Please complete the Prerequisite Skills on Page 548 #4-12

Bell Ringer

Please complete the Prerequisite Skills on Page 548 #4-12

Chapter 8:Rational Functions

Big ideas: Performing operations with rational expressions Solving rational equations

Lesson 1: Model Inverse & Joint Variation

Essential question

What are the differences between direct, inverse and

joint variation?

VOCABULARY

Inverse variation ~ The relationship of two variables x and y if there is a nonzero number a such that y =

Constant of variation ~ The nonzero constant a in a direct variation equation y=ax, an inverse variation equation y=, or a joint variation equation z=axy

Joint variation ~ A relationship that occurs when a quantity varies directly with the product of two or more other quantities.

EXAMPLE 1 Classify direct and inverse variation

Tell whether x and y show direct variation, inverse variation, or neither.

Given Equation Rewritten Equation Type of Variation

a. xy = 7 y =7

xInverse

b. y = x + 3 Neither

y

4c. = x y = 4x Direct

EXAMPLE 2 Write an inverse variation equation

The variables x and y vary inversely, and y = 7 when x = 4. Write an equation that relates x and y. Then find y when x = –2 .

y =a

xWrite general equation for inverse variation.

Substitute 7 for y and 4 for x. 7 =a

4

28 = a Solve for a.

The inverse variation equation is y =28x

When x = –2, y = 28–2

= –14.

ANSWER

Essential question

What are the differences between direct, inverse and joint variation?

~ y varies directly with x if y=ax for nonzero constant a

~ y varies inversely with x if xy = a for a nonzero constant a

~ z varies jointly with x and y if z = axy for nonzero constant a

Bell Ringer

Y varied directly with x. If y = 36 when x = 8, find y when x = 5

Lesson 4: Multiply & Divide Rational expressions

Essential question

What are the steps for multiplying & dividing rational expressions?

VOCABULARY

Simplified form of a rational expression ~ A rational expression in which the numerator and denominator have no common factors other than 1 and -1

Reciprocal ~ The multiplicative inverse of any nonzero number

EXAMPLE 1 Simplify a rational expression

x2 – 2x – 15x2 – 9

Simplify :

x2 – 2x – 15x2 – 9

(x +3)(x –5)(x +3)(x –3)= Factor numerator and denominator.

(x +3)(x –5)(x +3)(x –3)

= Divide out common factor.

Simplified form

SOLUTION

x – 5x – 3=

ANSWERx – 5x – 3

EXAMPLE 3 Standardized Test Practice

SOLUTION

8x 3y2x y2

7x4y3

4y56x7y4

8xy3= Multiply numerators

and denominators.

8 7 x x6 y3 y

8 x y3 = Factor and divide out

common factors.

7x6y= Simplified form

The correct answer is B.ANSWER

EXAMPLE 4 Multiply rational expressions

Multiply: x2 + x – 203x

3x –3x2

x2 + 4x – 5

x2 + x – 203x

3x –3x2

x2 + 4x – 5

3x(1– x)(x –1)(x +5)

=(x + 5)(x – 4)

3xFactor numerators and denominators.

3x(1– x)(x + 5)(x – 4)=

(x –1)(x + 5)(3x)Multiply numerators and denominators.

3x(–1)(x – 1)(x + 5)(x – 4)=

(x – 1)(x + 5)(3x)Rewrite 1– x as (– 1)(x – 1).

3x(–1)(x – 1)(x + 5)(x – 4)=

(x – 1)(x + 5)(3x)Divide out common factors.

SOLUTION

EXAMPLE 4 Multiply rational expressions

= (–1)(x – 4) Simplify.

= –x + 4 Multiply.

ANSWER –x + 4

EXAMPLE 5 Multiply a rational expression by a polynomial

Multiply: x + 2x3 – 27

(x2 + 3x + 9)

x + 2x3 – 27

(x2 + 3x + 9)

Write polynomial as a rational expression.

=x + 2x3 – 27

x2 + 3x + 91

(x + 2)(x2 + 3x + 9)

(x – 3)(x2 + 3x + 9)= Factor denominator.

(x + 2)(x2 + 3x + 9)

(x – 3)(x2 + 3x + 9)= Divide out common factors.

