4.02 Review: Two-step equations
Determine whether or not the given value of x is a solution to the following equations:
4 x - 2 = 18 where x = 5
6 x - 13 = 12 where x = 4
5 x + 12 = 31 where x = 4
3 x + 1 = 7 where x = 2
Solve:
7 p + 6 = 41
7 x + 17 = 45
4 x + 9 = 25
2 t + 13 = 35
7 m + 18 = 39
12 p + 3 = 51
2x + 14 = 20
6y + 4 = 34
3q + 9 = 12
-9 n + 8 = -55
- 5 t + 27 = 17
8u + 12 = -12
11x + 38 = 27
7y + 38 = 3
2 w + 9 = 37
p + 3.6 = 6.2
-r + 2.3 = -1.8
Solve:
13 x - 2 = 24
5 t - 15 = 15
29 m - 8 = 21
11 x - 22 = 22
2p - 17 = 3
9q - 4 = 23
6n - 29 = 7
-7s - 23 = 19
12m - 7 = -19
-4x - 29 = -13
Solve:
8 m + 9 = 65
8 x - 9 = 39
4 x - 8 = 0
5 x - 8 = 2
4 x + 36 = 40
7 x - 63 = 49
- 10 + 3 k = 5
- x - 7 = 7
- x + 5 = - 1
7 x + 14 = 0
- 63 - 9 y = 63
Solve:
\dfrac{x}{3} + 2 = 7
\dfrac{q}{2} + 18 = 25
\dfrac{s}{7} + 14 = 5
\dfrac{m}{5} + 13 = 22
\dfrac{p}{11} - 3 = 3
\dfrac{n}{8} - 5 = 4
\dfrac{p}{13} - 11 = -7
\dfrac{x}{12} + 2 = -6
Solve:
\dfrac{47 + m}{15} = 3
\dfrac{m+26}{17} = 3
\dfrac{t-5}{4} = 8
\dfrac{q-17}{3} = -4
\dfrac{m - 6}{4} = 4.25
Solve:
\dfrac{c - 4}{4} = 7
\dfrac{4}{9} \times n = 10
- 6 = \dfrac{y}{- 7}
\dfrac{x}{2} - 3 = 1
\dfrac{x}{2} + 8 = 10
\dfrac{x + 9}{7} = 4
\dfrac{x}{7} + 3 = - 1
- \dfrac{1}{8} x = 6
- 5 = \dfrac{x}{5}
\dfrac{1}{8} x = - 8
- \dfrac{1}{8} x = - 8
Solve:
24 + p = - 2
11 - \dfrac{x}{6} = 9
14 = 3 r - 7
\dfrac{-2 + n}{6} = -15
17 - 8 s = -7
10 = 2x + 6
-15 -5x = 10
11 = \dfrac{x+1}{2}