Page 55 - Math Course 1 (Book 2)
P. 55
Solving Linear Equations
No Solutions or Identity Standardized Test Example
1 8
Solve (21c – 56) = 3 (c – ) Find the value of x so that the f gures have the
7 3 same area.
A. true for all values of c
A. 1
B. no solution B. 2
C. 3
C. 1 D. 4
3
D. 0
Answer Answer
Skill Practice!
Justify each step.
1. 4k – 3 = 2k + 5 2. 2(8u + 2) = 3(2u – 7)
4k – 3 – 2k = 2k + 5 – 2k a.
2k – 3 = 5 b. 16u + 4 = 6u – 21 a.
2k – 3 + 3 = 5 + 3 c.
2k = 8 d. 16u + 4 – 6u = 6u – 21 – 6u b.
e.
2k 8 10u + 4 = –21 c.
=
2 2
k = 4 f. 10u + 4 – 4 = –21 – 4 d.
10u = –25 e.
10u –25
= f.
10 10
u = –2.5 g.
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