Question: I have a problem I haven't been able to solve for a class. A man took a trip in a car. He drove $70$ miles at a slower speed. Then, he went the next $300$ miles at a speed that was $40$ mph faster than earlier. The time he spent driving at the faster speed was twice the time spent at the slower speed. Find the two speeds. I think I could break this down to the following.

Let $s\_1$ be the slower speed, $s\_2$ the faster speed, $d\_1$ the shorter duration, and $d\_2$ the longer duration. Then

$d\_1=\frac{70}{s\_1}$

$d\_2=\frac{300}{s\_1+40}$

$d\_2=2d\_1$

and so

$2d\_1=\frac{300}{s\_1+40}$

I think in my head I can come up with $35$ mph and $75$ mph but that's just because I tried a bunch of numbers that seemed normal for driving. How can I solve this?

Answer: 
Let $t\_1$ be the time spent during the first part and $v\_1$ be the speed of the first part

We have : 

$t\_1 \times v\_1 = 70 $

$2t\_1 \times (v\_1 +40) = 300$

So by replacing $t\_1 \times v\_1$ in the second equation we deduce $ t\_1 = 2 $.

And then we find $v\_1 = 35$.
time = distance/speed

Let s be the slower speed.

Solve the following equation for s:

2\*(70/s) = (300/(s + 40))

s = 35 mph, s + 40 = 75 mph