Question 173612
Let {{{l}}}= length
Let {{{w}}}= width
Let {{{P}}}= perimeter
{{{P = l + l + w + w}}}
{{{P = 2l + 2w}}}
Given is {{{P = 62}}}
(1) {{{62 = 2l + 2w}}}
The problem says if I replace 
{{{l}}} with {{{l + 21}}}, and
{{{w}}} with {{{2w}}}, then
{{{P = 120}}}
{{{120 = 2*(l + 21) + 2*2w}}}
(2) {{{120 = 2l + 42 + 4w}}}
Subtract (1) from (2)
{{{58 = 2w + 42}}}
{{{2w = 16}}}
{{{w = 8}}}
Plug this result back into (1)
{{{62 = 2l + 2*8}}}
{{{2l = 62 - 16}}}
{{{2l = 46}}}
{{{l = 23}}}
The length is 23 m and the width is 8 m
check:
{{{120 = 2l + 42 + 4w}}}
{{{120 = 2*23 + 42 + 4*8}}}
{{{120 = 46 + 42 + 32}}}
{{{120 = 120}}}
OK
{{{62 = 2l + 2w}}}
{{{62 = 2*23 + 2*8}}}
{{{62 = 46 + 16}}}
{{{62 = 62}}}
OK