Single variable equations are some of the most common types of problems on the ACT math section. You must know how to set up, use, and manipulate these kinds of equations, as they are a foundational element of mathematics upon which more complicated expressions (multiple variable, quadratics, etc.) are based.
So make sure you are prepared to tackle the ins and outs of single variable equations (no matter how they are presented on the ACT), before you take on some of the more complicated elements of ACT math.
This guide will be your complete walk-through of single variable equations for the ACT--what they are, how you’ll see them on the test, and how to set up and solve them.
And the mystery unfolds.
What Are Single Variable Equations?
To understand a single variable equation, let us break it into its two components: the variable and the equation.
A variable is a symbolic placeholder for a number we do not yet know. It’s very common to see $x$ or $y$ used as a variable in math problems, but variables can be represented by any symbol or letter.
$x + 4 = 14$
In this case, $x$ is our variable. It represents a number that is currently unknown.
An equation sets two mathematical expressions equal to one another. This equality is represented with an equals sign (=) and each side of the expression can be as simple as a single integer or as complex as an expression with multiple variables, exponents, or anything else.
$({x +y^2})/14 - 65(x - 3) = 2$
The above is an example of an equation. Each side of the expression equals the other.
So if we put together our definitions, we know that:
A single variable equation is an equation in which there is only one variable used. (Note: the variable can be used multiple times and/or used on either side of the equation; all that matters is that the variable remains the same.)
${(x + 4)}/2 = 12$
$6x + 3 - 2x = 19$
$4y - 2 = y + 7$
These are all examples of single variable equations. You can see how some expressions used the variable multiple times or used the variable in both expressions (on either side of the equals sign).
No matter how many times the variable is used, these still count as single variable problems because the variable remains constant and there are no other variables.
Finding your missing variable is like finding that last missing piece of the puzzle.
Typical Single Variable Equations on the ACT
Single variable equations will fall into two broad categories on the ACT--given equations and word problems. Let’s look at each type.
Given Equations
A given equation will provide you with the equation you need to use to solve the problem. We will go through the exact processes needed to solve this kind of problem in the next section, but for now just understand that your goal is to isolate your variable.
(We will go through how to solve this question later in the guide)
As you can see from this problem, the isolated variable may not be your final answer. Sometimes the question will ask you to solve for $x$, sometimes the question will ask you to solve for $x$ to a different term (as in this case, where they ask you to find $2x$).
Always pay close attention to exactly what the question is asking you to find! You need to first isolate your $x$ to solve the problem, but if you stop there then you will get the final answer wrong.
Word Problems
A word problem describes a scene in which you must set up your own single variable equation to solve it. Again, your final answer may be the value of your variable ($x$ or $y$, etc.) or your variable taken to a different term ($2x$, $y/2$, etc.).
(We will go through how to solve this question later in the guide)
How to Manipulate a Single Variable Equation
In order to solve a single variable equation, we must isolate our variable on one side of the equation. And the way we do this is by shifting the rest of our terms to the other side of the equation.
In order to shift our terms (numbers), we must therefore cancel them out on their original side by performing the opposite function of the term.
Opposite function pairs are:
Addition and subtraction
Multiplication and division
So if we have a term on one side that has a plus sign (addition), we must subtract that same amount from both sides.
$x + 2 = 6$
$x + 2 - 2 = 6 - 2$
$x = 4$
If we have a term that is multiplied, we must divide that same amount from both sides.
$3x = 18$
${3x}/3 = 18/3$
$x = 6$
And so on.
Whatever you do on one side of the equation, you must do on the other. This cancels out like terms and essentially moves your terms from one side of the equation to the other.
Single variable equations are all about maintaining balance.
Steps to Solving a Single Variable Problem
Let us take a typical variable expression and break it into the steps needed to solve it.
$3y - 10 + 2y = 15$. Find $y$.
1) Combine like terms
If there is more than one term with a same variable, we must combine them in order to ultimately isolate that variable. We can add or subtract terms with a same variable in the same way we can any other numbers.
$3y - 10 + 2y = 15$
Here we have a $3y$ and a $2y$. They are both positive, so we add them together.
$3y + 2y = 5y$
So now our equation looks like this:
$5y - 10 = 15$
2) Isolate the term with your variable
Once we have combined our variables, we must isolate the variable term. If the term is simply the variable itself (e.g. $y$), then we can skip this step. But since our term her is $5y$, we must isolate the whole term first.
