## A Year of Nights Wondering – VI

We were working on an application of 2 by 2 systems of linear equations.  We had built the system from the statement of the problem and I was about to solve it using the addition method.  I had spent some time talking about the importance of units and how every equation is a story.  Out of nowhere, Jack asked, “What about the units?” My jaw dropped.  I was about to add tickets to dollars which makes no sense. I assured Jack and the rest of the class that “things” would work out if we ignored the units until we stated our solution and told Jack that I would get back to him.  I am getting back to him with this post.

Here is how to solve the problem keeping track of the units for all the steps.  $t$ stands for number of tickets, \$ stands for number of dollars, $A$ stands for the number of adult tickets  and $C$ stands for the number of children’s tickets.  Think of $A$ as a pure number with unit $(t)$ appended to mean number of tickets (adult). The expressions in parentheses stand for units but can be treated as usual like any other algebraic non-zero variable, for instance, $(\frac{\}{t})(t) = \frac{(\ t)}{(t)}=(\)$.

We are given $A(t)+C(t)=301(t)$  meaning that the number of adult tickets plus the number of children’s tickets sold total $301$ tickets and $3(\frac{\}{t})A(t)+1(\frac{\}{t})C(t) = 485(\)$ meaning that if we sell adult tickets for three dollars per ticket and sell children’s tickets for one dollar per ticket, our total take will be 485 dollars.  Note that the unit expressions in each of the left hand terms of the second equation simplify to $(\)$.  We solve this system by multiplying the first equation by $3(\frac{\}{t})$ and subtracting the second equation from it.  I am getting tired of typing all the latex codes so here  the rest of the algebra manipulation by hand from OneNote.

Solving with Units – Part one

What did we do?  We assumed that all tickets cost three dollars, both adult tickets and children’s tickets, then we subtracted the actual take from the adult tickets and the actual take from the children’s tickets and the total take from what it would have been if we had charged three dollars for children’s tickets.  This gave us the extra money we would have made by charging children two dollar more per ticket.  Now we divide each side by $2(\)$ the extra cost per ticket to get $C=209$, the number of children’s tickets we sold.  It is easy to use the equation $A(t)+C(t)=301(t)$ to get $A = 92$ and we have our solution.

Here is what happens if we use the substitution method.

Solve by Substitution using Units

$C$ can be found using the first equation similar what we did using the addition method.

So Jack, we can preserve our units and everything will work.  So let’s just ignore the units until the end to save ourselves some trouble.  And if we are getting into difficulties when we are trying to solve such a system, maybe looking at how the units are behaving will help us see our way more clearly.