Question

5. Three times the width of a certain rectangle exceeds twice its length by two inches. Four times its length is twelve more

than its perimeter. Write a system of equations that could be used to solve this problem. (hint: P = 2L + 2W)
A. 3W = 2L + 2
2L = 2W + 12

B. 3W + 2 = 2L
4L = P - 12

C.3W = 2L + 2
4L + 12 =P

D. 2W + 2 = 2L
4L = 12 + P

E. 3W + 2 = 2L
4L = 12 + P

F. 2L - 2 = 3W
P = 4L - 12
plsss help

Answer 1

Answer: A

Explanation:

L=Length W=Width

3W=2L+2

4L=2L+2W+12

2L=2W+12

Option A

Hope this helps!! :)

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