Question

A rectangle has a perimeter of 34 cm in an area of 52 cm squared it's length is 5 more than twice its width write and solve the system of equations to find the dimensions of the rectangle

Answer 1

Let $l$ be the length and $w$ be the width of the rectangle.

Given perimeter of the rectangle

$2(l+w)=34 \; l+w=17$

Area of the rectangle is

$lw=52$

Also length is 5 more than twice its width,

$l=2w+5$

From the above equations

$(l-w)^2=(l+w)^2-4lw=17^2-4*52=81\\ l-w=9,\; since \; l>w$

Now $2l=9+17=26\;,l=13,w=4$

The dimensions of the rectangle are length $l=13$, and width $w=4$.

Answer 2

L+W=17 W=4 is the answer.