INTRODUCTORY ALGEBRA APPEAL PRACTICE TEST
6/07
DO NOT WRITE ON THIS PAPER. RETURN IT TO THE OFFICE WHEN YOU ARE
READY TO TAKE THE OFFICIAL TEST.
Give all answers in simplest form.
-3
1
1. Add: 4 + 8
2. Divide and simplify:
3. Simplify
5  15

7 14
1
(6 x  18)  4(3x  2)
3
4. Simplify. Write all answers without negative or zero exponents.
18x4y3
-12x2y5
5. Solve for x: 18 – 4(2x – 3) + 5x = 41
6. Solve for x:
2
1
1
4
3x + 4 = 6x – 3
7. Solve the inequality
-2x + 3 < 7
8. Graph the solution for 5x – 1 < 3x + 7
M
9. Solve for d if A = fjd , given that A = 5, f = 2, j = 3, and M = 6.
3x  y
 5w , solve for y.
4
10.
In equation
11.
Find the missing value so that (1, __) is a solution for 3x – 2y = 7.
12.
Graph the line x + 2y = 7.
13.
Find the slope and y-intercept of the line whose equation is 2y – 3x = 10.
14.
Write the equation of the line passes that through the points (4 , -1) and (3 , -3).
15.
Write the equation of the line with slope -2 that passes through the point (-1 , -3).
Introductory Algebra Appeals Practice Test
16.
Solve the following system of equations
4x – y = 10
y = 3x + 2
17.
Simplify, writing the answer without negative or zero exponents. (6 xy 4 )(3xy)
18.
Joe bought some large frames for $30 each and some small frames for $16 each. If he
bought a total of 20 frames for $530, find how many of each type he bought.
19.
Simplify, writing the answer without negative or zero exponents: (x -4) 3
20.
Simplify the expression: 6 12 – 3 20 + 36
21.
Write the following decimal in scientific notation: 0.00519
22.
Multiply:
23.
Multiply and simplify:
(3x – 5)(x + 2)
24.
Multiply and simplify:
(x – 3)(x2 + x + 3)
25.
Factor completely:
3x3 + 11x2 – 4x
26.
Factor completely:
3x2 + 4x – 4
27.
Factor completely:
4x2 – 9
28.
Solve by factoring:
3x2 + x = 0
29.
Solve by factoring:
2x2 – 3x – 5 = 0
30.
Use the Pythagorean Theorem to find the missing side of the right triangle. Round your
answer to two decimal places if necessary.
x(3x2 + 5x – 2)
7
5
x
2
Introductory Algebra Appeals Practice Test
Practice Test Solutions:
3
1
6
1
5
1. - 4 + 8 = - 8 + 8 = - 8
2.
5 1 1 4 2 1 2
5  15 5 14
2

 


 

7 14
7  15 7 1  1 5 3 1  3  3
3.
2x – 6 – 12x – 8 = -10x – 14
18x4y3
-3x2
4. -12x2y5 = 2y2
5. 18 – 4 (2x – 3) + 5x = 41
18 – 8x +12 + 5x = 41
-3x + 30 = 41
-3x = 11
11
x =- 3
2
1 1
4
6. 3x + 4 = 6x – 3
2
1
1
-4
12(3x ) + 12(4 ) = 12(6x ) + 12( 3 )
8x + 3 = 2x – 16
6x = -19
-19
x= 6
7. Algebraic Solution:
-2x + 3 < 7
-2x + 3 – 3 < 7 – 3
-2x < 4
-2x 4
-2 > -2
3
Introductory Algebra Appeals Practice Test
x > -2
8. 5x – 1 < 3x – 7
+1
+1
5x
< 3x – 6
-3x
-3x
2x
-6
< 2
2
x < -3
-3
M
M
9. A = fjd ; Afjd = M; d = Afj
 3x  y 
4
  4(5w)
 4 
3x  y  20w
10.
d
0
6
6 1


5  2  3 30 5
y= 20w – 3x
-2y
4
11. 3x – 2y = 7; 3(1) – 2y = 7; 3 – 2y = 7; 3 – 3 – 2y = 7 – 3; -2y = 4; -2 = -2 ; y = -2
1
7
12. x + 2y = 7; 2y = -x + 7; y = -2x + 2
x
y
1
3
1
7 6
y = -2 (1) + 2 = 2 = 3
3
2
1
7 4
y = -2 (3) + 2 = 2 = 2
5
1
1
7 2
y = -2 (5) + 2 = 2 = 1
y
x
13.
2y = 3x + 10
2 y 3 x 10


2
2
2
slope:
3
2
y
3x
5
2
y-intercept: 5 or (0 , 5)
4
Introductory Algebra Appeals Practice Test
y2 – y1 -3 – -1 -3 + 1 -2
14. m = x – x = 3 – 4 = -1 = -1 = 2
2
1
y = mx + b -1 = 2(4) + b
b = -9
the equation of the line is:
y = 2x – 9
15. y = mx + b
-3 = (-2)(-1) + b
-3 = 2 + b
b = -5
The equation of the line is: y = -2x – 5
16.
4x – (3x + 2) = 10
4x – 3x – 2 = 10
x = 12
17. 18 x 2 y 3  18 x 2
18.
Then y = 3(12) + 2 and so y = 38.
The solution is (12 , 38).
1 18 x 2
 3
y3
y
x + y = 20
30x + 16y = 530
small frames: 5
large frames: 15
1
19. (x -4)3 = x -12 = x12
20. 6 12 – 3 20 + 36
6 4 3 – 3 4 5 +6
6.2 3 – 3.2 5 +6
12 3 – 6 5 + 6
21. 0.00519 = 5.19x10-3
22. x(3x2 + 5x – 2) = x . 3x2 + x . 5x + x . (-2) = 3x3 + 5x2 – 2x
23. (3x – 5)(x + 2) = 3x2 – 5x + 6x – 10 = 3x2 + x – 10
5
Introductory Algebra Appeals Practice Test
24. (x – 3)(x2 + x + 3) = x3 + x2 + 3x – 3x2 – 3x – 9 = x3 – 2x2 – 9
25. 3x3 + 11x2 – 4x = x(3x2 + 11x – 4)
3(-4) = 12
You want two numbers whose product is –12 and whose sum is 11.
12 . -1 = -12 and 12 + -1 = 11
x(3x2 – x + 12x – 4)
x [ x(3x – 1) + 4(3x – 1) ]
x(3x – 1)(x + 4)
26. 3x2 + 4x – 4 = (3x – 2)(x + 2)
27. 4x2 – 9 = (2x + 3)(2x – 3)
28.
3x2 + x = 0
x(3x + 1) = 0
x = 0
3x + 1 = 0
3x + 1 – 1 = 0 – 1
3x = -1
3x -1
3 = 3
-1
x = 3
29.
2x2 – 3x – 5 = 0
2x2 – 5x + 2x – 5 = 0
x(2x – 5) + 1(2x – 5) = 0
(2x – 5)(x + 1) = 0
2x – 5 = 0
x+1=0
5
x=2
x = -1
30. x2 + 52 = 72
x2 + 25 = 49
6
Introductory Algebra Appeals Practice Test
x2 + 25 – 25 = 49 – 25
x2 = 24
x=
24 = 4.90
7