System of linear equations (SOLE)

System of linear equations (SOLE) problems use multiple equations with multiple variables. For example, the two equations below use the variables $x$ and $y$. Taken together, they form a system of linear equations.

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

Your job is to find the value(s) of the variables that make both equations true at the same time.

Sidenote

SOLE is a challenging topic that is worth the effort

SOLE questions can feel challenging at first because you have to keep track of more than one equation at once. The good news is that the core ideas are straightforward: you apply the same algebra rules you already know, just in a more organized way.

On the GRE, many questions can be solved in more than one way. In this chapter, we focus on methods that are usually the simplest and fastest. It also helps to think of SOLE as a flexible tool: once you know the main techniques, you can use them in many different question types.

So 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 used for SOLE. Sometimes you’ll use them to solve for a specific variable (like $x$). Other times, you’ll use them to create an expression the question asks for (like $2x+y$).

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 idea behind isolate and substitute is:

• First, rewrite one equation so that one variable is alone on one side (for example, $x=\text{something}$).
• Then substitute that expression into the other equation.
• That leaves you with an equation in just one variable, which you can solve.

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:

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

Here, we subtracted $3y$ from both sides. Now $x$ is written in terms of $y$.

Next, substitute that expression for $x$ into the other equation:

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

Now the equation contains only $y$, so solve for $y$:

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

Now plug $y=21/8$ back into the first equation to solve for $x$:

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

So the solution is:

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

Now evaluate each answer choice:

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 is to combine the equations by adding or subtracting them. The goal is to eliminate one variable by making its coefficients cancel.

Here’s an example:

Solve for $y$:

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

Both equations contain $4x$, so subtracting one equation from the other will eliminate $x$.

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

When you subtract an equation, subtract every term (it’s the same as multiplying the entire equation by $-1$). Now solve for $y$:

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

The question only asks for $y$, so you can stop here. If you also wanted $x$, you could substitute $y=-2$ back into either equation:

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

Sometimes the question asks for an expression, not a single variable. You can still use the same elimination idea and stop as soon as you reach the expression you need.

What is the value of $3x+y$?

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

Subtract the second equation from the first:

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

Because the question asks for $3x+y$, you’re already done.

Sometimes you need to adjust one or both equations first so that a variable’s coefficients match.

Solve for $x$:

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

The $y$-coefficients are $10$ and $-5$. If you multiply the second equation by $2$, you get $-10y$, which will cancel with $+10y$.

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

Now add this new equation to the first equation:

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

So $x=8$.

SOLE technique #3: Multiply or divide the equations

The final technique is similar in spirit to elimination, but instead of adding or subtracting equations, you multiply or divide them.

Solve for $x$:

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

You could solve this by substitution, but first let’s show how multiplying the equations works.

When you multiply (or divide) two equations, you multiply (or divide) the left sides together and the right sides together. The key rule is consistency: whatever you do to one side, you do to the other side.

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 ($y^2$)

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

Now simplify:

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

This gives two possible values:

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

However, you must check which values actually work in the original system. From the first equation,

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

Since $y^2$ is always nonnegative, $y^2/5$ can’t be negative. So $x=-5$ is not possible here. The only valid solution is $x=5$.

Now let’s solve the same system using substitution.

Solve for $x$:

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

Instead of substituting for a single variable like $x$, you can substitute for any expression that makes the equation easier. Here, substitute $y^2=125/x$ into the first equation:

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

This again gives:

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

And just as before, you would reject $x=-5$ by checking against the original equations.

There are often multiple valid solution paths. Use the method that feels most natural, and if one approach gets messy, switch to another.

How to solve word problems with a system of equations

Sometimes a system of equations is hidden inside a word problem. Your job is to translate the situation into equations, then solve using the techniques above.

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?

Before solving, set up the equations.

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 solve using isolate and substitute. First, isolate $b$:

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

Now substitute into the ratio 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 find $b$:

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

So Sal bought $15$ apples and $21$ bananas. The question asks:

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?

Compute the difference:

$$21-15=6$$

There are $6$ more bananas than apples.

Bringing it all together: question walkthrough video

Here’s a video that walks through one of our practice questions and shows these ideas in action: