Question 71972
Yuo can use the distance formula: d = rt where: d = distance, r = rate (speed), and t = time of trip. In this problem, the distance out is equal to the distance back, so {{{d[1] = d[2]}}}
For the outbound trip, t = 4 hrs, and speed = {{{r[1]}}} so:
{{{d[1] = r[1](4)}}}
For the return trip, t = 3 hrs, and speed = {{{r[1] + 7}}} so:
{{{d[2] = (r[1]+7)(3)}}} but {{{d[1] = d[2]}}}, so:
{{{r[1](4) = (r[1]+7)(3)}}} Simplify and solve for {{{r[1]}}}
{{{4r[1] = 3r[1]+21}}} Subtract {{{3r[1]}}} from both sides.
{{{r[1] = 21}}}mph This is the outbound speed.
{{{r[2] = r[1]+7}}}
{{{r[2] = 21+7}}}
{{{r[2] = 28}}}mph This is the return speed.