System of linear equations (SOLE)

System of linear equations (SOLE) problems involve multiple equations that include multiple variables in each equation. For example, the two equations below using the variables $x$ and $y$, when taken together, are a system of linear equations.

$$\begin{array}{rl}2x-15& =2/3y\\ 4x-5/3y& =5\end{array}$$

Your job is to figure out the possible values of the variables that make the math equations correct.

Sidenote

SOLE is a challenging topic that is worth the effort

If math isn’t your strong point (or even if it is), you will likely find these types of questions to be challenging. However, like most things on the GRE, the core concepts are straightforward, and if you go slowly and take things one step at a time, you might be surprised at your ability to solve these.

Many questions on the GRE can be solved in multiple ways, and we tend to focus on the methods that are simplest and fastest, since things are easier when you pick the right tool for the job. That said, SOLE is like a multitool or Swiss Army knife, and can be used to solve a wide variety of question types. It’s a fantastic backup option to have in your toolkit, if you understand how to use it.

So in short, this chapter is worth reading and re-reading a few times!

How do you solve a system of linear equations?

There are three fundamental techniques that are used for SOLE. In some cases you’ll be using these to rearrange the equations to solve for a specific variable (e.g. the value of $x$), and in other cases you might be trying to arrive at a fragment of an equation that matches something in the question (e.g. the value of $2x+y$).

The three fundamental techniques used for SOLE equations are:

1. Isolate and substitute
2. Add or subtract the equations
3. Multiply or divide the equations

SOLE technique #1: Isolate and substitute

The core concept for the isolate and substitute technique is that we want to rewrite one of the equations so that we have one variable by itself, expressed in terms of the other variable(s), and then we plug that into the second equation so we’re left with only one variable at all. Then it’s just a matter of doing the math.

Let’s do an example question:

Which of the following could be true?

Given:
$x+3y=12$
$y+3x=15$

Select all that apply.
A. $x>y$
B. $y>x$
C. $y=x$
D. $21y=33x$
E. $21x=33y$

The first step is to isolate one of the variables. Let’s isolate $x$ in the first equation:

$$\begin{array}{rl}x+3y& =12\\ x& =12-3y\end{array}$$

Basically, we just took the original equation and subtracted $3y$ from each side. That left us with the value of $x$ expressed in terms of $y$ as $x=12-3y$.

The next step is to substitute that value into the other equation:

$$\begin{array}{rl}y+3x& =15\\ y+3(12-3y)& =15\end{array}$$

Now that we’ve substituted out $x$ for its value expressed in terms of $y$, our new equation only has the $y$ variable left. We can step through the math to find the value of $y$.

$$\begin{array}{rl}y+3(12-3y)& =15\\ y+36-9y& =15\\ 36-8y& =15\\ 36& =15+8y\\ 21& =8y\\ 21/8& =y\end{array}$$

And now that we have the value of $y$ as a normal number without any variables, we can plug that back into the first equation and solve for $x$:

$$\begin{array}{rl}x+3y& =12\\ x+3(21/8)& =12\\ x+63/8& =12\\ x& =12-63/8\\ x& =96/8-63/8\\ x& =33/8\end{array}$$

At this point we’ve figured out the values for both $x$ and $y$:

$$\begin{array}{rl}x& =33/8\\ y& =21/8\end{array}$$

And now it’s just a matter of solving the question using these values:

Select all that apply.
A. $x>y$
B. $y>x$
C. $y=x$
D. $21y=33x$
E. $21x=33y$

So we get:

Select all that apply.
A. $33/8>21/8$ TRUE
B. $21/8>33/8$ FALSE
C. $33/8=21/8$ FALSE
D. $21(21/8)=33(33/8)$ FALSE
E. $21(33/8)=33(21/8)$ TRUE

And that’s it!

SOLE technique #2: Add or subtract the equations

The second technique to solve a system of equations is to combine the equations by adding or subtracting one from the other. In the simplest form, one of the variables will have the same coefficient in both equations, and when you subtract one from the other, it will cancel out that variable. Here’s an example:

Solve for $y$:

$$\begin{array}{rl}4x-2y& =16\\ 4x+4y& =4\end{array}$$

See how both equations have a term of $4x$? That’s a hint to subtract one from the other.

$$\begin{array}{cccccc}& 4x& -2y& =& 16& \\ -(& 4x& +4y& =& 4& )\\ & & -6y& =& 12& \end{array}$$

Just be careful that you subtract the entire equation, i.e. it’s the same as multiplying the whole thing by $-1$ so it’s really $-4x$, $-4y$, and $-4$. And now we’re left with just a $y$ variable and can easily solve the remaining math:

$$\begin{array}{rl}-6y& =12\\ y& =-2\end{array}$$

The question only asks us for the value of $y$, so we’re done. But if we wanted to, we could plug this back into the first equation to find the value of $x$:

$$\begin{array}{rl}4x-2y& =16\\ 4x-2(-2)& =16\\ 4x+4& =16\\ 4x& =12\\ x& =3\end{array}$$

Sometimes, instead of asking us to solve for just a single variable like $y$, you might get a question that asks for a specific expression involving multiple variables. We can use the same technique as before, and essentially just stop solving partway through:

What is the value of $3x+y$?

Given:
$6x-7y=50$
$3x-8y=25$

This question works out nicely using the subtraction method:

$$\begin{array}{cccccc}& 6x& -7y& =& 50& \\ -(& 3x& -8y& =& 25& )\\ & 3x& +y& =& 25& \end{array}$$

Because the question specifically asks for the value of $3x+y$, we’re already done!

