SOLUTION: The area of a rectangle is 55 yd^2 , and the length of the rectangle is 4yd less than three times the width, what is the length & width

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Question 862818: The area of a rectangle is 55 yd^2 , and the length of the rectangle is 4yd less than three times the width, what is the length & width
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
Simple algebra applied to rectangle.

A = 55 yd^2
L = length
w = width
k = 4 yd
m = the factor, 3
-

L=-k%2Bmw.
A=wL.
Substitute for L in the A equation.
w%28mw-k%29=A
mw%5E2-kw=A
highlight_green%28mw%5E2-kw-A=0%29

WHAT NEXT?
Use the general solution of a quadratic equation to solve for w. Substitute the given values for k, A, and m and compute w. Use the now found value of w to compute L.