIn this video, we will solve problem involving quadra...
‼️FIRST QUARTER‼️🔴 GRADE 9: SOLVING PROBLEMS INVOLVING QUADRATIC EQUATIONS🔴 GRADE 9First Quarter: https://tinyurl.com/y5wjf97p Second Quarter: https ...
Math9_Q2-Mod3 - Free download as PDF File (.pdf), Text File (.txt) or read online for free. This document is a mathematics module that covers solving problems involving quadratic equations and rational algebraic equations. It includes two lessons that teach students how to solve word problems involving these types of equations. The module provides learning objectives, icons to guide students ...
In module 8, you have learned solving rational algebraic expressions reducible into quadratic equations. Mastery to this topic is a valuable tool to answer many real life problems which you will learn from this module. LEARNING COMPETENCY. The learners will be able to: solve problems involving quadratic equations and rational algebraic equations.
Solving Rational Equations. A rational equation is an equation containing at least one rational expression. Rational expressions typically contain a variable in the denominator. For this reason, we will take care to ensure that the denominator is not 0 by making note of restrictions and checking our solutions.
Number Problems. The algebraic setups of the word problems that we have previously encountered led to linear equations. When we translate the applications to algebraic setups in this section, the setups lead to quadratic equations. Just as before, we want to avoid relying on the "guess and check" method for solving applications.
Mathematical Thinking: Mathematically proficient students can apply the mathematics they know to solve problems arising in everyday life, society, and the workplace. 9.1 Properties of Radicals 9.2 Solving Quadratic Equations by Graphing 9.3 Solving Quadratic Equations Using Square Roots 9.4 Solving Quadratic Equations by Completing the Square 9.5 Solving Quadratic Equations Using the Quadratic ...
Solve Rational Equations. We have already solved linear equations that contained fractions. We found the LCD of all the fractions in the equation and then multiplied both sides of the equation by the LCD to "clear" the fractions. We will use the same strategy to solve rational equations. We will multiply both sides of the equation by the LCD.
High School Math Solutions - Quadratic Equations Calculator, Part 1 A quadratic equation is a second degree polynomial having the general form ax^2 + bx + c = 0, where a, b, and c... Enter a problem
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This document outlines a lesson plan on solving quadratic equations. The lesson begins with motivating students on the importance of solving quadratic equations to model real-world problems. Examples are then presented to illustrate how to translate word problems into quadratic equations and solve for unknown variables. Students practice working in groups to solve sample problems. Formative ...
You have already mastered how to solve quadratic equations in standard form, find the roots of an equation as well as the sum and product of the roots. But w...
In mathematics, a quadratic equation (from Latin quadratus 'square') is an equation that can be rearranged in standard form as [1] + + =, where x represents an unknown value, and a, b, and c represent known numbers, where a ≠ 0. (If a = 0 and b ≠ 0 then the equation is linear, not quadratic.)The numbers a, b, and c are the coefficients of the equation and may be distinguished by ...
Problem Solving Involving Quadratic Equations and Rational Algebraic Equations - Free download as Powerpoint Presentation (.ppt / .pptx), PDF File (.pdf), Text File (.txt) or view presentation slides online. Here are the step-by-step solutions to the problems: 1. The difference of the squares of two consecutive even integers is 68. A) Let x = the first even integer B) x + 2 = the second even ...
8. Mathematics. Quarter 1 - Module 5C: "Solving Problems Involving Rational Algebraic Expressions". ntroductory MessageThis Self-Learning Module (SLM) is prepared so that you, our dear learners, can continue your studies and. arn while at home. Activities, questions, directions, exercises, and discussions are carefully stated for you to und.
The process of solving each equation remains the same, but the added challenge of uncovering a hidden message makes the practice more enjoyable and rewarding. Another important aspect of these worksheets is that they often include a mix of equation types to ensure students are exposed to a broad range of problem-solving scenarios.
1. The document provides a lesson plan on solving quadratic equations that are not in standard form. It includes objectives, content, learning resources, procedures for instruction and practice activities, and a reflection on student learning. 2. Key procedures covered are transforming equations to standard form, using factoring, extracting square roots, completing the square, and the ...
Solve rational equations with real solutions containing factorable algebraic expressions algebraically and graphically. Algebraic expressions should be limited to linear and quadratic expressions. Verify possible solution(s) to rational equations algebraically, graphically, and with technology to justify the reasonableness of answer(s).