Question 230078
Let {{{s}}} = it's usual speed
Let {{{t}}} = it's usual time for 300 km journey
given:
(1) {{{300 = st}}} km
(2) {{{300 = (s + 5)*(t - 2)}}}
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This is 2 equations and 2 unknowns, so it's solvable
Expand (2)
(2) {{{300 = (s + 5)*(t - 2)}}}
(2) {{{300 = st + 5t - 2s - 10}}}
substitute (1) in (2)
{{{300 = 300 + 5t - 2s - 10}}}
{{{5t = 2s + 10}}}
{{{t = (2/5)*s + 2}}}
Plug this back into (1)
(1) {{{300 = st}}}
{{{300 = s*((2/5)*s + 2)}}}
{{{300 = (2/5)*s^2 + 2s}}}
{{{1500 = 2s^2 + 10s}}}
{{{2s^2 + 10s - 1500 = 0}}}
{{{s^2 + 5s - 750 = 0}}}
Solve using quadratic formula
{{{s = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
{{{a = 1}}}
{{{b = 5}}}
{{{c = -750}}}
{{{s = (-5 +- sqrt( 5^2-4*1*(-750) ))/(2*1) }}}
{{{x = (-5 +- sqrt( 25 + 3000 ))/2 }}}
{{{s = (-5 +- sqrt( 3025 ))/2 }}}
{{{s = (-5 +- 55)/2 }}}
{{{s = 25}}} (the negative answer can't be used)
It's usual speed is 25 km/hr
check answer:
(1) {{{300 = st}}} km
{{{300 = 25t}}}
{{{t = 12}}} hrs
(2) {{{300 = (s + 5)*(t - 2)}}}
{{{300 = (25 + 5)*(12 - 2)}}}
{{{300 = 30*10}}}
{{{300 = 300}}}
OK