Question 1102120
Let {{{ t }}} = time in hrs for 258 mi trip
{{{ 10 - t }}} = time in hrs for 208 mi trip
Let {{{ s }}} = speed in mi/hr for 258 mi trip
{{{ s + 9 }}} = speed in mi/hr for 208 mi trip
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(1) {{{ 258 = s*t }}}
(2) {{{ 208 = ( s + 9 )*( 10 - t ) }}}
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(2) {{{ 208 = 10s + 90 - s*t - 9t }}}
(2) {{{ 208 = 10s + 90 - t( s + 9 ) }}}
and
(1) {{{ t = 258/s }}}
Plug this into (2)
(2) {{{ 208 = 10s + 90 - ( 258/s )*( s + 9 ) }}}
(2) {{{ 118 = 10s - 258 - 2322/s }}}
(2) {{{ 376 =  10s - 2322/s }}}
(2) {{{ 376s = 10s^2 - 2322 }}}
(2) {{{ 10s^2 - 376s - 2322 = 0 }}}
(2) {{{ 5s^2 - 188s - 1161 = 0 }}}
Use quadratic formula
{{{ s = ( -b +-sqrt( b^2 - 4*a*c )) / ( 2a ) }}}
{{{ a = 5 }}}
{{{ b = -188 }}}
{{{ c = -1161 }}}
{{{ s = ( 188 +-sqrt( 35344 - 4*5*( -1161 ))) / 10 }}}
{{{ s = ( 188 +- sqrt( 3544 + 23220 )) / 10 }}}
{{{ s = ( 188 + sqrt( 26764 ) )/10 }}}
{{{ s = ( 188 + 163.6 ) / 10 }}}
{{{ s = 351.6/10 }}}
{{{ s = 35.16 }}}
and
{{{ s + 9 = 44.16 }}}
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He drove 35.16 mi/hr for 1st part and
44.16 mi/hr for 2nd part
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Get another opinion unless you know
what answer is. Check math too.