Question

The length of arectangle is 15 cm more than the width. A second rectangle whose perimeter is 72 cm is 5 cm wider but 2 cm shorter than the first rectrangle. What are the dimensions of reach rectangle?

Answer 1

L = W + 15

(2nd rectangle) 2L + 2W = 72

Substitution. 2(L-2) + 2(W+5) = 72

We can divide the 2

L - 2 + W + 5 = 36

L + W + 3 = 36

L + W = 33

From the first equation, (L = W + 15), we can substitute again

W + 15 + W = 33

2W = 18

W = 9

L = W + 15

L = 9 + 15 = 24

The dimensions of the first rectangle is 9 cm by 24 cm