1. Unit 3
Practice Test

2. Rational Equations Work Application1
Rational Equations Work Application
Gloria can clean a house in 4 hours and Joe can clean the house in 6 hours. About how many hours will it take them to clean the house, if they work together?
Gloria’s Rate
Joe’s Rate
Let x = Number of hours for both to clean house

3. Rational Equations Work Application1
Rational Equations Work Application
Let x = Number of hours for both to clean house
Gloria’s portion
of house cleaned
Joe’s portion
of house cleaned
Total
house cleaned + =
Gloria’s Rate 
Hours
Worked
Joe’s Rate 
Hours
Worked + = 1
LCD = 12

4. Rational Equations Work Application1
Rational Equations Work Application
Let x = Number of hours for both to clean house
LCD = 12 + = 1
12  +
12  =
12  1 + = 12
3x x =
5x =
Answer
2.4 hours
x =

5. 2 1. y = x2 + 5 0 = x2 + x – 2 (7 – x) = x2 + 5 0 = (x + 2)(x – 1)
What is the solution of the system of equations?
Method: Use substitution
Step 1: Substitute equation 2 into equation 1. Then solve.
y = x2 + 5
0 = x2 + x – 2
(7 – x) = x2 + 5
0 = (x + 2)(x – 1)
7 – x = x2 + 5
+ x x
x + 2 = 0
x – 1 = 0
7 = x2 + x + 5
x = –2
x = 1
– –7
0 = x2 + x – 2

6. 2 x = –2 x = 1 What is the solution of the system of equations?
Step 2: Find y-values, by substituting x-values into
either equation.
Use equation y = 7 – x
x = –2:
y = 7 – (–2)
x = 1:
y = 7 – (1)
y = 7 + 2
y = 6
y = 9
(1, 6)
(–2, 9)

7. 3 x + 3 > 0 2x – 5 > 0 –3 –3 +5 +5 x > –3 2x > 5
Find the domain of each function.
x + 3 > 0
2x – 5 > 0
–3 –3
x > –3
2x > 5

8. x2 + 8 = y – 6x y-intercept 4 x-intercept Let y = 0 x2 + 8 = y – 6x
Find the intercepts of the graphs of each equation.
x2 + 8 = y – 6x
x-intercept
y-intercept
Let y = 0
x2 + 8 = y – 6x
Let x = 0
x2 + 8 = 0 – 6x
x2 + 8 = y – 6x
x2 + 8 = –6x
(0)2 + 8 = y – 6(0)
+6x x
0 + 8 = y – 0
x2 + 6x + 8 = 0
(x + 2)(x + 4) = 0
8 = y
x + 2 = 0
x + 4 = 0
x = –2
x = –4

9. 4x + 2y = 12 x-intercept y-intercept 4 Let y = 0 Let x = 0
Find the intercepts of the graphs of each equation.
4x + 2y = 12
x-intercept
y-intercept
Let y = 0
Let x = 0
4x + 2y = 12
4x + 2y = 12
4x + 2(0) = 12
4(0) + 2y = 12
4x + 0 = 12
0 + 2y = 12
4x = 12
2y = 12
x = 3
y = 6

10. Horizontal compression by ¼5
Describe the transformation for each function.
Shift down 7 units
Horizontal compression by ¼

11. Reflection across y-axis5
Describe the transformation for each function.
Reflection across y-axis
Vertical compression by ⅓

12. Reflection across x-axis5
Describe the transformation for each function.
Shift 5 units right
Reflection across x-axis
Vertical stretch by 6

13. x2 – 1 = 0 (x + 1)(x – 1) = 0 x + 1 = 0 x – 1 = 0 x = –1 x = 16
Vertical Asymptote
x2 – 1 = 0
Let denominator equal 0
Solve for x.
(x + 1)(x – 1) = 0
x + 1 = 0
x – 1 = 0
Domain
Range
x = –1
x = 1
x  1
y  0
x  –1
Horizontal Asymptote
y = 0

14. Let denominator equal 0 x – 1 = 0 +1 +1 Solve for x. x = 1 x  1 y  46
Vertical Asymptote
Let denominator equal 0
x – 1 = 0
Solve for x.
x = 1
Domain
Range
Horizontal Asymptote
x  1
y  4
y = 4

15. Let denominator equal 0 x – 1 = 0 +1 +1 Solve for x. x = 1 x  16
Vertical Asymptote
Let denominator equal 0
x – 1 = 0
Solve for x.
x = 1
Domain
Range
Horizontal Asymptote
x  1
All real
numbers
None

16. x2 – 9 = 0 (x + 3)(x – 3) = 0 x + 3 = 0 x – 3 = 0 x = –3 x = 36
Vertical Asymptote
Let denominator equal 0
x2 – 9 = 0
Solve for x.
(x + 3)(x – 3) = 0
x + 3 = 0
x – 3 = 0
Domain
Range
x = –3
x = 3
x  3
y  3
x  –3
Horizontal Asymptote
y = 3

17. 7
Sketch the graph of each function.

18. 7
Sketch the graph of each function.

