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Systems of Equations on ACT Math: Algebra Strategies and Practice Problems

Posted by Courtney Montgomery | Sep 16, 2015 6:30:00 PM

ACT Math

 

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If you’ve already tackled your single variable equations, then get ready for systems of equations. Multiple variables! Multiple equations! (Whoo!) Even better, systems of equations questions will always have multiple methods with which to solve them, depending on how you like to work best.

So let us look not only at how systems of equations work, but all the various options you have available to solve them.

This will be your complete guide to systems of equations questions--what they are, the many different ways for solving them, and how you’ll see them on the ACT.


Before You Continue

You will never see more than one systems of equations question per test, if indeed you see one at all. Remember that quantity of questions answered (as accurately as possible) is the most important aspect of scoring well on the ACT, because each question is worth the same amount of points.

This means that you should prioritize understanding the more fundamental math topics on the ACT, like integers, triangles (coming soon!), and slopes. If you can answer two or three integer questions with the same effort as you can one question on systems of equations, it will be a better use of your time and energy.

With that in mind, the same principles underlying how systems of equations work are the same for other algebra questions on the test, so it is still a good use of your time to understand how they work.

 

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Let's go tackle some systems questions, then! Whoo!

 

What Are Systems of Equations?

Systems of equations are a set of two (or more) equations that have two (or more) variables. The equations relate to one another, and each can be solved only with the information that the other provides.

Most of the time, a systems of equations question on the ACT will involve two equations and two variables. It is by no means unheard of to have three or more equations and variables, but systems of equations are rare enough already and ones with more than two equations are even rarer than that.

It is possible to solve systems of equations questions in a multitude of ways. As always with the ACT, how you chose to solve your problems mostly depends on how you like to work best as well as the time you have available to dedicate to the problem.

The three methods to solve a system of equations problem are:

  1. Graphing

  2. Substitution

  3. Subtraction

Let us look at each method and see them in action by using the same system of equations as an example.

For the sake of our example, let us say that our given system of equations is:

$$3x + 2y = 44$$

$$6x - 6y = 18$$


Solving Method 1 - Graphing

In order to graph our equations, we must first put each equation into slope-intercept form. If you are familiar with your lines and slopes, you know that the slope-intercept form of a line looks like:

$y = mx + b$

If a system of equations has one solution (and we will talk about systems that do not later in the guide), that one solution will be the intersection of the two lines.

So let us put our two equations into slope-intercept form.

$3x + 2y = 44$

$2y = -3x + 44$

$y = {-3/2}x + 22$

And

$6x - 6y = 18$

$-6y = -6x + 18$

$y = x - 3$

Now let us graph each equation in order to find their point of intersection.


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Once we graphed our equation, we can see that the intersection is at (10, 7).

So our final results are $x = 10$ and $y = 7$


Solving Method 2 - Substitution

Substitution is the second method for solving a system of equations question. In order to solve this way, we must isolate one variable in one of the equations and then use that found variable for the second equation in order to solve for the remaining variable.

This may sound tricky, so let's look at it in action. For example, we have our same two equations from earlier,

$$3x + 2y = 44$$

$$6x - 6y = 18$$

So let us select just one of the equations and then isolate one of the variables.

In this case, let us chose the second equation and isolate our $y$ value. (Why that one? Why not!)

$6x - 6y = 18$

$-6y = -6x + 18$

$y = x - 3$

Next, we must plug that found variable into the second equation. (In this case, because we used the second equation to isolate our $y$, we need to plug in that $y$ value into the first equation.)

$3x + 2y = 44$

$3x + 2(x - 3) = 44$

$3x + 2x - 6 = 44$

$5x = 50$

$x = 10$

And finally, you can find the numerical value for your first variable ($y$) by plugging in the numerical value you found for your second variable ($x$) into either the first or the second equation.

$3x + 2y = 44$

$3(10) + 2y = 44$

$30 + 2y = 44$

$2y = 14$

$y = 7$

Or

$6x - 6y = 18$

$6(10) - 6y = 18$

$60 - 6y = 18$

$-6y = -42$

$y = 7$

Either way, you have found the value of both your $x$ and $y$.

