Question 869356
Assign variables.
s = length of square
x = rectangle length
y = rectangle width
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Find equations and expressions.
{{{4s=160}}}.
{{{2(x+y)=160}}}.
Area of square, {{{s^2}}}.
Area of rectangle, {{{xy}}}.
{{{s^2-xy=100}}}.


SYSTEM OF EQUATIONS:
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4s=160
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2x+2y=160
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s^2-xy=100
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The question is to find length of the rectangle.  No distinction is necessary about x or y for length; either one is length or width.  First, solve for s and obtain two equations in two unknowns.


{{{s=160/4}}}, {{{s=40}}}.
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{{{2x+2y=160}}}
{{{x+y=80}}} simplified perimeter equation for rectangle.
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{{{-xy=100-s^2}}}
{{{xy=s^2-100}}}
{{{xy=40^2-100}}}
{{{xy=1600-100}}}
{{{xy=1500}}} eliminated variable s and simplified area equation.


SIMPLIFIED SYSTEM:
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{{{highlight_green(xy=1500)}}} and {{{highlight_green(x+y=80)}}}.


Solve the simplified system.
{{{y=80-x}}}.
{{{x(80-x)=1500}}}
{{{80x-x^2-1500=0}}}
{{{x^2-80x+1500=0}}}
discriminant, {{{(80)^2-4*1500=6400-6000=400}}}.
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General solution for a quadratic equation:
{{{x=(80-sqrt(400))/2}}}
{{{x=(80-20)/2}}}
{{{highlight(x=30)}}}
OR
{{{x=(80+20)/2}}}
{{{highlight(x=50)}}}


Length of the rectangle can be either 30 OR 50.