How to calculate per unit cost? I need help understanding this math problem from my TEAS Mometrix Practice exam. It gives an explanation but I still don't understand. The question is:
Mandy can buy $4$ containers of yogurt and $3$ boxes of crackers for $\$9.55$. She can buy $2$ containers of yogurt and $2$ boxes of crackers for $\$5.90$. How much does one box of crackers cost?
The answer is $ \$2.25$.
I understand that I have to set up the problem like this:
\begin{align}
4x+3y &= 9.55 \\
2x+2y &= 5.90
\end{align}
but the explanation says that I have to multiply the bottom equation by $-2$; therefore, it becomes $-4x+ -4y= -11.80$
and then  I have to add the equations 
\begin{align}
4x+3y   &= 9.55\\
-4x+-4y &= -11.80
\end{align}
by top to bottom getting  either $\pm 2.25$
Can you explain why this is the answer? And also why does one have multiply by -2?
 A: 
"the explanation says that I have to multiply the bottom equation by $−2$". 

No, you don't have to. That is just one convenient way to solve the problem. 
Keep in mind that your goal is "How much does one box of crackers cost?", namely, the value of $y$. 


then  I have to add the equations 
  \begin{align}
4x+3y   &= 9.55\\
-4x+-4y &= -11.80
\end{align}
  by top to bottom getting  either $\pm 2.25$.

What you get should be $-y=-2.25$ which implies that $y=2.25$.


"Can you explain why this is the answer?"

$y=\pm 2.25$ is not an answer. $y=2.25$ is. If that is what is written, it must be a typo. 
A: One possible solution is the use of  Gaussian elimination. That is the solution you suggest in your question. Note that there are basically three operations that you can apply to a system, that don't change the solution.


*

*Type 1: Swap the positions of two equations.

*Type 2: Multiply an equation by a nonzero number.

*Type 3: Add to one equation to the multiple of another.


That these operations don't change the solution of the system is illustrated in this nice answer.
Let's start from the beginning with: 
\begin{align*}
4x+3y=&9.55 &&(I)\\
2x+2y=&5.90 &&(II)
\end{align*}


*

*Using an operation of Type 2, will result in:


\begin{align*}
4x+3y=&9.55   &&(I)\\
4x+4y=&11.80  &&(III) = 2\cdot(I)
\end{align*}


*Using an operation of Type 3, will result in:
\begin{align*}
4x+3y&=9.55 &&(I)\\
y&=2.25 &&(IV) = (III) - (I)
\end{align*}

*Now we can simply plug in $y=2.25$ into (I) and yield: 
$$ 4x=9.55 - 3\cdot 2.25= 2.8 \qquad ⇒ x = 0.7$$


I have to multiply the bottom equation by −2

That is not true. In the first step above, we multiplied equation (II) with 2. But we can also multiply the first one with $\frac{1}{2}$:
\begin{align*}
2x+1.5y=&4.775   &&(III)=\frac{1}{2}\cdot(I)\\
2x+2y=&5.90  &&(II)
\end{align*}
Then you can do the second step as well, right? 
The reason why the answer is given like that is, that Gaussian elimination is usually taught in the way, that the resulting equations form the so called row echelon form. The idea of the row echelon form is to use the three operations above to get a system that looks like this. So the last row contains one variable, the second to last row contains two variables etc. But, as always, it does not matter what way you choose, to get the result. 

There are also different ways to solve it, for example you can use (II) to express $y$ as a variable depending on $x$ 
$$ y = \frac{5.90 - 2x}{2}= \frac{5.90}{2} - x = 2.95 - x \qquad (*)$$
and then plug that into (I):
 \begin{align*}
&&4x+3y&= 9.55  \\
⇔&&  4x + 3\cdot(2.95-x)&=9.55 \\
⇔&&  4x - 3x &=9.55-8.85 \\
⇔&&  x &= 0.7
\end{align*}
Then using $(*)$ we get $y = 2.25$. 
A: There are several ways to solve this or similar problems we have 
$$4 \cdot x + 3 \cdot y = 9.55 \\
2 \cdot x  + 2 \cdot y = 5.90$$
One approch as given in your answer is to multiply the second equation by 2 or -2 you then have $4 \cdot x$ in both equations so can cancel for $x$ by adding the two equations or subtracting the equations as appropriate.  The reason this works is that by multiplying all terms by the same number you have created another true statement.
This is not the only approach however,  Another approach would be for example to solve one equation for say $y$ then substituting it in the other.
$$2 \cdot x + 2 \cdot y = 5.90 \Rightarrow y = \frac{5.90 - 2 \cdot x}{2}$$
Putting that in the other equation we have
$$4 \cdot x + 3 \cdot y = 4 \cdot x + 3 \cdot \frac{5.90 - 2 \cdot x}{2} = 4 \cdot x + 8.85 - 3 \cdot x= 9.55 \Rightarrow x = 9.55 - 8.85 = 0.7$$
Putting this back in the other equation
$$ 2 \cdot x + 2 \cdot y = 2 \cdot 0.7 +2 \cdot y = 5.90 \Rightarrow y = \frac{5.9 - 1.4}{2} = 2.25$$