Again, $x = 10$ and $y = 7$


Solving Method 3 - Subtraction

Subtraction is the last method for solving our systems of equations questions. In order to use this method, you must subtract out one of the variables completely so that you can find the value of the second variable.

Do take note that you can only do this if the variables in question are exactly the same. If the variables are NOT the same, then we can first multiply one of the equations--the entire equation--by the necessary amount in order to make the two variables the same.

In the case of our two equations, none of our variables are equal.

$$3x + 2y = 44$$

$$6x - 6y = 18$$

We can, however, make two of them equal. In this case, let us decide to subtract our $x$ values and cancel them out. This means that we must first make our $x$’s equal by multiplying our first equation by 2, so that both $x$ values match. So:

$3x + 2y = 44$

$6x - 6y = 18$

Becomes:

$2(3x + 2y = 44)$ => $6x + 4y = 88$ (The entire first equation is multiplied by 2.)

And

$6x - 6y = 18$ (The second equation remains unchanged.)

Now we can cancel out our $y$ values by subtracting the entire second equation from the first.

$6x + 4y = 88$

-

$6x - 6y = 18$

--------------------

$4y - -6y = 70$

$10y = 70$

$y = 7$

Now that we have isolated our $y$ value, we can plug it into either of our two equations to find our $x$ value.

$3x + 2y = 44$

$3x + 2(7) = 44$

$3x + 14 = 44$

$3x = 30$

$x = 10$

Or

$6x - 6y = 18$

$6x - 6(7) = 18$

$6x - 42 = 18$

$6x = 60$

$x = 10$

Our final results are, once again, $x = 10$ and $y = 7$.

 

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If this is all unfamiliar to you, don't worry about feeling overwhelmed! It may seem like a lot right now, but, with practice, you'll find the solution method that fits you best.

 

No matter which method we use to solve our problems, a system of equations will either have one solution, no solution, or infinite solutions.

In order for a system of equations to have one solution, the two (or more) lines must intersect at one point so that each variable has one numerical value.

In order for a system of equations to have infinite solutions, each system will be identical. This means that they are the same line.

And, in order for a system of equations to have no solution, the $x$ values will be equal when the $y$ values are each set to 1. This means that, for each equation, both the $x$ and $y$ values will be equal. The reason this results in a system with no solution is that it gives us two parallel lines. The lines will have the same slope and never intersect, which means there will be no solution.

For instance,

For which value of $a$ will there be no solution for the systems of equations?

$2y - 6x = 28$

$4y - ax = 28$

  1. -12

  2. -6

  3. 3

  4. 6

  5. 12

We can, as always use multiple methods to solve our problem. For instance, let us first try subtraction.

We must get the two $y$ variables to match so that we can eliminate them from the equation. This will mean we can isolate our $x$ variables to find the value of our $a$.

So let us multiply our first equation by 2 so that our $y$ variables will match.

$2(2y - 6x = 28)$ => $4y - 12x = 56$

Now, let us subtract our equations

$4y - 12x = 56$

-

$4y - ax = 28$

----------------------

$-12x - -ax = 28$

We know that our $-12x$ and our $-ax$ must be equal, since they must have the same slope (and therefore negate to 0), so let us equate them.

$-12x = -ax$

$a = 12$

$a$ must equal 12 for there to be no solution to the problem.

Our final answer is E, 12.

If it is frustrating or confusing to you to try to decide which of the three solving methods “best” fits the particular problem, don’t worry about it! You will almost always be able to solve your systems of equations problems no matter which method you choose.

For instance, for the problem above, we could simply put each equation into slope-intercept form. We know that a system of equations question will have no solution when the two lines are parallel, which means that their slopes will be equal.

Begin with our givens,

$2y - 6x = 28$

$4y - ax = 28$

And let’s take them individually,

$2y - 6x = 28$

$2y = 6x + 28$

$y = 3x + 14$

And

$4y - ax = 28$

$4y = ax + 28$

$y = {a/4}x + 7$

We know that the two slopes must be equal, so we will find $a$ by equating the two terms.

$3 = a/4$

$12 = a$

Our final answer is E, 12.

As you can see, there is never any “best” method to solve a system of equations question, only the solving method that appeals to you the most.


