Question 798724
Let x = width
Let y = length
Given used, {{{xy=39}}};
{{{2y=-5+3x}}}.

Rectangle Fact justifies {{{2x+2y=perimeter}}}


You first want to solve for x and y.  You can find perimeter when you know x and y.



SOLUTION:
More than one way to do this.  You'll solve for one variable from one of the "given" described equations, substitute this into the other equation, and solve for the single variable; then use the value to solve for the other variable.


A SOLUTION CHOICE:
{{{xy=39}}}
{{{y=39/x}}}.
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{{{2*(39/x)=-5+3x}}}, substituting into the other equation...
{{{78=-5x+3x^2}}}
{{{3x^2-5x-78=0}}}, QUADRATIC EQUATION
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Discriminant, {{{5^2-4*3(-78)=25+12*78=961}}}, to be neat, this would be a square value.
... LUCK!  {{{sqrt(961)=31}}}.
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Directly using general solution to a quadratic equation:
{{{highlight(x=(5+- sqrt(discriminant))/(2*3))}}}
We KNOW we want the positive value, and NOT the negative value.
{{{highlight(x=(5+31)/(2*3))}}}
{{{highlight(x=6)}}} and so {{{y=39/6=highlight(6&1/2)}}}


Perimeter is simply {{{2(6+6&1/2)=13}}} feet.