rectangle area= 99inches, length is 5 times more than 2/3 the width, what are the dimensions

Question

I can't figure how to solve for the dimesions if I know the area and that the length is 5 more than 2/3 the width. I keep setting it equal to 99 and trying to solve for it and it's not making any sense to me.

Answer 1

If the question is length is 5" more than 2/3 width, then length=2w/3+5 and area is w(2w/3+5)=99 sq in. Multiply through by 3 to get rid of fraction: w(2w+15)=297; 2w^2+15w-297=0=(w-9)(2w+33), so w=9 (the other value of w is negative and we can't have a negative length). So length=2/3w+5=6+5=11". Check 9*11=99.