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Some paths might make more sense to you, some might seem confusing or cumbersome. Either way, you will be able to solve your systems questions no matter what route you choose.

 

Typical Systems of Equations Questions

There are essentially two different types of system of equations questions you’ll see on the test. Let us look at each type.

Equation Question

As with our previous examples, many systems of equations questions will be presented to you as actual equations.

The question will almost always ask you to find the value of a variable for one of three types of solutions--the one solution to your system, for no solution, or for infinite solutions.

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(We will work through how to solve this question later in the guide.)


Word Problems

You may also see a systems of equations question presented as a word problem. Often (though not always), these types of problems on the ACT will involve money in some way.

In order to solve this type of equation, you must first define and write out your system so that you can solve it.

For instance,

A movie ticket is 4 dollars for children and 9 dollars for adults. Last Saturday, there were 680 movie-goers and the theater collected a total of 5,235 dollars. How many movie-goers were children on Saturday?

  1. 88

  2. 112

  3. 177

  4. 368

  5. 503

First, we know that there were a total of 680 movie-goers, made up of some combination of adults and children. So:

$a + c = 680$

Next, we know that adult tickets cost 9 dollars, children’s tickets cost 4 dollars, and that the total amount spent was 5,235 dollars. So:

$9a + 4c = 5,235$

Now, we can, as always, use multiple methods to solve our equations, but let us use just one for demonstration. In this case, let us use substitution so that we can find the number of children who attended the theater.

If we isolate our $a$ value in the first equation, we can use it in the second equation to solve for the total number of children.

$a + c = 680$

$a = 680 - c$

So let us plug this value into our second equation.

$9a + 4c = 5,235$

$9(680 - c) + 4c = 5235$

$6120 - 9c + 4c = 5235$

$-5c = -885$

$c = 177$

177 children attended the theater that day.

Our final answer is C, 177.


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You know what to look for and how to use your solution methods, so let's talk strategy.

Strategies for Solving Systems of Equations Questions

All systems of equations questions can be solved through the same methods that we outlined above, but there are additional strategies you can use to solve your questions in the fastest and easiest ways possible.


1) To begin, isolate or eliminate the opposite variable that you are required to find

Because the goal of most ACT systems of equations questions is to find the value of just one of your variables, you do not have to waste your time finding ALL the variable values. The easiest way to solve for the one variable you want is to either eliminate your unwanted variable using subtraction, like so:

Let us say that we have a systems problem in which we are asked to find the value of $y$. 

$$4x + 2y = 20$$

$$8x + y = 28$$

If we are using subtraction, let us eliminate the opposite value that we are looking to find (namely, $x$.)

$4x + 2y = 20$

$8x + y = 28$

First, we need to set our $x$ values equal, which means we need to multiply the entire first equation by 2. This gives us:

$8x + 4y = 40$

-

$8x + y = 28$

-------------------

$3y = 12$

$y = 4$

Alternatively, we can isolate the opposite variable using substitution, like so:

$4x + 2y = 20$

$8x + y = 28$

So that we don't waste our time finding the value of $x$ in addition to $y$, we must isolate our $x$ value first and then plug that value into the second equation. 

$4x + 2y = 20$

$4x = 20 - 2y$

$x = 5 - {1/2}y$

Now, let us plug this value for $x$ into our second equation.

$8x + y = 28$

$8(5 - {1/2}y) + y = 28$

$40 - 4y + y = 28$

$-3y = -12$

$y = 4$

As you can see, no matter the technique you choose to use, we always start by isolating or eliminating the opposite variable we want to find. 


2) Practice all three solving methods to see which one is most comfortable to you

You’ll discover the solving method that suits you the best when it comes to systems of equations once you practice on multiple problems. Though it is best to know how to solve any systems question in multiple ways, it is completely okay to pick one solving method and stick with it each time.

When you test yourself on systems questions, try to solve each one using more than one method in order to see which one is most comfortable for you personally.


3) Look extra carefully at any ACT question that involves dollars and cents

Many systems of equations word problem questions are easy to confuse with other types of problems, like single variable equations or equations that require you to find alternate expressions. A good rule of thumb, however, is that it is highly likely that your ACT math problem is a system of equations question if you are asked to find the value of one of your variables and/or if the question involves money in some way.

