Difficult word problem 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?
 A: 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 $v_1 = 35$.
A: 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 $s_1$:
$$2s_1 = 70 \quad \iff \quad s_1 = \frac{70}{2} = 35\text{ mph}.$$

So your results are correct: $s_2 = s_1 + 40 = 35 + 40 = 75$ mph.
A: 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
