<PRE><B>Question 763175</B>
Distance = rate x time.&lt;P&gt;
Set up two equations for this problem.  Call the unknown rate x and the unknown time t.&lt;P&gt;
600 = xt &lt;P&gt;
600/t = x Eq 1&lt;P&gt;
The distance is the same, so the first rate times time has to equal the second rate times time, and that provides the second equation.&lt;P&gt;
xt = (x-10)*(t+2) = xt +2x - 10t - 20&lt;P&gt;
xt&#39;s cancel.&lt;P&gt;
0 = 2x - 10t - 20 EQ 2&lt;P&gt;
Substitute the x value (in terms of t) from EQ1 into EQ2&lt;P&gt;
0 = 2(600/t) -10t -20 = 1200 - 10t^2 -20t &lt;P&gt;
Multiply all terms by -1, so the t^2 is positive.  Divide all terms by 10.&lt;P&gt;
0 = t^2 + 2t -120 = (t - 10)(t+12) so t=10 or t=-12.  The time of a trip cannot be negative, so discard -12.&lt;P&gt;
EQ1:  600/10 = x = 60 MPH is the speed of the original trip.&lt;P&gt;
The return trip was 10 MPH slower, which is 50MPH.

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