Question

dimension of a rectangle are such that its length is 11in. more than width. if the length were doubled and if the width were decreased by 5 in. the area would be increased by 130 in.^2 what r the length and width of the rectangle

Answer 1

put w = width, then
\[l=w+11\] so initial area is
\[w(w+11)\]
if you double the length you get
\[2l=2(w+11)\] degrease the width by 5 you get
\[w-5\] and the area of the changed figure is
\[2(w+11)(w-5)\] the second area is 130 larger than the first, so we can solve
\[w(w+11)+130=2(w+11)(w-5)\]

Answer 2

hard part was getting an equation to solve, it should be good from there. fairly sure this is correct because i think the problem was cooked to give a nice integer solution

Answer 3

thank you