1. Please complete the Prerequisite Skills on Page 548 #4-12
Bell Ringer
Please complete the Prerequisite Skills on Page 548 #4-12

2. Chapter 8: Rational Functions
Big ideas:
Performing operations with rational expressions
Solving rational equations

3. Lesson 1: Model Inverse & Joint Variation

4. What are the differences between direct, inverse and joint variation?
Essential question

5. VOCABULARY
Inverse variation ~ The relationship of two variables x and y if there is a nonzero number a such that y = 𝑎 𝑥
Constant of variation ~ The nonzero constant a in a direct variation equation y=ax, an inverse variation equation y= 𝑎 𝑥 , 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.

6. 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
y = 7 x
a. xy = 7
Inverse
b. y = x + 3
Neither y 4 c.
= x
y = 4x
Direct

7. Write general equation for inverse variation.
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 x
Write general equation for inverse variation.
7 = a 4
Substitute 7 for y and 4 for x.
28 = a
Solve for a.
The inverse variation equation is y = 28 x
When x = –2, y = –2
= –14.
ANSWER

8. 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
Essential question

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

10. Lesson 4: Multiply & Divide Rational expressions

11. What are the steps for multiplying & dividing rational expressions?
Essential question

12. 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

13. Factor numerator and denominator.
EXAMPLE 1
Simplify a rational expression
x2 – 2x – 15
x2 – 9
Simplify :
SOLUTION
x2 – 2x – 15
x2 – 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.
x – 5
x – 3 =
Simplified form
ANSWER
x – 5
x – 3

14. Multiply numerators and denominators.
EXAMPLE 3
Standardized Test Practice
SOLUTION
8x 3y
2x y2
7x4y3 4y
56x7y4
8xy3 =
Multiply numerators and denominators.
x x6 y3 y
8 x y3 =
Factor and divide out common factors.
7x6y =
Simplified form
The correct answer is B.
ANSWER

15. Factor numerators and denominators.
EXAMPLE 4
Multiply rational expressions
x2 + x – 20 3x
3x –3x2
x2 + 4x – 5
Multiply:
SOLUTION
x2 + x – 20 3x
3x –3x2
x2 + 4x – 5
3x(1– x)
(x –1)(x +5) =
(x + 5)(x – 4) 3x
Factor 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.

16. Simplify. Multiply. EXAMPLE 4 Multiply rational expressions
ANSWER
–x + 4

17. Write polynomial as a rational expression.
EXAMPLE 5
Multiply a rational expression by a polynomial
Multiply:
x + 2
x3 – 27
(x2 + 3x + 9)
SOLUTION
x + 2
x3 – 27
(x2 + 3x + 9) =
x + 2
x3 – 27
x2 + 3x + 9 1
Write polynomial as a rational expression.
(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 + 2
x – 3
Simplified form
ANSWER
x + 2
x – 3

18. Multiply numerators and denominators.
GUIDED PRACTICE
for Examples 3, 4 and 5
Multiply the expressions. Simplify the result.
3x5 y2
8xy
6xy2
9x3y 8.
SOLUTION
3x5 y2
2xy
6xy2
9x3y
18x6y4
72x4y2 =
Multiply numerators and denominators.
18 x4 y2 x2 y2
x4 y2 =
Factor and divide out common factors. =
x2y2 4
Simplified form

19. Factor numerators and denominators.
GUIDED PRACTICE
for Examples 3, 4 and 5
2x2 – 10x
x + 3 9.
x2– 25
2x2
SOLUTION
x + 3
2x2
2x2 – 10x
x2– 25
2x(x –5)
(x –5)(x +5) =
x + 3 2x
(x)
Factor numerators and denominators.
2x(x –5) (x + 3) =
(x –5)(x + 5)2x (x)
Multiply numerators and denominators. =
2x(x –5) (x + 3)
(x –5)(x + 5)2x (x)
Divide out common factors.
x + 3 =
x(x + 5)
Simplified form

