Question 157086
Find the dimensions of a rectangle whose perimeter is 26 meter and whose area is 40 square meters.

Solution:

let length=L
let width=w

Area A=LxW=40 .....EQ 1
Perimeter P=2(L+W)=26
L+W=13.............EQ 2

substituting the value of W from EQ 2 into EQ 1 we have
LX(13-L)=40
13L-L^2=40
REARRANGING
L^2-13L+40=0
BREAKING THE MIDDLE TERM
L^2-5L-8L+40=0
L(L-5)-8(L-5)=0
(L-5)(L-8)=0
THIS IMPLIES
L=5 0R L=8
W=8 0R W=5