Question 762860
Call the rectangles r1 and r2
Call the breadth of r1 {{{ x }}}
Call the length of r1 {{{ y }}}
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The breadth of r2 = {{{ x + 4 }}}
The length of r2 = {{{  y - 5 }}}
( if I made the signs the same, the area of r2
would be greater than the area of r1 )
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(1) {{{ 120 = x*y }}}
(2) {{{ 120 = ( x+4 )*( y -5 ) }}}
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(2) {{{ x*y + 4y - 5x - 20 = 120 }}}
(2) {{{ x*y + 4y - 5x = 140 }}}
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(1) {{{ y = 120/x }}}
By substitution:
(2) {{{ x*(120/x) + 4*(120/x) - 5x = 140 }}}
(2) {{{ 120 + 4*(120/x) - 5x = 140 }}}
(2) {{{ 4*(120/x) - 5x = 20 }}}
(2) {{{ 480/x - 5x = 20 }}}
(2) {{{ 480 - 5x^2 = 20x }}}
(2) {{{ 5x^2 + 20x = 480 }}}
(2) {{{ x^2 + 4x = 96 }}}
Complete the square
(2) {{{ x^2 + 4x + (4/2)^2 = 96 + (4/2)^2 }}}
(2) {{{ x^2 + 4x + 4 = 96 + 4 }}}
(2) {{{ ( x + 2 )^2 = 100 }}}
(2) {{{ (x + 2 )^2 = 10^2 }}}
(2) {{{ x + 2 = 10 }}}
(2) {{{ x = 8 }}}
and, since
(1) {{{ 120 = x*y }}}
(1) {{{ y = 120/8 }}}
(1) {{{ y = 15 }}}
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And, also,
(2) {{{ 120 = ( x+4 )*( y -5 ) }}}
{{{ x + 4 = 12 }}}
{{{ y - 5 = 10 }}}
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Rectangle r1 has dimensions 8 and 15
Rectangle r2 has dimensions 12 and 20
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check:
{{{ A1 = 8*15 }}}
{{{ A1 = 120 }}}
and
{{{ A2 = 12*10 }}}
{{{ A2 = 120 }}}
OK