20. Multiply numerators and denominators.
GUIDED PRACTICE
for Examples 3, 4 and 5
x + 5
10.
x2 +x + 1
x3– 1
SOLUTION
x2 +x + 1
x + 5
x3– 1 =
x2 +x + 1
x + 5
(x – 1) (x2 +x + 1) 1
Factor denominators. =
(x + 5) (x2 +x + 1)
(x – 1) (x2 +x + 1)
Multiply numerators and denominators.
(x + 5) (x2 +x + 1)
(x – 1) (x2 +x + 1) =
Divide out common factors.
x + 5 =
x – 1
Simplified form

21. Multiply by reciprocal.
EXAMPLE 6
Divide rational expressions
Divide : 7x
2x – 10
x2 – 6x
x2 – 11x + 30
SOLUTION 7x
2x – 10
x2 – 6x
x2 – 11x + 30 7x
2x – 10
x2 – 6x
x2 – 11x + 30 =
Multiply by reciprocal. 7x
2(x – 5) =
(x – 5)(x – 6)
x(x – 6)
Factor. =
7x(x – 5)(x – 6)
2(x – 5)(x)(x – 6)
Divide out common factors. 7 2 =
Simplified form
ANSWER 7 2

22. Multiply by reciprocal.
EXAMPLE 7
Divide a rational expression by a polynomial
Divide :
6x2 + x – 15
4x2
(3x2 + 5x)
SOLUTION
6x2 + x – 15
4x2
(3x2 + 5x)
6x2 + x – 15
4x2
3x2 + 5x = 1
Multiply by reciprocal.
(3x + 5)(2x – 3)
4x2 =
x(3x + 5) 1
Factor.
(3x + 5)(2x – 3) =
4x2(x)(3x + 5)
Divide out common factors.
2x – 3
4x3 =
Simplified form
ANSWER
2x – 3
4x3

23. Multiply by reciprocal.
GUIDED PRACTICE
for Examples 6 and 7
Divide the expressions. Simplify the result. 4x
5x – 20
x2 – 2x
x2 – 6x + 8
11.
SOLUTION 4x
5x – 20
x2 – 2x
x2 – 6x + 8 4x
5x – 20
x2 – 2x
x2 – 6x + 8 =
Multiply by reciprocal.
4(x)(x – 4)(x – 2)
5(x – 4)(x)(x – 2) =
Factor.
4(x)(x – 4)(x – 2)
5(x – 4)(x)(x – 2) =
Divide out common factors. 4 5 =
Simplified form

24. Multiply by reciprocal.
GUIDED PRACTICE
for Examples 6 and 7
2x2 + 3x – 5 6x
(2x2 + 5x)
12.
SOLUTION
2x2 + 3x – 5 6x
(2x2 + 5x) =
2x2 + 3x – 5 6x
(2x2 + 5x) 1
Multiply by reciprocal.
(2x + 5)(x – 1)
6x(x)(2 x + 5) =
Factor.
(2x + 5)(x – 1)
6x(x)(2 x + 5) =
Divide out common factors.
x – 1
6x2 =
Simplified form

25. What are the steps for multiplying & dividing rational expressions?
Multiply: multiply the numerators / multiply the denominators then simplify
Divide: multiply the first expression by the reciprocal of the second expression, then follow the rules for multiplication
Essential question

26. Find the least common multiple of 20 and 45.
Bell Ringer
Find the least common multiple of 20 and 45.

27. Lesson 5: Add & Subtract Rational Expressions

28. What are the steps for adding or subtracting rational expressions with different denominators?
Essential question

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

30. Add numerators and simplify result.
EXAMPLE 1
Add or subtract with like denominators
Perform the indicated operation. 7 4x + 3 a. 2x
x + 6 – 5 b.
SOLUTION 7 4x + 3 a. =
7 + 3 4x 10 4x = 5 2x =
Add numerators and simplify result. 2x
x + 6 5 – b.
x + 6
2x – 5 =
Subtract numerators.

