Question 956108
Let {{{ d }}} = distance in miles the slower plane flies in {{{5}}} hrs
Let {{{ d + 1590 }}} = distance in miles  the faster plane flies in {{{5}}} hrs
Let {{{ s }}} = the speed in mi/hr of the slower plane
{{{ 4s }}} = the speed in mi/hr of the faster plane
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Equation for slower plane:
(1) {{{ d = s*5 }}}
Equation for faster plane:
(2) {{{ d + 1590 = 4s*5 }}}
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Substitute (1) into (2)
(2) {{{ 5s + 1590 = 20s }}}
(2) {{{ 15s = 1590 }}}
(2) {{{ s = 106 }}}
and
{{{ 4s = 424 }}}
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The average speed of the slower plane is 106 mi/hr
The average speed of the faster plane is 424 mi/hr
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check:
(1) {{{ d = 106*5 }}}
(1) {{{ d = 530 }}}
and
(2) {{{ d + 1590 = 20*106 }}}
(2) {{{ d = 2120 - 1590 }}}
(2) {{{ d = 530 }}}
OK