Question 512112
You have the right idea
To picture the situation, pretend you start a stopwatch
when Stella leaves.
Both will run different distances but for the same amount of time
Call this time {{{ t }}}
Call the distance Stella runs {{{ d }}}
-------------------
given:
Stella's equation:
(1) {{{ d = 14.3t }}}
Tess's equation:
(2) {{{ d - 50 = 12.5t }}}
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(b)
Substitute (1) into (2)
(2) {{{ 14.3t - 50 = 12.5t }}}
(2) {{{ 1.8t = 50 }}}
(2) {{{ t = 27.778 }}} sec
and
(1) {{{ d = 14.3*27.778 }}}
(1) {{{ d = 397.22 }}} ft
(c)
{{{ 27.778 }}} sec
--------------------
check answers:
(1) {{{ d = 14.3t }}}
(1) {{{ 397.22 = 14.3*27.778 }}}
(1) {{{ 397.22 = 397.22 }}}
and
(2) {{{ d - 50 = 12.5t }}}
(2) {{{ 397.22 - 50 = 12.5*27.778 }}}
(2) {{{ 347.22 = 347.22 }}}
OK