§ 6.6 Rational Equations. Blitzer, Intermediate Algebra, 4e – Slide #87 Solving a Rational EquationEXAMPLE Solve: SOLUTION Notice that the variable x

§ 6.6

Rational Equations

Blitzer, Intermediate Algebra, 4e – Slide #87

Solving a Rational Equation

EXAMPLEEXAMPLE

Solve:

SOLUTIONSOLUTION

Notice that the variable x appears in two of the denominators. We must avoid any values of the variable that make a denominator zero.

.4

5

9

5

47

xx

x

This denominator would equal zero if x = 0.

xx

x 4

5

9

5

47


Therefore, we see that x cannot equal zero.



The denominators are 5x, 5, and x. The least common denominator is 5x. We begin by multiplying both sides of the equation by 5x. We will also write the restriction that x cannot equal zero to the right of the equation.

0,4

5

9

5

47

x

xx

x

CONTINUECONTINUEDD

This is the given equation.

xx

x

xx

4

5

95

5

475 Multiply both sides by 5x,

the LCD.

x

xx

x

xx 4

1

5

5

9

1

5

5

47

1

5

Use the distributive property.



CONTINUECONTINUEDD

45947 xx Divide out common factors in the multiplications.

20947 xx Multiply.

2024 x Subtract.

x216 Add.

x8 Divide.

The proposed solution, 8, is not part of the restriction . It should check in the original equation.

0x



CONTINUECONTINUEDD Check 8:

xx

x 4

5

9

5

47

8

4

5

9

85

487

8

4

5

9

40

456

8

4

5

9

40

52

8

4

5

940

40

5240

8

440

5

94052

459852

207252

5252

This true statement verifies that the solution is 8 and the solution set is {8}.

?

?

?

?

?

?

?

true



Solving Rational Equations1) List restrictions on the variable. Avoid any values of the variable that make a denominator zero.

2) Clear the equation of fractions by multiplying both sides by the LCD of all rational expressions in the equation.

3) Solve the resulting equation.

4) Reject any proposed solution that is in the list of restrictions on the variable. Check other proposed solutions in the original equation.



EXAMPLEEXAMPLE

Solve:

SOLUTIONSOLUTION

1) List restrictions on the variable.

.72

56

4

13

x

x

x

x


This denominator would equal zero if x = 3.5.

The restrictions are

72

56

4

13

x

x

x

x

.5.3 and 4 xx



2) Multiply both sides by the LCD. The denominators are x – 4 and 2x – 7. Thus, the LCD is (x – 4)(2x + 7).

5.3 ,4 ,72

56

4

13

xxx

x

x

x

CONTINUECONTINUEDD


72

56724

4

13724

x

xxx

x

xxx Multiply both sides by

the LCD.

Simplify. 5641372 xxxx



4) Check the proposed solution in the original equation. Notice, there is no proposed solution. And of course, -7 = -20 is not a true statement. Therefore, there is no solution to the original rational equation. We say the solution set is , the empty set.

CONTINUECONTINUEDD

This is the equation cleared of fractions.

5641372 xxxx

Use FOIL on each side.201967196 22 xxxx

Subtract from both sides.2019719 xx

207

26x

Subtract 19x from both sides.




EXAMPLEEXAMPLE

Solve:

SOLUTIONSOLUTION

1) List restrictions on the variable. By factoring denominators, it makes it easier to see values that make the denominators zero.

.2

1

4

2

82

122

xxxx

x

This denominator is zero if x = -4 or x = 2.

This denominator would equal zero if x = -4.

The restrictions are .2 and 4 xx

2

1

4

2

24

12

xxxx

x




2) Multiply both sides by the LCD. The factors of the LCD are x + 4 and x – 2. Thus, the LCD is (x + 4)(x - 2).

CONTINUECONTINUEDD


Multiply both sides by the LCD.

Use the distributive property.

2 ,4 ,2

1

4

2

24

12

xxxxxx

x

2

124

4

2

24

1224

xxx

xxx

xxx

2

124

4

224

24

1224

xxx

xxx

xx

xxx




CONTINUECONTINUEDD

Simplify. 42212 xxx

42212 xxx This is the equation with cleared fractions.

44212 xxx Use the distributive property.

454 xx Combine like terms.

453 x Subtract x from both sides.

93 x Add 5 to both sides.

3x Divide both sides by 3.



4) Check the proposed solutions in the original equation. The proposed solution, 3, is not part of the restriction that Substitute 3 for x, in the given (original) equation. The resulting true statement verifies that 3 is a solution and that {3} is the solution set.

CONTINUECONTINUEDD

and 4x.2x


EXAMPLEEXAMPLE

Rational functions can be used to model learning. Many of these functions model the proportion of correct responses as a function of the number of trials of a particular task. One such model, called a learning curve, is

where f (x) is the proportion of correct responses after x trials. If f (x) = 0, there are no correct responses. If f (x) = 1, all responses are correct. The graph of the rational function is shown on the next page. Use the function to solve the following problem.

1.09.0

4.09.0

x

xxf




CONTINUECONTINUEDD

A Learning Curve

00.10.20.30.40.50.60.70.80.9

11.11.2

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Number of Learning Trials

Pro

po

rtio

n o

f C

orr

ec

t R

es

po

ns

es

1.09.0

4.09.0

x

xxf



CONTINUECONTINUEDD

SOLUTIONSOLUTION

How many learning trials are necessary for 0.5 of the responses to be correct? Identify your solution as a point on the graph.

Substitute 0.5, the proportion of correct responses, for f (x) and solve the resulting rational equation for x.

1.09.0

4.09.05.0

x

xThe LCD is 0.9x + 0.1.

1.09.0

4.09.01.09.05.01.09.0

x

xxx Multiply both sides by the

LCD.

4.09.01.09.05.0 xx Simplify.



CONTINUECONTINUEDD

4.09.005.045.0 xx Use the distributive property on the left side.

4.005.045.0 x Subtract 0.9x from both sides.

45.045.0 x Subtract 0.05 from both sides.

1x Divide both sides by -0.45.

The number of learning trials necessary for 0.5 of the responses to be correct is 1.

The solution is identified as a point on the graph at the beginning of the problem.

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§ 6.6 Rational Equations. Blitzer, Intermediate Algebra, 4e – Slide #87 Solving a Rational EquationEXAMPLE Solve: SOLUTION Notice that the variable x