=x + 2x – 3

Simplified form

SOLUTION

ANSWERx + 2x – 3

GUIDED PRACTICE for Examples 3, 4 and 5

Multiply the expressions. Simplify the result.

3x5 y2

8xy

6xy2

9x3y8.

3x5 y2

2xy

6xy2

9x3y

18x6y4

72x4y2= Multiply numerators

and denominators.

Factor and divide out common factors.

Simplified form

18 x4 y2 x2 y2

18 4 x4 y2 =

=x2y2

4

SOLUTION


9.

x + 3

2x2

2x2 – 10xx2– 25

Factor numerators and denominators.

2x(x –5) (x + 3)=

(x –5)(x + 5)2x (x)Multiply numerators and denominators.

x + 3=

x(x + 5)

Divide out common factors.

x + 3

2x2

2x2 – 10xx2– 25

2x(x –5)(x –5)(x +5)

=x + 3

2x (x)

= 2x(x –5) (x + 3)(x –5)(x + 5)2x (x)

Simplified form

SOLUTION


10.

Factor denominators.

Multiply numerators and denominators.

x + 5=x – 1


x2 +x + 1x + 5x3– 1

Simplified form

x2 +x + 1x + 5x3– 1

=x2 +x + 1x + 5

(x – 1) (x2 +x + 1) 1

=(x + 5) (x2 +x + 1)

(x – 1) (x2 +x + 1)

(x + 5) (x2 +x + 1)

(x – 1) (x2 +x + 1)=

SOLUTION

EXAMPLE 6 Divide rational expressions

Divide : 7x2x – 10

x2 – 6xx2 – 11x + 30

7x2x – 10

x2 – 6xx2 – 11x + 30

7x2x – 10 x2 – 6x

x2 – 11x + 30= Multiply by reciprocal.

7x2(x – 5)

=(x – 5)(x – 6)

x(x – 6)

=7x(x – 5)(x – 6)

2(x – 5)(x)(x – 6) Divide out common factors.

Factor.

7

2= Simplified form

SOLUTION

ANSWER 7

2

EXAMPLE 7 Divide a rational expression by a polynomial

Divide : 6x2 + x – 154x2

(3x2 + 5x)

6x2 + x – 154x2

(3x2 + 5x)

6x2 + x – 154x2 3x2 + 5x

= 1 Multiply by reciprocal.

(3x + 5)(2x – 3)

4x2=

x(3x + 5)1 Factor.


Simplified form2x – 34x3

=

(3x + 5)(2x – 3)=

4x2(x)(3x + 5)

SOLUTION

ANSWER2x – 34x3

GUIDED PRACTICE for Examples 6 and 7

Divide the expressions. Simplify the result.

4x5x – 20

x2 – 2xx2 – 6x + 8

11.

4x5x – 20

x2 – 2xx2 – 6x + 8

Multiply by reciprocal.


Factor.

Simplified form

4x5x – 20 x2 – 2x

x2 – 6x + 8=

4(x)(x – 4)(x – 2)

5(x – 4)(x)(x – 2)=

4(x)(x – 4)(x – 2)

5(x – 4)(x)(x – 2)=

45

=

SOLUTION


2x2 + 3x – 5

6x(2x2 + 5x)12.

2x2 + 3x – 5

6x(2x2 + 5x)

Multiply by reciprocal.


Factor.

Simplified form

=2x2 + 3x – 5

6x (2x2 + 5x)1

(2x + 5)(x – 1)

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

(2x + 5)(x – 1)

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

x – 1

6x2=

SOLUTION

Essential question

What are the steps for multiplying & dividing rational expressions?

Multiply: multiply the numerators / multiply the denominators then

simplifyDivide: multiply the first

expression by the reciprocal of the second expression, then follow the

rules for multiplication

Bell Ringer

Find the least common multiple of 20 and 45.

Lesson 5: Add & Subtract Rational Expressions

Essential question

What are the steps for adding or subtracting

rational expressions with different denominators?

VOCABULARY

Complex fraction ~ A fraction that contains a fraction in its numerator or denominator.

EXAMPLE 1 Add or subtract with like denominators

Perform the indicated operation.

74x

+3

4xa. 2x

x + 6– 5

x + 6b.