Sometimes it might be necessary to manipulate one or both of the equations before adding or subtracting, since for that to work, the coefficients of the variables need to be equal. Here’s an example:

Solve for $x$:

$$\begin{array}{rl}7x+10y& =2\\ -3x-5y& =3\end{array}$$

We have $7x$ and $-3x$, and $10y$ and $-5y$. Neither of these match! But, we have an easy path forward since we can multiply the second equation by $2$ to get $-10y$, which will cancel out the $10y$ in the first equation.

$$\begin{array}{rl}-3x-5y& =3\\ 2(-3x-5y)& =2(3)\\ -6x-10y& =6\end{array}$$

Just be careful that when you multiply an equation like this, you need to multiply both the left and right sides. Now we can add our modified second equation to the first, and they will cancel out nicely:

$$\begin{array}{cccccc}& 7x& +10y& =& 2& \\ +(& -6x& -10y& =& 6& )\\ & x& & =& 8& \end{array}$$

And there you have it!

SOLE technique #3: Multiply or divide the equations

The final technique we’ll cover is similar to the previous one. Instead of adding or subtracting to combine our equations, we can also multiply or divide them.

Solve for $x$:

$$\begin{array}{rl}{y}^2/5& =x\\ 125/x& =y^2\end{array}$$

You might notice that we’ve already isolated both $x$ and $y^2$, and we could substitute either into the other equation to solve this. We’ll walk through that approach in a moment, but first, let’s illustrate how we could solve this by multiplying the two equations.

When you multiply (or divide) two equations, you basically just match up the two sides and multiply/divide down. In this example, we’ll:

• Multiply the left side of the first equation ($y^2/5$) by the left side of the second equation ($125/x$)
• Multiply the right side of the first equation ($x$) by the right side of the second equation ($y^2$)

You just need to be consistent: multiply both or divide both sides - no mixing and matching.

$$\begin{array}{ccccc}& y^2/5& =& x& \\ \times (& 125/x& =& y^2& )\\ & (y^2/5)(125/x)& =& (x)(y^2)& \end{array}$$

And now we can simplify and solve:

$$\begin{array}{rl}(y^2/5)(125/x)& =(x)(y^2)\\ 125y^2/5x& =x{y}^2\\ 25y^2/x& =x{y}^2\\ 25y^2/x& =x{y}^2\\ 25y^2& =x^2{y}^2\\ 25& =x^2\end{array}$$

This leaves us with two cases:

$$\begin{array}{rl}x& =5\\ x& =-5\end{array}$$

However, we’re not done yet. There is some odd behavior that we need to be careful of with the square root we use for $25=x^2$ since we get two possible values: $5$ and $-5$. At least one of these values will work, and sometimes both will, but not always. We need to plug them back into the original equation to make sure that they check out.

$$\begin{array}{rl}{y}^2/5& =x\\ y^2/5& \ne -5\end{array}$$

There’s no way that $-5$ could be the solution, since squaring a number will always result in a positive number. The only solution that is valid in this case is $x=5$.

There are usually many different ways to solve math questions. Let’s do this same one again using substitution.

Solve for $x$:

$$\begin{array}{rl}{y}^2/5& =x\\ 125/x& =y^2\end{array}$$

Previously we’ve shown how we can isolate and substitute a single variable. This technique is generally applicable - you can isolate and substitute anything that might make your equation easier to solve. This time, instead of substituting in $x$, let’s see how it works when we substitute in $y^2=125/x$.

$$\begin{array}{rl}{y}^2/5& =x\\ (125/x)/5& =x\\ 25/x& =x\\ 25& =x^2\end{array}$$

This leaves us with the same two possibilities:

$$\begin{array}{rl}x& =5\\ x& =-5\end{array}$$

Even though we used a different method to solve this, you can see that we ended up with the same outcome. Use whichever methods are the most natural for you! And if one method doesn’t work, just try again using a different approach.

How to solve word problems with a system of equations

Sometimes system of equations problems will be disguised in the format of a word problem. Your job is to translate the situation described into equations, and then apply the techniques you’ve learned to solve them.

Here is a very simple example:

Sal bought a total of 36 fruits. If the ratio of fruits bought was 7 bananas for every 5 apples, how many more bananas were bought than apples?

We’ll walk through it in a moment, but can you set up the equations on your own?

Use $a$ for the number of apples, and $b$ for the number of bananas.

Try solving it on your own, and then double-check your work below!

We’ll explain this using the isolate and substitute method. First, let’s isolate $b$:

$$\begin{array}{rl}a+b& =36\\ b& =36-a\end{array}$$

And then let’s substitute it into the other equation:

$$\begin{array}{rl}7/5& =b/a\\ 7/5& =(36-a)/a\\ 7a/5& =36-a\\ 7a& =180-5a\\ 12a& =180\\ a& =15\end{array}$$

Now that we’ve figured out the number of apples, getting the number of bananas is simple:

$$\begin{array}{rl}a+b& =36\\ 15+b& =36\\ b& =21\end{array}$$

They bought $15$ apples and $21$ bananas. Let’s look at the question one final time:

Sal bought a total of 36 fruits. If the ratio of fruits bought was 7 bananas for every 5 apples, how many more bananas were bought than apples?

How many more bananas than apples?

$$21-15=6$$

There are 6 more bananas than apples. Easy 😄

Bringing it all together: question walkthrough video

Here’s a video going through one of our practice questions to demonstrate these ideas in action: