Problema Solution
A rectangle's length is one-and-a-half times its width. The length is increased by 4 inches and its width by 3 inches. The resulting area is 97 square inches more than the original rectangle. What were the original dimensions?
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let
l = the original length of the rectangle
w = the original width of the rectangle
A = l*w is the area of the rectangle
length is one-and-a-half times its width
l = 1.5*w
length is increased by 4 inches: l + 4
width by 3 inches: w + 3
resulting area: (l + 4)(w + 3)
resulting area is 97 square inches more than the original rectangle
(l + 4)(w + 3) = 97 + lw
3l + 4w + lw + 12 = 97 + lw
3l + 4w = 97 - 12 + lw - lw
3l + 4w = 85
by solving the system of equations
l = 1.5*w
3l + 4w = 85
we find
l = 15 in
w = 10 in
click here to see the step by step solution of the system of equations
the original dimensions of the rectangle are 15 inches length and 10 inches width.