Question 981472
Let {{{ s }}} = his speed going back home in mi/hr
{{{ s + 9 }}} = his speed going away from home in mi/hr
Let {{{ t }}} = his time going back home in hrs
{{{ 8 - t }}} = his time in hrs going away from home
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Equation for going away from home:
(1) {{{ 160 = ( s + 9 )*( 8 - t ) }}}
Equation for going back home:
(2) {{{ 160 = s*t }}}
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(1) {{{ 160 = 8s + 72 - s*t - 9t }}}
and
(2) {{{ s = 160/t }}}
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Substitute (2) into (1)
(1) {{{ 160 = 8*( 160/t ) + 72 - ( 160/t )*t - 9t }}}
Multiply both sides by {{{ t }}}
(1) {{{ 160t = 1280 + 72t - 160t + 9t^2 }}}
(1) {{{ 9t^2 - 248t + 1280 = 0 }}}
Use quadratic equation
{{{ t = ( -b +- sqrt( b^2 - 4*a*c )) / (2*a) }}}
{{{ a = 9 }}}
{{{ b = -248 }}}
{{{ c = 1280 }}}
{{{ t = ( -(-248) +- sqrt( (-248)^2 - 4*9*1280 )) / (2*9) }}}
{{{ t = ( 248 +- sqrt( 61504 - 46080 )) / 18 }}}
{{{ t = ( 248 +- sqrt( 15424 )) / 18 }}}
You can finish- find {{{ t }}}, then {{{ s }}}
check my math -I'm not sure of calculations
I think the method is OK