Question 1035397
Let {{{ r }}} = rate in dollars/hr for the job
paying less per hour 
{{{ r + 2.25 }}} = rate in dollars/hr for he job
paying more per hour
Let {{{ t }}} = time to earn $900 at lower paying job
{{{ t - 10 }}} = time to earn $1,000 at higher paying job
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(1) {{{ 1000 = ( r + 2.25 )*( t - 10 ) }}}
(2) {{{ 900 = r*t }}}
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(2) {{{ t = 900/r }}}
and
(1) {{{ 1000 = r*t + 2.25t - 10r - 22.5 }}}
(1) {{{ 1000 = t*( r + 2.25 ) - 10r - 22.5 }}}
(1) {{{ 1000 = ( 900/r )*( r + 2.25 ) - 10r - 22.5 }}}
(1) {{{ 1000 = 900 +  2025/r - 10r - 22.5 }}}
(1) {{{ 122.5 = 2025/r - 10r }}}
Multiply both sides by {{{ r }}}
(1) {{{ 122.5r = 2025 - 10r^2 }}}
(1) {{{ 10r^2 + 122.5r - 2025 = 0 }}}
(1) {{{ 100r^2 + 1225r - 20250 = 0 }}}
(1) {{{ 4r^2 + 49r - 810 = 0 }}}
Solve with quadratic formula
{{{ r = ( -b +- sqrt( b^2 - 4*a*c )) / (2*a) }}} 
{{{ a = 4 }}}
{{{ b = 49 }}}
{{{ c = -810 }}}
{{{ r = ( -49 +- sqrt( 49^2 - 4*4*(-810) )) / (2*4) }}}  
{{{ r = ( -49 +- sqrt( 2401 + 12960 )) / 8 }}}  
{{{ r = ( -49 +- sqrt( 15361 )) / 8 }}}  
{{{ r = ( -49 + 123.94 ) / 8 }}}
{{{ r = 74.94 / 8 }}}
{{{ r = 9.37 }}}
and
{{{ r + 2.25 = 20.99 }}}
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Lower paying rate is $9.37 / hr
Higher paying rate is $20.99 / hr
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check answers:
(2) {{{ 900 = r*t }}}
(2) {{{ 900 = 9.37*t }}}
(2) {{{ t = 96.05 }}}
and
(1) {{{ 1000 = ( r + 2.25 )*( t - 10 ) }}}
(1) {{{ 1000 = ( 9.37 + 2.25 )*( t - 10 ) }}}
(1) {{{ 1000 = 11.62*( t - 10 ) }}}
(1) {{{ 1000 = 11.62t - 116.2 }}}
(1) {{{ 1116.2 = 11.62t }}}
(1) {{{ t = 96.06 }}}
close enough -hope I got it!