Question 654575
Let {{{ s }}} = the car's actual speed in mi/hr
Let {{{ t }}} = the car's actual time in hours
given:
(1) {{{ 1280 = s*t }}}
(2) {{{ 1280 = ( s + 16 )*( t -4 ) }}}
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(2) {{{ 1280 = s*t - 4s + 16t - 64 }}}
(2) {{{ 1344 = s*t - 4s + 16t }}}
Substitute (1) into (2) 
(2) {{{ 1344 = 1280 - 4s + 16t }}}
(2) {{{ 16t = 4s + 64 }}}
And, from (1)
(1) {{{ s = 1280/t }}}
Plug this back into (2)
(2) {{{ 16t = 4*( 1280/t ) + 64 }}}
(2) {{{ 16t^2 = 5120 + 64t }}}
(2) {{{ t^2 - 4t - 320 = 0 }}}
I can solve this by inspection
(2) {{{ ( t - 20 )*( t + 16 ) }}}
{{{ t = 20 }}} ( ignore {{{ t = -16 }}}, can't have negative time )
Plug this result back into (1)
(1) {{{ 1280 = s*20 }}}
(1) {{{ s = 64 }}}
The car's speed is 64 mi/hr
check:
(2) {{{ 1280 = ( 64 + 16 )*( 20 -4 ) }}}
(2) {{{ 1280 = 80*16 }}}
(2) {{{ 1280 = 1280 }}}
OK