Question 1018183
They are doing something a little bit illegal here
and you should be aware of it. They are saying
an area ( in square inches ) is equal to a perimeter
( in inches ). 
You can only do this if you say the VALUES are equal,
but you can't make an equation and say 
[ inches ] = [ square inches ]
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Let the length of the rectangle = {{{ L }}}
Let the width of the rectangle = {{{ W }}}
(1) {{{ L = 2W + 1 }}}
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Let the area of the rectangle  = {{{ A }}}
Let the perimeter of the rectangle  = {{{ P }}}
{{{ A = L*W }}}
and
{{{ P = 2L + 2W }}}
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They are saying:
{{{ A = P + 5 }}} ( even though the units don't match )
(2) {{{ L*W = 2L + 2W + 5 }}}
Substitute (1) into (2)
(2) {{{ ( 2W + 1 )*W = 2*( 2W + 1 ) + 2W + 5 }}}
(2) {{{ 2W^2 + W = 4W + 2 + 2W + 5 }}}
(2) {{{ 2W^2 + W = 6W +7 }}}
(2) {{{ 2W^2 - 5W - 7 = 0 }}}
Use the quadratic formula to solve
{{{ W = ( -b +- sqrt( b^2 - 4*a*c )) / (2*a) }}} 
{{{ a = 2 }}}
{{{ b = -5 }}}
{{{ c = -7 }}}
{{{ W = ( -(-5) +- sqrt( (-5)^2 - 4*2*(-7) )) / (2*2) }}} 
{{{ W = ( 5 +- sqrt( 25 + 56 )) /4 }}} 
{{{ W = ( 5 +- sqrt( 81 )) /4 }}} 
{{{ W = 5/4 + 9/4 }}}
{{{ W = 14/4 }}}
{{{ W = 7/2 }}} ( note that I can't use the negative square root )
The width is 3.5 in
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check:
(1) {{{ L = 2W + 1 }}}
(1) {{{ L = 2*(7/2) + 1 }}}
(1) {{{ L - 8 }}}
and
(2) {{{ L*W = 2L + 2W + 5 }}}
(2) {{{ L*(7/2) = 2L + 2*(7/2) + 5 }}}
(2) {{{ (7/2)*L = 2L + 12 }}}
(2) {{{ (7/2)*L - (4/2)*L = 12 }}}
(2) {{{ (3/2)*L = 12 }}}
(2) {{{ L = (2/3)*12 }}}
(2) {{{ L = 8 }}}
OK