10-7 Solving Rational Equations Warm Up Warm Up Lesson Presentation Lesson Presentation California Standards California StandardsPreview.

10-7 Solving Rational Equations

Warm UpWarm Up

Lesson Presentation

California Standards

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Warm Up

1. Find the LCM of x, 2x2, and 6.

2. Find the LCM of p2 – 4p and p2 – 16.

Multiply. Simplify your answer.

3. 4.

5.


Preparation for 15.0 Students apply algebraic techniques to solve rate problems, work problems, and percent mixture problems.

California Standards


rational equationextraneous solutions

Vocabulary


A rational equation is an equation that contains one or more rational expressions. If a rational equation is a proportion, it can be solved using the Cross Product Property.


Additional Example 1: Solving Rational Equations by Using Cross Products

Use cross products.5x = (x – 2)(3)

5x = 3x – 6

2x = –6

Solve . Check your answer.

x = –3

Distribute 3 on the right side.

Subtract 3x from both sides.

Divide both sides by 2.


Additional Example 1 Continued

Check

–1 –1

Substitute –3 for x in the original equation.


Check It Out! Example 1a


Use cross products.


Subtract n from both sides.

3n = (n + 4)(1)

3n = n + 4

2n = 4

n = 2 Divide both sides by 2.


Check It Out! Example 1a Continued

Check

Substitute 2 for n in the original equation.


Check It Out! Example 1b


4h = (h + 1)(2)

4h = 2h + 2

2h = 2

h = 1

Use cross products.


Subtract 2h from both sides.



Check It Out! Example 1b Continued

Check

Substitute 1 for h in the original equation.

10-7 Solving Rational EquationsCheck It Out! Example 1c


21x = (x – 7)(3)

21x = 3x – 21

18x = –21

x =

Use cross products.


Subtract 3x from both sides.



Check

Check It Out! Example 1c Continued

Substitute for x in the

original equation.


Some rational equations contain sums or differences of rational expressions. To solve these, you must find a common denominator for all the rational expressions in the equation.


Additional Example 2A: Solving Rational Equations by Using the LCD

Solve each equation. Check your answer.

Step 1 Find the LCD.

2x(x + 1) Include every factor of the denominator.

Step 2 Multiply both sides by the LCD.

Distribute on the left

side.


Additional Example 2A Continued

Step 3 Simplify and solve.

(2x)(2) + 6(x + 1) = 5(x + 1)

4x + 6x + 6 = 5x + 5

10x + 6 = 5x + 5

5x = –1

Divide out common factors.

Simplify.

Distribute and multiply.

Combine like terms.

Subtract 5x and 6 from both sides.



Additional Example 2A Continued

Check


Additional Example 2B: Solving Rational Equations by Using the LCD



(x2) Include every factor of the denominator.


Distribute on the left side.


Additional Example 2B Continued

Step 3 Simplify and solve.


4x – 3 = x2

–x2 + 4x – 3 = 0

x2 – 4x + 3 = 0

(x – 3)(x – 1) = 0

x = 3 or 1

Simplify.Subtract x2 from both

sides.

Factor.

Solve.

Multiply by – 1.


Additional Example 2B Continued

Check

10-7 Solving Rational EquationsCheck It Out! Example 2a



a(a +1) Include every factor of the denominator.





Step 3 Simplify and solve.Divide out

common factors.

3a = 4(a + 1)

3a = 4a + 4

–4 = a

Simplify.

Distribute the 4.

Subtract the 4 and 3a from both sides.



Check





2j(j +2) Include every factor of the denominator.






12j – 10(2j + 4) = 4j + 8

12j – 20j – 40 = 4j + 8

–12j = 48

j = –4

Simplify.

Distribute 10.

Combine like terms.


Divide both sides by –12.



Check


When you multiply each side of an equation by the LCD, you may get an extraneous solution. An extraneous solution is a solution to a resulting equation that is not a solution to the original equation. Because of extraneous solutions, it is especially important to check your answers.


Additional Example 3: Extraneous Solutions


2(x2 – 1) = (x + 1)(x – 6)

2x2 – 2 = x2 – 5x – 6

x2 + 5x + 4 = 0

(x + 1)(x + 4) = 0

x = –1 or x = –4

Use cross products.

Distribute 2 on the left side. Multiply the right side.

Subtract x2 from both sides. Add 5x and 6 to both sides.

Factor the quadratic expression.

Use the Zero Product Property.Solve.

10-7 Solving Rational EquationsAdditional Example 3 Continued


Because and

are undefined, –1 is

not a solution.

–1 is an extraneous solution. The only solution is –4.

Check


Check It Out! Example 3a

Solve. Check your answer.

(x – 2)(x – 7) = 3(x – 7) Use cross products.

Distribute 3 on the right side. Multiply the left side.

x2 – 9x + 14 = 3x – 21

x2 – 12x + 35 = 0 Subtract 3x from both sides. Add 21 to both sides.

(x – 7)(x – 5) = 0

x = 7 or x = 5

Factor the quadratic expression.Use the Zero Product Property.Solve.



7 is an extraneous solution. The only solution is 5.

Because and

are undefined, 7 is

not a solution.

Check



Solve. Check your answer.

(x + 1)(x – 3) = 4(x – 2) Use cross products.

Distribute 4 on the right side. Multiply the left side.

x2 – 2x – 3 = 4x – 8

x2 – 6x + 5 = 0 Subtract 4x from both sides. Add 8 to both sides.

(x – 1)(x – 5) = 0

x = 1 or x = 5

Factor the quadratic expression.Use the Zero Product Property.Solve.

10-7 Solving Rational EquationsCheck It Out! Example 3b Continued

There are no extraneous solutions. The solutions are 1 and 5.

1 and 5 are

solutions.

Check

10-7 Solving Rational EquationsCheck It Out! Example 3c

Solve. Identify any extraneous solutions.

6(x2 + 2x) = 9(x2) Use cross products.

Distribute 6 on the left side. Multiply the right side.

3x2 – 12x = 0

3x(x – 4) = 0

3x = 0, or x – 4 = 0

x = 0 or x = 4

Factor the quadratic expression.

Use the Zero Product Property.Solve.

Subtract 9x2 from both sides.Multiply through with – 1.

6x2 + 12x) = 9x2


Check It Out! Example 3c Continued

Check

0 is an extraneous solution. The only solution is 4.

Because and

are undefined, 0 is

not a solution.


Lesson Quiz


3.

2.1. 24 –4, 3

4. Solve .

Check your answer.

–5

10-7 Solving Rational Equations Warm Up Warm Up Lesson Presentation Lesson Presentation California Standards California StandardsPreview.

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