Question

The perimeter of a rectangle is 52 m. If the width were doubled and the length were increased by 17 m, the perimeter would be 96 m. What is the length of the original rectangle?

Answer 1

Let $w,h$ be the dimensions of the rectangle. If the original perimeter is 52, we deduce that

$2(w+h)=52 \iff w+h=26$

Doubling the width and increasing the length by 17 leads to a new perimeter of

$96=2(2w+(l+17))=4w+2l+34 \iff 4w+2l=62$

From the first equation, we know that (for example) $w=26-l$

Plug this in the second equation to get

$4(26-l)+2l=62 \iff 104-4l+2l=62 \iff -2l=-42 \iff l=21$

So, the original length was 21, which implies that the original width was

$w=26-l=26-21=5$