Rowing on a lake in 1.5 hours 
Sue Chin has just enough money to rent a canoe for 1.5 hours. How far out on a lake can she paddle and return on time if she paddles out at 2 km/h and back at 4 km/h?

My work:
$$d=rt$$
$$d=2x+4(1.5-x)$$
$$d=-2x+6$$
This is where I am stuck…
 A: Let $x$ be the time she spends paddling at $2$ km/h. Then $\frac{3}{2}-x$ is the time she spends paddling at $4$ km/h. Notice that $2x$ is the distance she travels one way and $4(1.5-x)$ is the distance she travels to return to her original point. These two distances are the same, so you have the equation $2x = 4(\frac{3}{2}-x)$. Solve this for $x$. The distance she travels out on the lake is $2x$ (or $4(\frac{3}{2}-x)$), which is what you want to find. Note that the question asks you how far on the lake she travels, not the total distance she travels.
A: Hint:
Distance "out" and distance "in" must be the same. When she paddles out her speed is 2km/h and when she paddles in her speed is 4km/h. Now:
$d_1=d_2$, use $d=rt$
$2t_1=4t_2$
On the other hand we know that $t_1+t_2=1.5$ hours. Now solve this system for $t_1$ and $t_2$ and then it is easy to obtain $d_1$ which represents how far she can paddle.
A: The distances travelled outbound and inbound are the same; call this length $d$. Since time is distance over speed:
$$\frac d2+\frac d4=1.5$$
$$\frac{3d}4=1.5$$
$$d=1.5×\frac43=2$$
So she can row out at most 2 kilometres.