SOLUTION

74x

+3

4xa. =

7 + 34x

104x

=52x= Add numerators and

simplify result.

x + 6 2x – 5=2x

x + 65

x + 6–b. Subtract numerators.

GUIDED PRACTICE for Example 1

Perform the indicated operation and simplify.

16x

a. 712x

+5

12x=

7 – 512x

= 212x

= Subtract numerators and simplify results .

1 x2

b. 2 3x2

+1

3x2=

2 + 13x2 =

33x2

= Add numerators and simplify results.

c. 4x x–2

–x

x–2=

4x–xx–2

= 3xx–2

= Subtract numerators.

d. 4xx2+1

+ 2

x2+12x2+2x2+1

=2(x2+1)

x2+1= = 2

Factor numerators and simplify results .

3x3x – 2

EXAMPLE 2 Find a least common multiple (LCM)

Find the least common multiple of 4x2 –16 and 6x2 –24x + 24.

SOLUTION

STEP 1

Factor each polynomial. Write numerical factors asproducts of primes.

4x2 – 16 = 4(x2 – 4) = (22)(x + 2)(x – 2)

6x2 – 24x + 24 = 6(x2 – 4x + 4) = (2)(3)(x – 2)2

EXAMPLE 2 Find a least common multiple (LCM)

STEP 2

Form the LCM by writing each factor to the highest power it occurs in either polynomial.

LCM = (22)(3)(x + 2)(x – 2)2 = 12(x + 2)(x – 2)2

EXAMPLE 3 Add with unlike denominators

Add: 9x2

7+

x3x2 + 3x

SOLUTION

To find the LCD, factor each denominator and write each factor to the highest power it occurs. Note that 9x2 = 32x2 and 3x2 + 3x = 3x(x + 1), so the LCD is 32x2 (x + 1) = 9x2(x 1 1).

Factor second denominator.7

9x2

x3x2 + 3x

=7

9x2 +3x(x + 1)

x+

79x2

x + 1x + 1

+3x(x + 1)

x3x3x

LCD is 9x2(x + 1).

EXAMPLE 3 Add with unlike denominators

Multiply.

Add numerators.3x2 + 7x + 79x2(x + 1)

=

7x + 79x2(x + 1)

3x29x2(x + 1)

+=

EXAMPLE 4 Subtract with unlike denominators

Subtract: x + 22x – 2

–2x –1x2 – 4x + 3

–

SOLUTION

x + 22x – 2

–2x –1x2 – 4x + 3

–

x + 22(x – 1)

– 2x – 1(x – 1)(x – 3)

–= Factor denominators.

x + 22(x – 1)

=x – 3x – 3

– – 2x – 1(x – 1)(x – 3) 2

2 LCD is 2(x 1)(x 3).

x2 – x – 62(x – 1)(x – 3)

– 4x – 22(x – 1)(x – 3)

–= Multiply.

EXAMPLE 4 Subtract with unlike denominators

x2 – x – 6 – (– 4x – 2)2(x – 1)(x – 3)

= Subtract numerators.

x2 + 3x – 42(x – 1)(x – 3)

= Simplify numerator.

Factor numerator. Divide out common factor.

Simplify.

=(x –1)(x + 4)

2(x – 1)(x – 3)

x + 42(x –3)

=


Find the least common multiple of the polynomials.

5. 5x3 and 10x2–15x

STEP 1


5x3 = 5(x) (x2)

10x2 – 15x = 5(x) (2x – 3)

STEP 2


LCM = 5x3 (2x – 3)


Find the least common multiple of the polynomials.

6. 8x – 16 and 12x2 + 12x – 72

STEP 1


8x – 16 = 8(x – 2) = 23(x – 2)

12x2 + 12x – 72 = 12(x2 + x – 6) = 4 3(x – 2 )(x + 3)

STEP 2


= 24(x – 2)(x + 3)

LCM = 8 3(x – 2)(x + 3)


4x3

–717.

SOLUTION

4x3

–71

4x3

77

71–

4x4x

LCD is 28x

4x7(4x)

214x(7)

–Multiply

Simplify21 – 4x

28x=


13x2

+x

9x2 – 12x8.

SOLUTION

13x2

+x

9x2 – 12x

=1

3x2+

x3x(3x – 4)

=1

3x2

3x – 4 3x – 4

+ x3x(3x – 4)

x x

3x – 4 3x2 (3x – 4)

+x2

3x2 (3x – 4 )=

Factor denominators

LCD is 3x2 (3x – 4)

Multiply


3x – 4 + x2 3x2 (3x – 4)

x2 + 3x – 4 3x2 (3x – 4)

Add numerators

Simplify


xx2 – x – 12

+ 512x – 48

9.

SOLUTION

xx2 – x – 12

+ 512x – 48

=x

(x+3)(x – 4)+

5

12 (x – 4)

x(x + 3)(x – 4)

=1212

+5

12(x – 4)

x + 3

x + 3

5(x + 3)

12(x + 3)(x – 4)12x

12(x + 3)(x – 4)= +

Factor denominators

LCD is 12(x – 4) (x+3)

Multiply


12x + 5x + 1512(x + 3)(x – 4)

=

17x + 1512(x +3)(x + 4)

=

Add numerators

Simplify


x + 1x2 + 4x + 4

– 6x2 – 4

10.

SOLUTION

x + 1x2 + 4x + 4

– 6x2 – 4

=x + 1

(x + 2)(x + 2)

6

(x – 2)(x + 2) –

x + 1

(x + 2)(x + 2)=

x – 2x – 2

–6

(x – 2)(x + 2)x + 2x + 2

=x2 – 2x + x – 2

(x + 2)(x + 2)(x – 2) –

6x + 12

(x – 2)(x + 2)(x + 2)

Factor denominators

LCD is (x – 2) (x+2)2

Multiply

GUIDED PRACTICE for Example 2, 3 and 4

=x2 – 2x + x – 2 – (6x + 12)

(x + 2)2(x – 4)

=x2 – 7x – 14

(x + 2)2 (x – 2)

Subtract numerators

Simplify

EXAMPLE 6 Simplify a complex fraction (Method 2)

Simplify:

5x + 4

1x + 4

+ 2x

SOLUTION

The LCD of all the fractions in the numerator and denominator is x(x + 4).

5x + 4

1x + 4

+ 2x

5x + 4

1x + 4

+ 2x

=x(x+4)

x(x+4)Multiply numerator and denominator by the LCD.

x + 2(x + 4)5x

= Simplify.

Simplify.5x

3x + 8=


x

6

x

3–

x

57

10–

– 5x 3 (2x – 7)

=

Multiply numberator and denominator by the LCD

Simplify

x

6

x

3–

x

57

10–

11.

x

6

x

3–

x

57

10–

3030

=


2

x–

2

x+

4

3

12.

2 – 4x2 + 3x

=

2 (1 – 2x )

2 + 3x= Simplify

Simplify

2

x–

2

x+

4

3=

2

x–

2

x+

4

3

xx Multiply numberator and

denominator by the LCD


3x + 5

2x – 3

+ 1x + 5

13.

3x – 33x + 7

=

3(x – 3)

3x + 7=

Multiply numberator and denominator by the LCD

Simplify

Simplify

3x + 5

2x – 3

+ 1x + 5

=

3x + 5

2x – 3

+ 1x + 5

(x + 5)(x – 3)

(x + 5)(x – 3)

Essential question

What are the steps for adding or subtracting rational expressions with

different denominators?

1. Find the least common denominator2. Rewrite each fraction using the

common denominator3. Add or subtract

4. Simplify

Bell Ringer

Solve the equation: =

Lesson 6: Solving Rational Equations

Essential question

What are the steps for solving rational equations?

VOCABULARY

Cross mulitplying ~ A method for solving a simple rational equation for which each side of the equation is a single rational expression

Extraneous solution

EXAMPLE 1 Solve a rational equation by cross multiplying

Solve: 3x + 1

= 94x + 1

3x + 1

= 94x + 1

Write original equation.

3(4x + 5) = 9(x + 1) Cross multiply.

12x + 15 = 9x + 9 Distributive property

3x + 15 = 9 Subtract 9x from each side.

3x = – 6 Subtract 15 from each side.

x = – 2 Divide each side by 3.

The solution is –2. Check this in the original equation.

