1. Lesson 11-5 Warm-Up

2. “Solving Rational Equations” (11-5)
How do you solve rational equation containing one or more rational expressions?
To solve a rational equation with one or more rational expressions (i.e. unlike denominators containing variables), multiply both sides of the equation by the LCD to eliminate the rational expressions. Example:

3. Solve + = . Check the solution.
Solving Rational Equations
LESSON 11-5
Additional Examples
Solve + = . Check the solution. 3 8 4 x 14 2x 3 8 4 x 14 2x
+ =
The denominators are 8, x, and 2x. The LCD is 8x.
8x = 8x 3 8 4 x + 14 2x
Multiply each side by 8x. This means
you multiply every term in the equation by 8x. 1 1 4
8x = 8x 3 8 4 x + 14 2x 8x
Use the Distributive Property. 1 1 1
No rational expressions (fractions), so now you can solve.
3x + 32 = 56
Subtract 32 from each side.
3x = 24
Simplify.
+ 0
Divide each side by 3.
x = 8
Simplify

4. Check: + Substitute 8 for x. = (continued) 3 8 4 (8) 14 2(8) 7 8 14 16
Solving Rational Equations
LESSON 11-5
Additional Examples
(continued)
Check: 3 8 4
(8) 14
2(8) +
Substitute 8 for x. 7 8 14 16 7 8 =

5. Solve = – 1. Check the solution.
Solving Rational Equations
LESSON 11-5
Additional Examples 6 x2 5 x
Solve = – 1. Check the solution. 5 x
– 1 6 x2 =
Multiply each side by the LCD, x2. This
Means multiply every term in the equation
by x2. 5 x
– x2 (1) 6 x2 =
Use the Distributive Property. 1 x 1 1
6 = 5x – x2
Simplify.
x2 – 5x + 6 = 0
Collect like terms on one side.
(x – 3)(x – 2) = 0 Factor the quadratic expression.
x – 3 = 0 or x – 2 = 0 Use the Zero-Product Property.
Isolate the x’s.
x = 3 or x = 2 Solve.

6. – 1 – 1 Check: = = (continued) 6 (3)2 5 (3) 6 (2)2 5 (2) 2 3 3 2
Solving Rational Equations
LESSON 11-5
Additional Examples
(continued) 6
(3)2 5
(3)
– 1 6
(2)2 5
(2)
– 1
Check: 2 3 = 3 2 =

7. Person Work Rate Time Worked Part of (part of job/min.) (min) Job Done
Solving Rational Equations
LESSON 11-5
Additional Examples
Renee can mow the lawn in 20 minutes. Joanne can do the same job in 30 minutes. How long will it take them if they work together?
Define: Let n = the time to complete the job if they work together (in minutes).
Person Work Rate Time Worked Part of
(part of job/min.) (min) Job Done
Renee n
Joanne n 1 20 30 n
Words: Renee’s part done + Joanne’s part done = complete job.

8. Multiply each side by the LCD, 60.
Solving Rational Equations
LESSON 11-5
Additional Examples
(continued) n 20 30 +
Translate: = 1 60
= 60(1)
Multiply each side by the LCD, 60.
3n + 2n = 60 Use the Distributive Property.
5n = 60 Combine like terms.
n = 12 Divide both sides by 5.
It will take two of them 12 minutes to mow the lawn working together. 1 20
Check: Renee will do • = of the job, and Joanne will do of the job. Together, they will do = 1, or the whole job. 12 3 5 2 30 • = +

9. “Solving Rational Equations” (11-5)
How can you solve a rational equation in the form of a proportion (equal fractions)?
To solve a rational equation in the form of a proportion, or equal fractions, simply cross multiply as you would a proportion with rational numbers. Example: Given Note: The process of cross multiplying may give an extraneous solution, or a solution that solves the new equation, but not the original one. Therefore, always check your solutions, because they may not make not work in the original equation.

10. Solve = . Check the solution.
Solving Rational Equations
LESSON 11-5
Additional Examples 4 x2 1
x + 8
Solve = Check the solution. 4 x2 1
x + 8 =
4(x + 8) = x2(1) Write cross products.
4x + 32 = x2 Use the Distributive Property.
x2 – 4x – 32 = 0 Collect all of the terms on one side of the equal sign.
(x – 8)(x + 4) = 0 Factor the quadratic expression.
x – 8 = 0  or  x + 4 = Use the Zero-Product Property.
x = 8  or x = –4 Solve.

11. Check: = Substitute 8 and -4 for x. = = (continued) 4 x2 1 x + 8 4
Solving Rational Equations
LESSON 11-5
Additional Examples
(continued)
Check: 4 x2 1
x + 8 = 4
(8)2 1
(8) + 8 4
(–4)2 1
(–4) + 8
Substitute 8 and -4 for x. = 4 64 1 16 4 16 1 =

12. (x + 3) (x – 1) = 4(x – 1) Write the cross products.
Solving Rational Equations
LESSON 11-5
Additional Examples
x + 3
x – 1 4
x – 1
Solve = .
x + 3
x – 1 4 =
(x + 3) (x – 1) = 4(x – 1) Write the cross products.
x2 – x + 3x – 3 = 4x – 4 Use the Distributive Property to multiply
x2 + 2x – 3 = 4x – 4 Combine like terms.
x2 – 2x + 1 = Collect all of the terms on one side of the equal sign.
(x – 1) (x – 1) = 0 Factor.
x – 1 = 0 Use the Zero-Product Property.
x = 1 Simplify.

13. Undefined! There is no division by 0.
Solving Rational Equations
LESSON 11-5
Additional Examples
(continued)
Check:
1 + 3
(1) – 1 4
Substitute 1 for x. 4 =
Undefined! There is no division by 0.
The equation has no solution because 1 makes a denominator equal 0.