Question 956226
The dimensions of a retangle are such that its length is 3 inches more than its width . IF the length were doubled and if the width were decreased by 1 inch the area would be increased by 66iniches^2 what are the length and width
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let x=width of original rectangle
x+3=length of original rectangle
x(x+3)=area of original rectangle
..
x-1=width of larger rectangle
2(x+3)=length of larger rectangle
(x-1)(2(x+3))=area of  larger rectangle
..
area of larger rectangle-area of original rectangle=66 inches^2
2(x-1)((x+3))-x(x+3)=66
2(x^2+2x-3)-x^2-3x=66
2x^2+4x-6-x^2-3x=66
x^2+x-72=0
(x+9)(x-8)=0
x=8
x+3=11
width of original rectangle=8 inches
length of original rectangle=11 inches