Question 682925
Let {{{ s }}} = her speed in mi/hr in the city
{{{ s + 20 }}} = her speed in mi/hr on the highway
Let {{{ t }}} = time in hrs she drove in the city
{{{ 4 - t }}} = time in hrs she drove on the highway
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Harriet's equation for the city:
(1) {{{ 90 = s*t }}}
Harriet's equation for the highway:
(2) {{{ 130 = ( s + 20 )*( 4 - t ) }}}
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(1) {{{ t = 90/s }}}
and
(2) {{{ 130 = 4s + 80 - s*t - 20t }}}
(2) {{{ 130 = 4s + 80 - s*( 90/s ) - 20*( 90/s ) }}}
(2) {{{ 130 = 4s + 80 - 90 - 1800/s }}}
(2) {{{ 130s = 4s^2 - 10s - 1800 }}}
(2) {{{ 4s^2 - 140s - 1800 = 0 }}}
(2) {{{ s^2 - 35s - 450 = 0 }}}
Use the quadratic formula
{{{ s = ( -b +- sqrt( b^2 - 4*a*c )) / (2*a) }}}
{{{ a = 1 }}}
{{{ b = -35 }}}
{{{ c = -450 }}}
{{{ s = ( -(-35) +- sqrt( (-35)^2 - 4*1*(-450) )) / (2*1) }}}
{{{ s = ( 35 +- sqrt( 1225 + 1800 )) / 2 }}}
{{{ s = ( 35 +- sqrt( 3025 )) / 2 }}}
{{{ s = ( 35 + 55 ) / 2 }}}
{{{ s = 90/2 }}}
{{{ s = 45 }}}
and, since {{{ s + 20 }}} = her speed on the highway,
{{{ s + 20 = 45 + 20 }}}
{{{ s + 20 = 65 }}}
She drove 65 mi/hr on the highway
check:
(2) {{{ 130 = ( s + 20 )*( 4 - t ) }}}
(2) {{{ 130 = ( 45 + 20 )*( 4 - t ) }}}
(2) {{{ 130 = 65*( 4 - t ) }}}
(2) {{{ 130 = 260 - 65t }}}
(2) {{{ 65t = 130 }}}
(2) {{{ t = 2 }}} hrs
and
(1) {{{ t = 90/s }}}
(1) {{{ t = 90/45 }}}
(1) {{{ t = 2 }}} hrs
OK