# How do you solve 13=5-13+3a?

Mar 24, 2015

There is only one correct answer, but there are many ways to get to that answer.
(There is only one London, England, but many ways to get there.)

Here's one:
Look for things the can be combined on the left of the equal sign. No, there's only a 13 there.
Look for things that can be combined on the right of the equal sign. Yes, I can do 5-13. So we write:

$13 = 5 - 13 + 3 a$ is true if and only if:

$13 = - 8 + 3 a$.

Often the words "is true if and only if" clutter up our solution, so we leave them out. I'll leave them out from here on.

Now, I want an $a$ that makes it true that $13 = - 8 + 3 a$.
Our goal is to get $a$ alone on one side $=$ some number on the other side. $a = \text{some number}$
A lot of people like to put the unknown ($a$) on the left.
There are two ways of doing this. I think the simpler is to use the fact that $\text{this" = "that}$ exactly when $\text{that" = "this}$
So now we want:

$- 8 + 3 a = 13$

Look at the left side. Order of operations: if I picked a number for $a$ what arithmetic would I do in what order? First I'd multiply by 3. The I'd add $- 8$ (or subtract $8$).

To get $a$ alone, we'll undo those operations. First we'll add $+ 8$, then we'll divide by $3$.

The whole process, without words explaining it looks like this:

$13 = 5 - 13 + 3 a$
$13 = - 8 + 3 a$
$- 8 + 3 a = 13$
.
$\textcolor{red}{3 a = 13 + 8}$
.
$3 a = 21$
$\frac{3 a}{3} = \frac{21}{3}$
$a = 7$

The solution to the last equation is evident. The solution is $7$.
The solution to the first equation is also $7$

I left some blank lines. For many people it becomes tiresome to include the steps below in red: (The middle 2 lines.)
$- 8 + 3 a = 13$
$\textcolor{red}{\left(- 8 + 3 a\right) + 8 = 13 + 8}$
$\textcolor{red}{- 8 + 3 a + 8 = 21}$
$3 a = 21$

Here's a slightly different looking solution

$13 = 5 - 13 + 3 a$

$5 - 13 + 3 a = 13$

$3 a = 13 - 5 + 13$

$3 a = 21$

$a = \frac{21}{3} = 7$

Again, it is obvious that the solution to the last equation is $7$.

The solution to the first equation is also $7$