Question 860476
If the room measures {{{x}}} meters by {{{y}}} meters,
its perimeter (in meters) would be
{{{2(x+y)=48}}} ,
and its surface area (in square meters) would be
{{{x*y=128}}} .
 
{{{2(x+y)=48}}} <---> {{{x+y=48/2}}} <---> {{{x+y=24}}}
Since {{{128=2^7=16*8}}} and {{{16+8=24}}} ,
it is obvious that the room could be {{{highlight(16)}}} meters long by {{{highlight(8)}}} meters wide.
 
The teacher may expect that you solve {{{x+y=24}}} for {{{y}}} ,
{{{x+y=24}}} <---> {{{y=24-x}}} ,
and then substitute into {{{x*y=128}}} ,
to get a quadratic equation:
{{{x*(24-x)=128}}} --> {{{24x-x^2=128}}}  --> {{{x^2-24x+128=0}}}
You can solve  {{{x^2-24x+128=0}}} to get the two solutions
{{{x=8}}}-->{{{y=24-8=16}}} ,
and
{{{x=16}}}-->{{{y=24-16=8}}} ,
which tell you that the dimensions of that rectangular room
are {{{highlight(16)}}} meters (length) and {{{highlight(8)}}} meters (width).