19. Sketch the graph of x – 2 = 0 x = 2 y = 1 8 Vertical Asymptote
Horizontal Asymptote
y = 1

20. Sketch the graph of x – 2 = 0 x = 2 y = 1 x  2 y  1 8
Vertical Asymptote
x – 2 = 0
x = 2
Horizontal Asymptote
y = 1
Domain
Range
x  2
y  1

21. Sketch the graph of x + 3 = 0 x = –3 y = 0 8 Vertical Asymptote
Horizontal Asymptote
y = 0

22. Sketch the graph of x + 3 = 0 x = –3 y = 0 x  –3 y  0 8
Vertical Asymptote
x + 3 = 0
x = –3
Horizontal Asymptote
y = 0
Domain
Range
x  –3
y  0

23. a2 = a2 +3a + 9 0 = 3a + 9 –9 = 3a –3 = a 9 Square both sides
Subtract a2 on both sides
0 = 3a + 9
–9 = 3a
Subtract 9 on both sides
–3 = a
Final Answer: No Solution

24. 9 Square both sides Subtract a2 on both sides
Final Answer: No Solution

25. 5 – x = (x + 1)(x + 1) 5 – x = x2 + 2x + 1 –5 + x –5 + x10
5 – x = (x + 1)(x + 1)
5 – x = x2 + 2x + 1
–5 + x
–5 + x
= x2 + 3x – 4

26. 0 = x2 + 3x – 4 0 = (x – 1)(x + 4) x – 1 = 0 x + 4 = 0 x = 1 x = –410
0 = x2 + 3x – 4
0 = (x – 1)(x + 4)
x – 1 = 0
x + 4 = 0
x = 1
x = –4
Answer: x = 1

27. 11 LCD = 3x 10 4 Solve the equation = + 2. 3 x
(3x) = (3x) + 2(3x) 10 3 4 x
Multiply each term by the LCD, 3x.
Simplify.
10x = x
Combine like terms.
4x = 12
Solve for x.
x = 3

28. 12 LCD = (x – 4)(x + 4) 16 x2 – 16 2 x – 4 = Solve the equation.
Multiply each term by the LCD, (x – 4)(x +4).
= 2(x + 4)
Simplify.
= 2x + 8
Solve for x.
= 2x
4 = x
No solution

29. x – 5 = 0 x = 5 3(x – 5) = (12)(1) 3x – 15 = 12 x = 5 3x = 27 x = 913
Solve
Step 1A
Step 1B
Let denominator = 0
x – 5 = 0
x = 5
3(x – 5) = (12)(1)
Endpoints
3x – 15 = 12
x = 5
3x = 27
x = 9
x = 9

30. 5 < x < 9 2 –4 > 3 F 6 12 > 3 T 11 2 > 3 F Solve A B C13
Solve
Step 2 A B C 5 9
Step 3
Interval
Test
Number
Test of
Inequality
True / False A B C 2
–4 > 3 F 6
12 > 3 T 11
2 > 3 F
Step 4
5 < x < 9

31. x – 8 = 0 x = 8 3(x – 8) = (6)(1) 3x – 24 = 6 x = 8 3x = 30 x = 1014
Solve
Step 1A
Step 1B
Let denominator = 0
x – 8 = 0
x = 8
3(x – 8) = (6)(1)
Endpoints
3x – 24 = 6
x = 8
3x = 30
x = 10
x = 10

32. x < 8 or x > 10 7 –6 < 3 T 9 6 < 3 F 11 2 < 3 T Solve A14
Solve
Step 2 A B C 8 10
Step 3
Interval
Test
Number
Test of
Inequality
True / False A B C 7
–6 < 3 T 9
6 < 3 F 11
2 < 3 T
Step 4
x < 8 or
x > 10

33. 15 Subtract . 2x2 + 64 x2 – 64 – x + 8 x – 4 2x2 + 64 (x – 8)(x + 8) –
Factor the denominators.
The LCD is (x – 8)(x + 8), so multiply by
x – 4
x + 8
(x – 8)
2x2 + 64
(x – 8)(x + 8) –
x + 8
x – 4
x – 8
2x – (x – 4)(x – 8)
(x – 8)(x + 8)
Subtract the numerators.
2x – (x2 – 12x + 32)
(x – 8)(x + 8)
Multiply the binomials in the numerator.

34. 15 2x2 + 64 – x2 + 12x – 32 Distribute the negative sign.
Write the numerator in standard form.
(x + 8)(x + 4)
(x – 8)(x + 8)
Factor the numerator.
x + 4
x – 8
Divide out common factors.

35. 16

36. 17

37. x2 + 3x + 2 = 0 (x + 1)(x + 2) = 0 x + 1 = 0 x + 2 = 0 x = –1 x = –218
Identify any x-values for which
the expression is undefined.
Let denominator = 0
x2 + 3x + 2 = 0
(x + 1)(x + 2) = 0
x + 1 = 0
x + 2 = 0
–1 –1
–2 –2
x = –1
x = –2

38. 19
Simplify.
LCD = 2x
Multiply every fraction of the complex fraction by the LCD.

39. 19

40. Substitute 30 for y and -6 for x.20
Given: y varies directly as x, and y = 30 when x = –6. Write the direct variation function.
y = kx
y varies directly as x.
30 = k(–6)
Substitute 30 for y and -6 for x.
30 k(-6) =
Solve for the constant of variation k.
–5 = k
y = –5x
Write the variation function by using the value of k.

41. Substitute 8 for y and 4 for x.21
Given: y varies inversely as x, and y = 8
when x = 4. Write the inverse variation function.
y varies inversely as x.
Substitute 8 for y and 4 for x.
k = 32
Solve for k.
Write the variation formula.