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 $d\_1$ be the time spent driving at a slower speed, and $s\_1$ be the speed during that leg.

We can use your calculations: $$d\_1=\frac{70}{s\_1}\tag{1}$$ $$2d\_1=\frac{300}{s\_1+40}\tag{2}$$

Multiplying clearing the denominator in each of your equations gives:

$$d\_1 \cdot s\_1 = 70 \tag{1}$$

$$2d\_1 \cdot (s\_1 +40) = 2(d\_1 \cdot s\_1) + 40\cdot2d\_1 = 300 \tag{2}$$

Substituting $(1)$ into $(2)$ gives us:

$$2(70) + 80d\_1 = 140 + 80d\_1=300$$

$$80d\_1 = 300 - 140 = 160\quad \iff \quad d\_1 = \frac{160}{80} = 2.$$

Using the value $d\_1$ in equation $(1)$ to solve for $s\_1$:

$$2s\_1 = 70 \quad \iff \quad s\_1 = \frac{70}{2} = 35\text{ mph}.$$

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So your results are correct: $s\_2 = s\_1 + 40 = 35 + 40 = 75$ mph.
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 $v\_1 = 35$.