31. Subtract numerators and simplify results .
GUIDED PRACTICE
for Example 1
Perform the indicated operation and simplify. a. 7
12x + 5 =
7 – 5
12x = 2
12x = 1 6x
Subtract numerators and simplify results . b. 2
3x2 + 1 =
2 + 1
3x2 = 3
3x2 = 1 x2
Add numerators and simplify results. c. 4x
x–2 – x =
4x–x
x–2 = 3x
x–2 = 3x
3x – 2
Subtract numerators. d. 4x
x2+1 + 2
2x2+2
x2+1 =
2(x2+1)
x2+1 =
Factor numerators and simplify results .
= 2

32. 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 as products of primes.
4x2 – 16 = 4(x2 – 4) = (22)(x + 2)(x – 2)
6x2 – 24x + 24 = 6(x2 – 4x + 4) = (2)(3)(x – 2)2

33. 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

34. Factor second denominator.
EXAMPLE 3
Add with unlike denominators
Add:
9x2 7 + x
3x2 + 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). 7
9x2 x
3x2 + 3x = +
3x(x + 1)
Factor second denominator. 7
9x2
x + 1 +
3x(x + 1) x 3x
LCD is 9x2(x + 1).

35. Multiply. Add numerators. EXAMPLE 3 Add with unlike denominators
9x2(x + 1)
3x2 + =
Multiply.
3x2 + 7x + 7
9x2(x + 1) =
Add numerators.

36. Factor denominators. LCD is 2(x  1)(x  3). Multiply. EXAMPLE 4
Subtract with unlike denominators
Subtract:
x + 2
2x – 2
–2x –1
x2 – 4x + 3 –
SOLUTION
x + 2
2x – 2
–2x –1
x2 – 4x + 3 –
x + 2
2(x – 1)
– 2x – 1
(x – 1)(x – 3) – =
Factor denominators.
x + 2
2(x – 1) =
x – 3 –
– 2x – 1
(x – 1)(x – 3) 2
LCD is 2(x  1)(x  3).
x2 – x – 6
2(x – 1)(x – 3)
– 4x – 2 – =
Multiply.

37. Factor numerator. Divide out common factor.
EXAMPLE 4
Subtract with unlike denominators
x2 – x – 6 – (– 4x – 2)
2(x – 1)(x – 3) =
Subtract numerators.
x2 + 3x – 4
2(x – 1)(x – 3) =
Simplify numerator. =
(x –1)(x + 4)
2(x – 1)(x – 3)
Factor numerator. Divide out common factor.
x + 4
2(x –3) =
Simplify.

38. GUIDED PRACTICE
for Examples 2, 3 and 4
Find the least common multiple of the polynomials.
5. 5x3 and 10x2–15x
STEP 1
Factor each polynomial. Write numerical factors as products of primes.
5x3 = 5(x) (x2)
10x2 – 15x = 5(x) (2x – 3)
STEP 2
Form the LCM by writing each factor to the highest power it occurs in either polynomial.
LCM = 5x3 (2x – 3)

39. GUIDED PRACTICE
for Examples 2, 3 and 4
Find the least common multiple of the polynomials.
6. 8x – 16 and 12x2 + 12x – 72
STEP 1
Factor each polynomial. Write numerical factors as products of primes.
8x – 16 =
8(x – 2) =
23(x – 2)
12x2 + 12x – 72 =
12(x2 + x – 6) =
4 3(x – 2 )(x + 3)
STEP 2
Form the LCM by writing each factor to the highest power it occurs in either polynomial.
LCM = 8 3(x – 2)(x + 3)
= 24(x – 2)(x + 3)

40. LCD is 28x Multiply Simplify GUIDED PRACTICE for Examples 2, 3 and 4– 7 1 7.
SOLUTION 4x 3 – 7 1 4x 3 7 1 –
LCD is 28x 4x
7(4x) 21
4x(7) –
Multiply
21 – 4x
28x =
Simplify

41. Factor denominators LCD is 3x2 (3x – 4) Multiply GUIDED PRACTICE
for Examples 2, 3 and 4 1
3x2 + x
9x2 – 12x 8.
SOLUTION 1
3x2 + x
9x2 – 12x = 1
3x2 + x
3x(3x – 4)
Factor denominators = 1
3x2
3x – 4 + x
3x(3x – 4)
LCD is 3x2 (3x – 4)
3x – 4
3x2 (3x – 4) + x2
3x2 (3x – 4 ) =
Multiply

42. Add numerators Simplify GUIDED PRACTICE for Examples 2, 3 and 4
3x – 4 + x2
3x2 (3x – 4)
Add numerators
x2 + 3x – 4
3x2 (3x – 4)
Simplify

