Question

the dimensions of a rectangle are such that its length is 5 in more than its width. If the length were doubled and the width were decreases by 2 in, the area would be increased 136 in^2,what are the length and width of the rectangle

Answer 1

Initially
w=width
w+5=length
w(w+5)=original area

Subsequently:
w-2=width
2(w+5)=length
(w-2)*2(w+5)=2*original area=w(w+5)=area

Therefore
2(w-2)(w+5)=w(w+5)

Solve for w:
2(w-2)=w (w cannot be equal to -5)
2w-4=w
w=4

Answer 2

original width -- x
original length -- x+5
area = x(x+5)

new width = x-2
new length = 2x+10
new area = (2x+10)(x-2)

(2x+10)(x-2) - x(x+5) = 136
2x^2 + 6x - 20 - x^2 - 5x - 136 = 0
x^2 + x - 156 = 0
(x-12)(x+13) = 0
x = 12 or x = -13

ignoring the negative,

<b>width = 12
length = 17</b>

check:
original area = 12(17) = 204

new width = 10
new length = 34
new area = 10(34) = 340

increase in area = 340-204 = 136 , YEahh!

Answer 3

Good catch!
Thank you!