Algebra Rules Review Sheet Page 12

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12
REVIEW OF ALGEBRA
EXAMPLE 24
Solve
3x
2
4
.
SOLUTION
By Properties 4 and 6 of absolute values,
3x
2
4
is equivalent to
3x
2
4
or
3x
2
4
2
In the first case,
3x
2
, which gives
x
. In the second case,
3x
6
, which gives
3
x
2
. So the solution set is
[
)
2
2
x x
2 or x
,
2
,
3
3
EXERCISES
3
2
2
39.
40.
t
1
4t
9s
Click here for answers.
Click here for solutions.
A
S
2
3
41.
4t
12t
9
42.
x
27
1–16
Expand and simplify.
3
2
3
2
43.
44.
x
2x
x
x
4x
5x
2
2
4
1.
6ab 0.5ac
2.
2x
y
xy
3
2
3
2
45.
46.
x
3x
x
3
x
2x
23x
60
3.
4.
2x x
5
4
3x x
3
2
3
2
47.
48.
x
5x
2x
24
x
3x
4x
12
5.
6.
2 4
3a
8
4
x
2
2
7.
4 x
x
2
5 x
2x
1
49–54
Simplify the expression.
2
8.
5 3t
4
t
2
2t t
3
2
2
x
x
2
2x
3x
2
49.
50.
2
2
9.
10.
x
3x
2
x
4
4x
1 3x
7
x x
1 x
2
2
3
2
x
1
x
5x
6x
2
2
11.
12.
2x
1
2
3x
51.
52.
2
2
x
9x
8
x
x
12
4
13.
y
6
y 5
y
1
1
53.
2
14.
t
5
2 t
3 8t
1
2
x
3
x
9
2 2
2
15.
16.
1
2x x
3x
1
1
x
x
x
2
54.
2
2
x
x
2
x
5x
4
17–28
Perform the indicated operations and simplify.
55–60
Complete the square.
2
8x
9b
6
17.
18.
2
3b
2
2
55.
56.
x
2x
5
x
16x
80
1
2
1
1
2
2
57.
58.
x
5x
10
x
3x
1
19.
20.
x
5
x
3
x
1
x
1
2
2
59.
60.
4x
4x
2
3x
24x
50
u
2
3
4
21.
22.
u
1
2
2
u
1
a
ab
b
61–68
Solve the equation.
x y
x
23.
24.
z
y z
2
2
61.
62.
x
9x
10
0
x
2x
8
0
2
2r
s
a
b
2
2
63.
64.
x
9x
1
0
x
2x
7
0
25.
26.
s
6t
bc
ac
65.
2
66.
2
3x
5x
1
0
2x
7x
2
0
1
3
3
2
67.
68.
1
x
2x
1
0
x
3x
x
1
0
c
1
1
27.
28.
1
1
1
1
1
69–72
Which of the quadratics are irreducible?
c
1
1
x
2
2
69.
70.
2x
3x
4
2x
9x
4
29–48
Factor the expression.
2
2
71.
72.
3x
x
6
x
3x
6
29.
3
30.
2x
12x
5ab
8abc
2
2
31.
32.
73–76
x
7x
6
x
x
6
Use the Binomial Theorem to expand the expression.
2
2
33.
34.
x
2x
8
2x
7x
4
6
7
73.
74.
a
b
a
b
2
2
35.
36.
9x
36
8x
10x
3
75.
2
4
76.
2 5
x
1
3
x
2
2
37.
38.
6x
5x
6
x
10x
25

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