Question 164604
Let call the first square side a and the second square side b.
The perimeters are,
{{{P[a]=4a}}}
{{{P[b]=4b}}}
and the areas are,
{{{A[a]=a^2}}}
{{{A[b]=b^2}}}
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.
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"perimeter of one square exceeds that of another by 16" 
{{{P[a]=P[b]+16}}}
1.{{{4a=4b+16}}}
its area is 44 less than 4 times the area of the other
{{{A[a]=4A[b]-44}}}
2.{{{a^2=4b^2-44}}}
Use eq. 1 to come up with an expression for a in terms of b.
1.{{{4a=4b+16}}}
{{{a=b+4}}}
Now substitute that expression for a in eq. 2 and solve for b.
2.{{{a^2=4b^2-44}}}
{{{(b+4)^2=4b^2-44}}}
{{{b^2+8b+16=4b^2-44}}}
{{{3b^2-8b-60=0}}}
{{{(3b+10)(b-6)=0}}}
We'll only solve for the positive b solution since a negative length doesn't make sense.
{{{b-6=0}}}
{{{highlight(b=6)}}}
From eq. 1,
{{{a=b+4}}}
{{{a=6+4}}}
{{{highlight(a=10)}}}
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{{{P[a]=4a=40}}}
{{{P[b]=4b=24}}}
{{{A[a]=a^2=100}}}
{{{A[b]=b^2=36}}}