Question 859489
x for length, and y for width.


Original Rectangle
{{{x=y+4}}}; {{{xy=A}}} which is {{{(y+4)y=A}}}.
This area simplifies to {{{highlight_green(y^2+4y=A)}}}.


Changed Rectangle
{{{(x+8)(y-4)=A}}}; substituting as for the variables assigned, {{{((y+4)+8)(y-4)=A}}}.  
This area simplies:
{{{(y+12)(y-4)}}}
{{{highlight_green(y^2+8y-48=A)}}}.


A used in both rectangles because area is specified as equal for both.
This equality means {{{highlight(y^2+4y=y^2+8y-48)}}}.
Do the rest.  This equation becomes linear in y.... See how?