Question 946751
ASSIGN VARIABLES
w, width of rectangle
L, length of rectangle
p, perimeter of rectangle
A, area of rectangle
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A=102
w=p-40
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UNKNOWNS:  w and L.


{{{system(w=p-40,2w+2L=p,wL=A)}}}


Note that although p is unknown, it can be eliminated using the w equation
and the p equation.


{{{w=p-40}}}
{{{w+40=p}}}
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Equate the two expressions for p;
{{{w+40=2w+2L}}}
Simplify
{{{40=w+2L}}}
{{{w+2L=40}}}


The simpler equivalent system without p is
{{{system(w+2L=40,wL=A)}}}


Substitute for w in the area A equation:
{{{(40-2L)L=A}}}
simplify
{{{40L-2L^2-A=0}}}
{{{-40L+2L^2+A=0}}}
{{{highlight_green(2L^2-40L+A=0)}}}


Substituting the given value now for A would be convenient.
{{{2L^2-40L+102=0}}}
{{{L^2-20L+51=0}}}
FACTORABLE.
{{{(L-3)(L-17)=0}}}
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EITHER {{{highlight(L=3)}}}  OR  {{{highlight(L=17)}}}.


Find width, w, using the earlier found formula from the process,
w=40-2L:
EITHER {{{highlight(w=34)}}}  OR  {{{highlight(w=6)}}}



SUMMARY OF SOLUTIONS
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The dimensions are either   3 & 34, or  17 & 6.
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