Question 214899
I write 2 equations, 1 for each
leg of the trip.
Buffalo to Boston:
{{{d[1] = r[1]*t[1]}}}
{{{d[2] = r[2]*t[2]}}}
given:
{{{d[1] = 720}}} km
{{{d[2] = 720}}} km
{{{r[2] = r[1] + 10}}} km/hr
{{{t[1] + t[2] = 17}}} Hr
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I can rewrite the equations
{{{720 = r[1]*t[1]}}}
{{{t[1] = 720/r[1]}}}
and
{{{720 = (r[1] + 10)*t[2]}}}
{{{720 = (r[1] + 10)*(17 - t[1])}}}
{{{720 = 17r[1] + 170 - r[1]t[1] - 10t[1]}}}
By substitution:
{{{720 = 17r[1] + 170 - 720 - 10t[1]}}}
{{{720 + 550 = 17r[1] - 10*(720/r[1])}}}
Multiply both sides by {{{r[1]}}}
{{{1270r[1] = 17*(r[1])^2 - 7200}}}
{{{17*(r[1]^2) - 1270r[1] - 7200 = 0}}}
Using quadratic equation:
{{{r[1] = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
{{{a = 17}}}
{{{b = -1270}}}
{{{c = -7200}}}
{{{r[1] = (-(-1270) +- sqrt( (-1270)^2-4*17*(-7200) ))/(2*17) }}}
{{{r[1] = ( 1270 +- sqrt( 1612900 + 489600 ))/ 34 }}}
{{{r[1] = ( 1270 +- sqrt( 2102500 ))/ 34 }}}
{{{r[1] = ( 1270 +- 1450)/ 34 }}}
{{{r[1] = 2720/34}}}
{{{r[1] = 80}}} km/hr
His speed from Buffalo to Boston was 80 km/hr
check:
{{{r[2] = r[1] + 10}}} km/hr
{{{r[2] = 90}}} km/hr
{{{720 = r[1]*t[1]}}}
{{{720 = 80*t[1]}}}
{{{t[1] = 9}}} hr
{{{t[2] = 17 - 9}}}
{{{t[2] = 8}}}
and
{{{720 = 90*8}}}
{{{720 = 720}}}
OK
It looks like you had a sign wrong