Algebraic word problem involving auxiliary variable When Mr. and Mrs. Smith took the airplane, they had together 94 pounds of baggage.
He paid 1.50 and she paid 2.00 for excess weight. If Mr. Smith made the trip by himself with the combined baggage of both of them, he would have to pay $13.50. How many pounds of baggage can one person take along without being charged.
The variable in the problem is not explicitly stated and I have tried multiple ways of starting it. I realized i understand what the question is asking and can figure out how to answer it, but not algebraically. How can I answer it algebraically? Any help would be appreciated. Thanks!
 A: They want you to find how many pounds of baggage each person can take on the plane for free. Let
$x = $ the number of pounds someone can bring for free.
We also need to introduce a variable that represents the price per pound for excess baggage, which isn't mentioned in the word problem. This is called our auxiliary variable.
$y = $ the price per pound for excess baggage.
Here's an specific case of what's going on for a visual. In this case, we have a 42 pound weight limit. 

Now, think about what happens if Mr. Smith goes alone with both bags. He would have to pay $ \$13.50$ in extra fees. Looking at the diagram, we know both bags have a combined excess fee of $ \$3.50$. This means he would end up paying $ \$10$ for the bag that Mrs. Smith would of brought for free. This translates to 
$$ xy = 10.$$
This doesn't really help us yet, so let's think about this some more. As an extreme case, suppose nothing was free, and everything over $0$ pounds was an excess fee. If Mr. Smith went alone with both bags, we would have to pay the original excess fees of $ \$3.50$ plus $ \$10$ for Mrs. Smith's previously free baggage plus another $ \$10$ for his previously free baggage ("free" is the same on the bar graph). This means he would have to pay $ \$23.50$ in excess fees for 94 pounds. 
BUT... this would mean each excess pound of baggage would cost ....? Finding this will give you $y$. Then use the equation we made to find $x$. 
Once you find these, try making a diagram like we did above and see if it makes sense with the information you have. $y$ and the excess fees for each person will tell you how many pounds they went over. 
A: An exclusively algebraic solution follows. Generally, the cost a passenger should pay if he/she has excess baggage, is $$c=(w_t-w_f)\; q,$$where $w_t$ is the weight the passanger shows over the counter, $w_f$ is the maximum free weight, and $q$ is the amount per pound in excess.
The couple showed 94 pounds. Husband paid 1.50 and wife paid 2.00, individually. If Mr Smith made the trip by himself with the combined baggage of both of them, he would paid 13.50.
Let $w_w$ be the wife's weight, and $h_w$ the husband's weight. Then
\begin{align*}
2&=(w_w-w_f)\;q\\
1.5&=(w_h-w_f)\;q
\end{align*}
There is a relation between his weight and her weight: $w_w=94-w_h$. Then we can write\begin{align*}
2&=(94-w_h-w_f)\;q\\
1.5&=(w_h-w_f)\;q
\end{align*}
Now we add both equations and simplify. We have $$3.5=q(94-2w_f).$$
A second equation comes from the chance they both travelled but as a single passanger: $(94-w_f)\;q=13.50$. Now we have two equations in $q$ and $w_f$. Solving the system, we get $w_f=40$ pounds, and $q=0.25$ money units.
