SOLUTION: The dimensions of a rectangle are such that its length is
9
in. more than its width. If the length were doubled and if the width were decreased by
4
in., the area would be
Algebra.Com
Question 903644: The dimensions of a rectangle are such that its length is
9
in. more than its width. If the length were doubled and if the width were decreased by
4
in., the area would be increased by
168
in.2.
What are the length and width of the rectangle?
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
The dimensions of a rectangle are such that its length is
9in. more than its width. If the length were doubled and if the width were decreased by 4in., the area would be increased by 168 in^2.What are the length and width of the rectangle?
***
let x=original width of rectangle
x+9=original length of rectangle
2(x+9)=new length of rectangle
x-4=new width of rectangle
area=length*width
..
2(x+9)(x-4)-x(x+9)=168
2(x^2+5x-36)-x^2-9x=168
2x^2+10x-72-x^2-9x=168
x^2+x-240=0
(x+16)(x-15)=0
x=-16 (reject)
or
x=15
x+9=24
width of rectangle=15 in
length of rectangle=24 in
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