43. Factor denominators LCD is 12(x – 4) (x+3) Multiply GUIDED PRACTICE
for Examples 2, 3 and 4 x
x2 – x – 12 + 5
12x – 48 9.
SOLUTION x
x2 – x – 12 + 5
12x – 48 = x
(x+3)(x – 4) + 5
12 (x – 4)
Factor denominators x
(x + 3)(x – 4) = 12 + 5
12(x – 4)
x + 3
LCD is 12(x – 4) (x+3)
5(x + 3)
12(x + 3)(x – 4)
12x = +
Multiply

44. Add numerators Simplify GUIDED PRACTICE for Examples 2, 3 and 4
12x + 5x + 15
12(x + 3)(x – 4) =
Add numerators
17x + 15
12(x +3)(x + 4) =
Simplify

45. Factor denominators LCD is (x – 2) (x+2)2 Multiply GUIDED PRACTICE
for Examples 2, 3 and 4
x + 1
x2 + 4x + 4 – 6
x2 – 4
10.
SOLUTION
x + 1
x2 + 4x + 4 – 6
x2 – 4 =
x + 1
(x + 2)(x + 2) 6
(x – 2)(x + 2) –
Factor denominators
x + 1
(x + 2)(x + 2) =
x – 2 – 6
(x – 2)(x + 2)
x + 2
LCD is (x – 2) (x+2)2 =
x2 – 2x + x – 2
(x + 2)(x + 2)(x – 2) –
6x + 12
(x – 2)(x + 2)(x + 2)
Multiply

46. Subtract numerators Simplify GUIDED PRACTICE for Example 2, 3 and 4=
x2 – 2x + x – 2 – (6x + 12)
(x + 2)2(x – 4)
Subtract numerators =
x2 – 7x – 14
(x + 2)2 (x – 2)
Simplify

47. Multiply numerator and denominator by the LCD.
EXAMPLE 6
Simplify a complex fraction (Method 2) 5
x + 4 1 + 2 x
Simplify:
SOLUTION
The LCD of all the fractions in the numerator and denominator is x(x + 4). 5
x + 4 1 + 2 x 5
x + 4 1 + 2 x =
x(x+4)
Multiply numerator and denominator by the LCD.
x + 2(x + 4) 5x =
Simplify. 5x
3x + 8 =
Simplify.

48. Multiply numberator and denominator by the LCD
GUIDED PRACTICE
for Examples 5 and 6 x 6 3 – 5 7 10
11. x 6 3 – 5 7 10 x 6 3 – 5 7 10 30 =
Multiply numberator and denominator by the LCD
– 5x
3 (2x – 7) =
Simplify

49. Multiply numberator and denominator by the LCD
GUIDED PRACTICE
for Examples 5 and 6 2 x – + 4 3
12. 2 x – + 4 3 = 2 x – + 4 3
Multiply numberator and denominator by the LCD
2 – 4x
2 + 3x =
Simplify
2 (1 – 2x )
2 + 3x =
Simplify

50. Multiply numberator and denominator by the LCD
GUIDED PRACTICE
for Examples 5 and 6 3
x + 5 2
x – 3 + 1
13. 3
x + 5 2
x – 3 + 1 = 3
x + 5 2
x – 3 + 1
(x + 5)(x – 3)
Multiply numberator and denominator by the LCD
3x – 3
3x + 7 =
Simplify
3(x – 3)
3x + 7 =
Simplify

51. What are the steps for adding or subtracting rational expressions with different denominators?
1. Find the least common denominator
2. Rewrite each fraction using the common denominator
3. Add or subtract
4. Simplify
Essential question

52. Bell Ringer
Solve the equation: 4 𝑥 = 6 15

53. Lesson 6: Solving Rational Equations

54. What are the steps for solving rational equations?
Essential question

55. 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

56. Write original equation.
EXAMPLE 1
Solve a rational equation by cross multiplying
Solve: 3
x + 1 = 9
4x + 1 3
x + 1 = 9
4x + 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

57. ( ) Write original equation. Multiply each side by the LCD, 4x.
EXAMPLE 3
Standardized Test Practice
SOLUTION 5 x + 7 4 = – 9
Write original equation. – 9 x 4x 5 + 7 4 ( ) =
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.