$5y - 10 = 15$
So we must add 10 to either side of our equation. Why? Because we have a negative 10 and addition is the opposite of subtraction. And we must do it to either side to cancel out the 10 on the first expression in order to isolate our variable.
$5y - 10 + 10 = 15 + 10$
$5y = 25$
3) Isolate your variable
Now that we’ve isolated our term ($5y$), we can further isolate the variable itself.
Again, we perform an opposite function of the term. In this case, we have $5y$, which uses multiplication. In order to isolate the variable, we must therefore use division (the opposite of multiplication) by dividing on both sides.
$5y = 25$
${5y}/5 = 25/5$
$y = 5$
4) Double-check your variable by plugging it back in
Now that we’ve solved for our variable, let us check to make sure it is correct by plugging it back into the original equation.
$y = 5$
$3y - 10 + 2y = 15$
$3(5) - 10 + 2(5) = 15$
$15 - 10 + 10 = 15$
$15 = 15$
Success! We have correctly isolated the variable and found its value.
5) And, finally, double-check to make sure you are answering the right question!
In this case, we are done, because our initial question asked us to find the value of $y$. But you must always double-check to make sure you are answering the right question. If they had asked us the value for $5y$ or $y/3$, then we would have gotten the answer wrong if we had stopped here at $y = 5$.
Always double-check that your variable is correct and that you are answering the question the test is asking you to answer.
Now let’s try it again with our problem from earlier:
We have $7 + 3x = 22$ and we must isolate our variable in order to ultimately find $2x$
Step 1, combine like terms:
There are no like terms to combine, so we can skip step 1.
Step 2, isolate variable term:
$7 + 3x = 22$
$7 - 7 + 3x = 22 - 7$
$3x = 15$
Step 3, isolate variable:
$3x = 15$
${3x}/3 = 15/3$
$x = 5$
Step 4, double-check answer:
$7 + 3(5) = 22$
$7 + 15 = 22$
$22 = 22$
Success. But wait! We’re not done just yet.
Step 5, look at what the final question is asking:
We must finish the question by finding $2x$
$x = 5$
$2(5) = 10$
So our final answer is G, $2x = 10$
It may appear that performing a single variable equation requires a lot of steps, but the more you practice, the easier and more instinctive this process will become.
Test Your Knowledge
1)
2)
3)
4)
5)
Answers: C, G, B, G, E
Answer Explanations:
1) Ms. Lewis begins by driving 900 miles at 50 miles per hour and we want to find out how much faster she must go to travel the same amount of miles in three hours less time. Because she is driving the same amount, we can set these terms equal.
We are also only working with the variable of miles per hour, so this is a single variable equation.
Now, the two sides of the equation are dealing with miles and miles per hour. The first half of our equation will look like this:
$(900/50) - 3$
Why? Because Ms. Lewis is driving 900 miles at 50 miles per hour, so we need to divide the miles by mph in order to find out her travel time. And then we must reduce that amount by 3 because we are told that her new travel time will be 3 miles less than that.
This means that the other half of our equation will look like this:
$900/x$
Why? Because we know that the number of miles she drives will be the same, but our unknown is her miles per hour.
Now let's put them together and solve for our variable.
$(900/50) - 3 = 900/x$
$18 - 3 = 900/x$
$15 = 900/x$
Now we must isolate our $x$ value. Because it is acting as a denominator, we must multiply both sides of the equation by $x$.
$x * 15 = (900/x) * x$
$15x = 900$
Now, we can divide both sides by 15 in order to isolate our $x$ value.
$15x = 900$
${15x}/15 = 900/15$
$x = 60$
Finally, let us plug this value back into our original equation to double-check our answer.
$(900/50) - 3 = 900/x$
$(900/50) - 3 = 900/60$
$15 = 15$
We have successfully found our $x$ value, which is the new mileage per hour that Ms. Lewis must travel.
But wait, we're not done yet! The question asked us to find out how much faster she must drive, not the new miles per hour at which she must travel. This means we must take the difference of the original miles per hour and the new miles per hour.
$60 - 50 = 10$
She must drive 10 miles per hour faster in order to drive the same amount in three hours less time.
So our final answer is C, 10.
2) Here we have two cable companies and we are told that we must solve for when their rates are equal after an equal number of months. That means we have a single variable (the number of months) and we have an equation because we are setting each side equal (since the question specifies that their prices will be equal after an unknown number of months).
Uptown Cable has a flat fee of 120 dollars and an additional fee of 25 dollars per month. The flat fee will be unchanged (it only happens once), but the 25 dollars will be affected by the number of months. Since the number of months is our unknown variable, let’s give it a value of $x$.
