SOLUTION: If the numerical value of the area of a rectangle is 19 more than 2 times the perimeter, and the length is 2 more than the width, what is the width?
Area = L * W
2x + 19
w
Algebra.Com
Question 395421: If the numerical value of the area of a rectangle is 19 more than 2 times the perimeter, and the length is 2 more than the width, what is the width?
Area = L * W
2x + 19
w = X
l = X + 2
X(X + 2) = 2x + 19 ?
x^2 + 2x = 2x + 19 + x + 2
Answer by jrfrunner(365) (Show Source): You can put this solution on YOUR website!
known: Area=L*W Perimeter =2*(L+W)
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given : A=2P+19 or A=4*(L+W)+19 and L=2+W
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set both expression of area equal to each other
L*W=4L+4W+19
(2+W)*W=4*(2+W)+4W+19 substitute L=2+W
2W+W^2=8+8W+19
W^2-6W-27=0
(W-9)(W+3)=0
so W=9 or W=-3 measurements must be positive so eliminatte W=-3
solution, W=9
--
L=2+W=2+9+11
==
validate the answer by substituting back into Area equations
A=L*W=11*9=99
A=4*(11+9)+19=80+19=99 check
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