Question 1022258
{{{w=(3/5)L}}}


Changes to dimensions:
Width to w+2, length to L-2;
{{{w+2=L-2}}}


Solve the system {{{system(w=(3/5)L,w+2=L-2)}}}.


{{{w-L=-4}}}
{{{L-w=4}}}
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{{{5w=3L}}}
{{{3L-5w=0}}}
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New system to have the same solution can be  {{{system(L-w=4,3L-5w=0)}}}.


From that ,  {{{system(3L-3w=12,3L-5w=0)}}}
{{{3L-3w-3L-(-5w)=12-0}}}
{{{2w=12}}}
{{{highlight(w=6)}}} and {{{L=(5/3)w=(5/3)*6=highlight(10=L)}}}.



Full use of the substitution method would have also been good.  I arranged the two system equation to start with elimination method.