So our first expression will look like this:
$120 + 25x$
Now Downtown Cable has a 60 dollar flat fee (occurs only once) and a 35 dollar per month fee. We are trying the find the equal number of months for a Downtown Cable package and an Uptown Cable package, so our variable, $x$, will remain the same. So our second expression will look like this:
$60 + 35x$
Now we set the two expressions equal to one another. (Why? Because we are told that the prices will be equal after a certain number of months.)
$120 + 25x = 60 + 35x$
Now we solve by shifting the terms on each side of the equation. First, let us combine our variable terms by subtracting 25x from each side.
$120 + 25x - 25x = 60 + 35x - 25x$
$120 = 60 + 10x$
Now, let us subtract 60 from each side.
$120 - 60 = 60 - 60 + 10x$
$60 = 10x$
And finally, let us isolate our variable.
$60/10 = {10x}/10$
$6 = x$
So our final answer is G, in exactly 6 months, the prices of each cable package will be equal.
3) This question relies on manipulating fractions. If this process is unfamiliar to you, definitely check out our guide to ACT fractions and ratios. If this is familiar to you, then let’s keep going.
${1/3}k + {1/4}k =1$
We must find a common denominator of the two fractions in order to combine our like terms. In this case, the least common factor of 3 and 4 is 12. (For more on this process, check out our guide to ACT fractions and ratios.)
${4/12}k + {3/12}k = 1$
${7/12}k = 1$
Now we have a number (7) being divided by another number (12). We know that division is the opposite of multiplication, so we must multiply each side by 12.
$12 * {7/12}k = 1 * 12$
$7k = 12$
And finally, we must divide each side by 7 to isolate our variable.
$7k = 12$
${7k}/7 = 12/7$
$k = 12/7$
So our final answer is B, $12/7$
4) We have a consultant with a flat (one time) fee of 30 dollars and an additional fee of 45 dollars per hour. Because the 45 dollars is hourly, it changes based on our variable (the number of hours). We do not know the number of hours she works, but we do know that her final earnings were 210 dollars. So let’s set this up as an equation.
$30 + 45x = 210$
There are no like terms, so we can start isolating our variable.
$30 - 30 + 45x = 210 - 30$
$45x = 180$
${45x}/45 = 180/45$
$x = 4$
So our final answer is G, she worked 4 hours to earn 210 dollars.
5) This is a single variable problem that can be solved in one of two ways--you can either distribute first and then solve, or you can solve without the need to distribute. We’ll go through both ways here.
Solve with distributing:
$9(x - 9) = -11$
First, distribute your 9 across the expression $(x - 9)$
$9(x) - 9(9) = -11$
$9x - 81 = -11$
Now, isolate your variable term as usual.
$9x - 81 + 81 = -11 + 81$
$9x = 70$
And finally, isolate your variable.
$9x = 70$
${9x}/9 = 70/9$
So our final answer is E, 70/9.
Alternatively, you can solve this problem without the need to distribute your 9 across the expression (x - 9)
Solve without distributing:
$9(x - 9) = -11$
Divide each side by 9
${9(x - 9)}/9 = -11/9$
$x - 9 = -11/9$
Now, we must add 9 to each side.
$x - 9 + 9 = -11/9 + 9$
$x = -11/9 + 9$
In order to add $-11/9$ and 9, we must give them a common denominator. Again, check out the guide on fractions and ratios if this process is unfamiliar to you.
$x = -11/9 + 9/1(9/9)$
$x = -11/9 + 81/9$
$x = 70/9$
So again, our answer is E, 70/9.
Phew! I think this calls for dessert.
The Take-Aways
Single variations make up the backbone of many other ACT problems. By knowing how to manipulate these kinds of expressions, you’ll be able to build on these techniques to solve much more complex problems and equations.
Just remember to always perform the same act to each side of the equation and save isolating your variable for last. Now take your single variable knowledge and conquer the rest of our math guides. You’ve got this.
What’s Next?
You’ve build up your mathematical foundation and now you’re raring to take on more. Before you start in on another guide to an ACT math topic, make sure you have a good idea of all the topics covered on the ACT math.
Think you might need a tutor? Check out the best ways to shop around for a tutor who suits your needs, whether online or in person.
Taken a practice test and don’t know how you match up for schools? Make sure you have a good idea of what your ideal score truly is.
And if you feel like you’ve got a handle on the math itself, but struggle with the timing, then be sure to check out on our article on how to stop running out of time on the ACT.
