TITLE: Find common arrival time, given a system of equations
QUESTION [1 upvotes]: Ginny and Jenna are 20 miles from home. They have one pair of roller skates.  Jenna walks 4mph and skates 9mph. Ginny walks 3 mph and skates 8mph. They start for home at the same time. 
First, Ginny has the skates and Jenna walks. Ginny skates for a while, then takes the roller skates off and starts walking.When Jenna reaches the roller blades, she puts them on and starts skating home. 
If they both start at 4:00 and arrive home at the same time, what time is it when they get home?

My solution,
I assumed that the time that Ginny skates is $a$ hours and walks for $b$ hours. And hence Jenna skates for $b$ hours and walks for $a$ hours. And since total distance covered is 20 for both, I got the following 2 equations.
$$
\begin{align}
8a + 3b &= 20 \\ 
4a + 9b &= 20
\end{align}
$$
I solved this system of equations by elimination to get $b = \dfrac{4}{3}$ and $a = 2$, and $a + b = \dfrac{10}{3}$. This doesn't check out with the required solution which is $4$ and arriving at $8$ pm.
I have checked the simultaneous equation, so I have probably made a logic error. Any ideas where I went wrong. Thanks.

REPLY [0 votes]: Let $x$ be the distance Ginny skates and $y$ be the distance she walks.
$x+y=20 \tag{1}$
Ginny's journey time is 
\begin{align*}
\frac{x}{8}+\frac{y}{3}= t
\end{align*}
and Jenna's journey time which also the same as Ginny's (as they arrive together) is
\begin{align*}
\frac{x}{4}+\frac{y}{9}= t
\end{align*}
Equating the two journey times gives
\begin{align*}
\frac{x}{8}+\frac{y}{3}&=\frac{x}{4}+\frac{y}{9}\\
y &= \frac{9}{16}x \tag{2}
\end{align*}
We can solve the system consisting of equations (1) and (2) to find the distances Ginny skates and walks. We can then use the distances to find the time of arrival.
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