Question 238505
I can write 2 equations, 1 for each vehicle
{{{d[c] = r[c]*t[c]}}}
and
{{{d[b] = r[b]*t[b]}}}
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given:
{{{r[c] = 2r[b] - 30}}} mi/hr
{{{t[c] = 2}}} hrs
{{{t[b] = 2}}} hrs
{{{d[c] = d[b] + 20}}} mi
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Now I plug in the substitutions
{{{d[c] = r[c]*t[c]}}}
{{{d[b] + 20 = (2r[b] - 30)*2}}}
{{{d[b] + 20 = 4r[b] - 60}}}
(1) {{{4 r[b] - d[b] = 80}}}
and
{{{d[b] = r[b]*t[b]}}}
(2) {{{d[b] = 2r[b]}}}
Substitute (2) in (1)
(1) {{{4 r[b] - 2r[b] = 80}}}
{{{2r[b] = 80}}}
{{{r[b] = 40}}} mi/hr
and, since
{{{r[c] = 2r[b] - 30}}}
{{{r[c] = 2*40 - 30}}}
{{{r[c] = 50}}} mi/hr
The rate of the car is 50 mi/hr
check:
{{{d[b] + 20 = (2r[b] - 30)*2}}}
{{{d[b] + 20 = (2*40 - 30)*2}}}
{{{d[b] = (80 - 30)*2 - 20}}}
{{{d[b] = 100 - 20}}}
{{{d[b] = 80}}} mi
and
{{{d[c] = r[c]*t[c]}}}
{{{d[c] = 50*2}}}
{{{d[c] = 100}}}
In 2 hours, the car is 20 mi ahead of the bus
OK