Question 182513
For 1st leg:
(a) {{{d[1] = r[1]*t[1]}}}
For 2nd leg:
(b) {{{d[2] = r[2]*t[2]}}}
given:
{{{t[1] + t[2] = 1}}} hrs
{{{d[1] = 30}}} mi
{{{d[1] + d[2] = 50}}} mi
{{{r[2] = r[1] + 15}}} mi/hr
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Now rewrite (a) and (b)
(a) {{{d[1] = r[1]*t[1]}}}
(a) {{{30 = r[1]*t[1]}}}
and
(b) {{{d[2] = r[2]*t[2]}}}
(b) {{{50 - d[1] = (r[1] + 15)*(1 - t[1])}}}
(b) {{{50 - d[1] = r[1] + 15 - r[1]*t[1] - 15t[1]}}}
Note that {{{r[1]*t[1] = d[1]}}} and {{{d[1] = 30}}}
(b) {{{20 = r[1] - 15t[1] - 15 }}}
(b) {{{15t[1] = r[1] - 35}}}
From (a)
(a) {{{r[1] = 30/t[1]}}}
(c) {{{15t[1] = 30/t[1] - 35}}}
Multiply both sides by {{{t[1]}}}
(c) {{{15(t[1])^2 = 30 - 35t[1]}}}
(c) {{{3(t[1])^2 + 7t[1] - 6 = 0}}}
I'll say {{{t[1] = x}}} for now and use quadratic formula
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
{{{a = 3}}}
{{{b = 7}}}
{{{c = -6}}}
{{{x = (-7 +- sqrt( 7^2-4*3*(-6) ))/(2*3) }}} 
{{{x = (-7 +- sqrt( 49 + 72 ))/6 }}} 
{{{x = (-7 +- sqrt( 121 ))/6 }}} 
{{{x = (-7 +- 11)/6 }}}
{{{x = -3}}} (I can't use a negative answer)
{{{x = 2/3}}}
and, since {{{x = t[1]}}},
{{{t[1] = 2/3}}}
(a) {{{30 = r[1]*t[1]}}}
{{{30 = r[1]*(2/3)}}}
{{{r[1] = (3/2)*30}}}
{{{r[1] = 45}}} mi/hr answer
check answer:
(a) {{{30 = r[1]*t[1]}}}
(b) {{{50 - d[1] = (r[1] + 15)*(1 - t[1])}}}
----------------------
(a) {{{30 = 45*(2/3)}}}
(a) {{{30 = 30}}}
(b) {{{50 - 30 = (45 + 15)*(1 - 2/3)}}}
(b) {{{20 = (45 + 15)*(1 - 2/3)}}}
(b) {{{20 = 60*(1/3)}}}
(b) {{{20 = 20}}}
OK