Question 956760
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If the difference in area between 2 squares is 17 square inches and the sum of the perimeters of the 2 squares is 68 inches, how long is a side of the larger square?
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{{{a[S]}}}=side of small square; {{{a[L]}}}=side of large square; {{{P[B]}}}= Sum of perimeter of both squares
{{{4a[S]+4a[L]=68in}}}
{{{4(a[L]+a[S])=68in}}} Divide each side by 4.
{{{a[L]+a[S]=17in}}} Subtract {{{a[S]}}} from each side.
{{{a[L]=17in-a[S]}}} Use this to substitute.
{{{a[L]^2-a[S]^2=17in^2}}} Substitute for {{{a[L]}}}.
{{{(17in-a[S])^2-a[S]^2=17in^2}}}
{{{a[S]^2-34a[S]+289-a[S]^2=17in^2}}} Subtract 289 from each side.
{{{-34a[S]=-272}}} Divide each side by -34.
{{{a[S]=8}}} The side of the shorter square is 8 inches.
{{{4a[S]+4a[L]=68in}}}
{{{4(8in)+4a[L]=68in}}} 
{{{32in+4a[L]=68in Subtract 32in from each side.
{{{4a[L]=36in}}} Divide each side by 4.
{{{a[L]=9in}}} ANSWER: The side of the larger square is 9 inches
CHECK:
Difference in area is 17 square inches.
{{{4a[L]^2-4a[S]^2=17in^2}}}
{{{4(9^2in^2)-4(8^2in^2)=17in^2}}}
{{{81in^2-64in^2=17in^2}}}
{{{17in^2=17in^2}}}