SOLUTION: The length of a rectangle is 9.7cm more than 4 times the width. If the perimeter of the rectangle is 91.4cm what are the dimensions?

Question 655232: The length of a rectangle is 9.7cm more than 4 times the width. If the perimeter of the rectangle is 91.4cm what are the dimensions?
Found 2 solutions by checkley79, floresn:
Answer by checkley79(3341) (Show Source): You can put this solution on YOUR website!
L=4W+9.7
P=2L+2W
91.4=2(4W+9.7)+2W
91.4=8W+19.4+2W
91.4=10W+19.4
10W=91.4-19.4
10W=72
W=72/10
W=7.2 ANS. FOR THE WIDTH.
L=4*7.2+9.7
L=28+9.7
L=37.7 ANS FOR THE LENGTH.
PROOF
91.4=2*37.7+2*8
91.4=75.4+16
91.4=91.4

Answer by floresn(1) (Show Source): You can put this solution on YOUR website!
For perimeter of this rectangle, the equation is
P = 2L + 2W
where L = 4W + 9.7
and P = 91.4, therefore...
91.4 = 2(4W + 9.7) + 2W
91.4 = 8W + 19.4 + 2W
91.4 = 10W + 19.4, and in an effort to get "W" by itself, you subtract 19.4 from both sides of the equation, arriving at...
10W = 91.4 - 19.4
10W = 72
W = 7.2
If W = 7.2cm, then taking this and placing it in the equation L = 4W + 9.7, you get...
L = 4*(7.2) + 9.7
L = 28.8 + 9.7
L = 38.5
Proof:
So, 2L + 2W = P and...
2*(38.5) + 2*(7.2) = (77) + (14.4) = 91.4

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