algebra
posted by Anonymous .
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

Initially
w=width
w+5=length
w(w+5)=original area
Subsequently:
w2=width
2(w+5)=length
(w2)*2(w+5)=2*original area=w(w+5)=area
Therefore
2(w2)(w+5)=w(w+5)
Solve for w:
2(w2)=w (w cannot be equal to 5)
2w4=w
w=4 
original width  x
original length  x+5
area = x(x+5)
new width = x2
new length = 2x+10
new area = (2x+10)(x2)
(2x+10)(x2)  x(x+5) = 136
2x^2 + 6x  20  x^2  5x  136 = 0
x^2 + x  156 = 0
(x12)(x+13) = 0
x = 12 or x = 13
ignoring the negative,
width = 12
length = 17
check:
original area = 12(17) = 204
new width = 10
new length = 34
new area = 10(34) = 340
increase in area = 340204 = 136 , YEahh! 
Good catch!
Thank you!