ANSWER


SOLUTION

5x

+74

= – 9x

– 9x

4x4x 5x

+74( )=


Multiply each side by the LCD, 4x.

20 + 7x = –36 Simplify.

7x = – 56 Subtract 20 from each side.

x = – 8 Divide each side by 7.


The correct answer is B.

ANSWER

EXAMPLE 4 Solve a rational equation with two solutions

Solve: 1 – 8x – 5

=3x

1 – 8x – 5

=3x


x(x – 5) 1– 8x – 5( )= x(x – 5) 3

xMultiply each side by the LCD, x(x–5).

x(x –5) – 8x = 3(x – 5) Simplify.

x2 – 5x – 8x = 3x – 15 Simplify.

x2 – 16x +15 = 0 Write in standard form.

(x – 1)(x – 15) = 0Factor.

x = 1 or x = 15 Zero product property

EXAMPLE 4 Solve a rational equation with two solutions

The solutions are 1 and 15. Check these in the original equation.

ANSWER

EXAMPLE 5 Check for extraneous solutions

Solve: 6x – 3 =

8x2

x2 – 9– 4x

x + 3

SOLUTION

Write each denominator in factored form. The LCD is (x + 3)(x – 3).

=8x2

(x + 3)(x – 3)–

4xx + 3

6x –3

x + 3(x + 3)(x – 3) 6

x –3= (x + 3)(x – 3) 8x2

(x + 3)(x – 3)(x + 3)(x – 3) 4x

6(x + 3) = 8x2 – 4x(x – 3)

6x + 18 = 8x2 – 4x2 + 12x

EXAMPLE 5 Check for extraneous solutions

0 = 4x2 + 6x –18

0 = 2x2 + 3x – 9

0 = (2x – 3)(x + 3)

2x – 3 = 0 or x + 3 = 0

x = 32

or x = –3

You can use algebra or a graph to check whether either of the two solutions is extraneous.

The solution checks, but the apparent solution –3 is extraneous, because substituting it in the equation results in division by zero, which is undefined.

32

Algebra


Solve the equation by using the LCD. Check for extraneous solutions.

7

2+

3

x= 35.

SOLUTION

Write each denominator in factored form. The LCD is 2x

7

2+

3

x= 3

2x + 2x = 2x 37

2

3

x

7x + 6 = 6x

x = – 6


2

x+

4

3= 26.

SOLUTION


2

x+

4

3= 2

3x + 3x = 3x 22

x

4

3

6 + 4x = 6x

6 = 2x

x = 3


3

7+

8

x= 17.

SOLUTION


3

7+

8

x= 1

7x + 7x = 7x 13

7

8

x

3x + 56 = 7x

56 = 4x

x = 14


8. 3

2+ 4

x –1 x –1x +1=

SOLUTION

Write each denominator in factored form. The LCD is 2( x – 1)

3

2+ 4

x –1 x –1x +1=

(x – 1 )(2) + (x – 1)(2) = (x – 1)(2) 32

4x –1

x + 1x 1

3x – 3 + 8= 2x + 2

x = – 3


3xx +1

– 52x

= 32x

9.

SOLUTION

Write each denominator in factored form. The LCD is (x + 1)(2x)

3x 5 3x +1

–2x

=2x

32 x

2x (x + 1) – 2x (x +1) = 2x (x +1) 3xx +1

52 x


6x2 – 5x – 5 = 3x + 3

0 = 3x + 3 – 6x2 +5x + 5

0 = – 6x2 + 8x + 8

0 = (3x +2) (x – 2)

x – 2 = 0

x = 2or

3x + 2 = 0

x = 2

3–


5xx –2

= 7 +10x –2

10.

SOLUTION

Write each denominator in factored form. The LCD is x – 2

5xx –2

= 7 +10x –2

x – 2 = (x – 2) 7 + (x – 2)

10x – 2

5xx –2

5x = 7x – 14 + 10

4 = 2x

x = 2

x=2 results in no solution.

Essential question

What are the steps for solving rational equations?

If a proportion = find the cross product

If not a proportion – find the product by using LCD of each expression, simplify, check for

extraneous solutions

Documents

Please complete the Prerequisite Skills on Page 548 #4-12