58. EXAMPLE 3
Standardized Test Practice
ANSWER
The correct answer is B.

59. ( ) Write original equation. Multiply each side by the LCD, x(x–5).
EXAMPLE 4
Solve a rational equation with two solutions
1 – 8
x – 5 = 3 x
Solve:
1 – 8
x – 5 = 3 x
Write original equation.
x(x – 5) 1– 8
x – 5 ( ) = 3 x
Multiply 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) = 0
Factor.
x = 1 or x = 15
Zero product property

60. EXAMPLE 4
Solve a rational equation with two solutions
ANSWER
The solutions are 1 and 15. Check these in the original equation.

61. EXAMPLE 5
Check for extraneous solutions 6
x – 3 =
8x2
x2 – 9 – 4x
x + 3
Solve:
SOLUTION
Write each denominator in factored form. The LCD is (x + 3)(x – 3). =
8x2
(x + 3)(x – 3) – 4x
x + 3 6
x –3
x + 3
(x + 3)(x – 3) 6
x –3 =
8x2
(x + 3)(x – 3) 4x
6(x + 3) = 8x2 – 4x(x – 3)
6x + 18 = 8x2 – 4x2 + 12x

62. 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 = 3 2 or
x = –3
You can use algebra or a graph to check whether either of the two solutions is extraneous.
Algebra
The solution checks, but the apparent solution –3 is extraneous, because substituting it in the equation results in division by zero, which is undefined. 3 2

63. GUIDED PRACTICE
for Examples 3, 4 and 5
Solve the equation by using the LCD. Check for extraneous solutions. 7 2 + 3 x
= 3 5.
SOLUTION
Write each denominator in factored form. The LCD is 2x 7 2 + 3 x
= 3
2x x = 2x 3 7 2 3 x
7x + 6 = 6x
x = – 6

64. GUIDED PRACTICE
for Examples 3, 4 and 5 2 x + 4 3
= 2 6.
SOLUTION
Write each denominator in factored form. The LCD is 3x 2 x + 4 3
= 2
3x x = 3x 2 2 x 4 3
6 + 4x = 6x
6 = 2x
x = 3

65. GUIDED PRACTICE
for Examples 3, 4 and 5 3 7 + 8 x
= 1 7.
SOLUTION
Write each denominator in factored form. The LCD is 7x 3 7 + 8 x
= 1
7x x = 7x 1 3 7 8 x
3x + 56 = 7x
56 = 4x
x = 14

66. GUIDED PRACTICE
for Examples 3, 4 and 5 8. 3 2 + 4
x –1
x +1 =
SOLUTION
Write each denominator in factored form. The LCD is 2( x – 1) 3 2 + 4
x –1
x +1 =
(x – 1 )(2) (x – 1)(2) = (x – 1)(2) 3 2 4
x –1
x + 1
x  1
3x – 3 + 8= 2x + 2
x = – 3

67. GUIDED PRACTICE
for Examples 3, 4 and 5 3x
x +1 – 5 2x = 3 9.
SOLUTION
Write each denominator in factored form. The LCD is (x + 1)(2x) 3x 5 3
x +1 – 2x = 3
2 x
2x (x + 1) – 2x (x +1) = 2x (x +1) 3x
x +1 5

68. GUIDED PRACTICE
for Examples 3, 4 and 5
6x2 – 5x – 5 = 3x + 3
0 = 3x + 3 – 6x2 +5x + 5
0 = – 6x2 + 8x + 8
0 = (3x +2) (x – 2)
3x + 2 = 0
x = 2 3 –
x – 2 = 0
x = 2 or

69. GUIDED PRACTICE
for Examples 3, 4 and 5 5x
x –2 = 7 + 10
10.
SOLUTION
Write each denominator in factored form. The LCD is x – 2 5x
x –2 = 7 + 10
x – = (x – 2) (x – 2) 10
x – 2 5x
x –2
5x = 7x –
4 = 2x
x = 2
x=2 results in no solution.

70. 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
Essential question