Again, not all money questions are systems of equations and not all systems of equation word problem questions involve money, but the two have a high correlation on the ACT. When you see a dollar sign or a mention of currency, keep your eyes sharp.



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Ready to tackle your systems problems?

 

Test Your Knowledge

Now let us test your system of equation knowledge on more ACT math questions.

1) The sum of real numbers $a$ and $b$ is 20 and their difference is 6. What is the value of $ab$?

A. 51

B. 64

C. 75

D. 84

E. 91


2)

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3) What is the value of $x$ in the following systems of equations?

$$3x - 2y - 7 = 18$$

$$-x + y = -8$$

A. -1

B. 3

C. 8

D. 9

E. 18


Answers: E, B, D

Answer Explanations:

1) We are given two equations involving the relationship between $a$ and $b$, so let us write them out. 

$a + b = 20$

$a - b = 6$

(Note: we do not actually know which is larger--$a$ or $b$. But also notice that it doesn't actually matter. Because we are being asked to find the product of $a$ and $b$, it does not matter if $a$ is the larger of the two numbers or if $b$ is the larger of the two numbers; $a * b$ will be the same either way.)

Now, we can use whichever method we want to solve our systems question, but for the sake of space and time we will only choose one. In this case, let us use substitution to find the value of one of our variables. Let us begin by isolating $a$ in the first equation. 

$a + b = 20$

$a = 20 - b$

Now let's replace this $a$ value in the second equation. 

$a - b = 6$

$(20 - b) - b = 6$

$-2b = -14$

$b = 7$

Now we can replace the value of $b$ back into either equation in order to find the numerical value for $a$. Let us do so in the first equation. 

$a + b = 20$

$a + 7 = 20$

$a = 13$

We have found the numerical values for both our unknown variables, so let us finish with the final step and multiply them together. 

$a = 13$ and  $b = 7$

$(13)(7)$

$91$

Our final answer is E, 91. 

 

2) We know that a system has infinite solutions only when the entire system is equal. Right now, our coefficients (the numbers in front of the variables) for $x$ and $y$ are not equal, but we can make them equal by multiplying the first equation by 3. That way, we can transform this pairing:

$2x - y = 8$

$6x - 3y = 4a$

Into:

$6x - 3y = 24$

$6x - 3y = 4a$

Now that we have made our $x$ and $y$ values equal, we can set our variables equal to one another as well. 

$24 = 4a$

$a = 6$

In order to have a system that has infinite solutions, our $a$ value must be 6. 

Our final answer is B, 6. 

 

3) Before we decide on our solving method, let us combine all of our similar terms. So,

$3x - 2x - 7 = 18$ => $3x - 2y = 25$

Now, we can again use any solving method we want to, but let us choose just one to save ourselves some time. In this case, let us use subtraction. 

So we have:

$3x - 2y = 25$

$-x + y = -8$

Because we are being asked to find the value of $x$, let us subtract out our $y$ values. This means we must multiply the second equation by 2.

$2(-x + y = -8)$

$-2x + 2y = -16$

Now, we have a $-2y$ in our first equation and a $+2y$ in our second, which means that we will actually be adding our two equations instead of subtracting them. (Remember: we are trying to eliminate our $y$ variable completely, so it must become 0.)

$3x - 2y = 25$

+

$-2x + 2y = -16$

---------------------

$x = 9$

We have successfully found the value for $x$. 

Our final answer is D, 9. 

 

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Good job! The tiny turtle is proud of you.

 

The Take Aways

As you can see, there is a veritable cornucopia of ways to solve your systems of equations problems, which means that you have the ability to be flexible with them more than many other types of problems. So take heart that your choices are many for how to proceed, and practice to learn the method that suits you the best.


What’s Next?

Ready to take on more math topics? Of course you are! Luckily, we've got your back, with math guides on all the different math topics you'll see on the ACT. From circles to polygons, angles to trigonometry, we've got guides for your needs. 

Bitten by the procrastination bug? Learn why you're tempted to procrastinate and how to beat the urge

Want to skip to the most important math guides? If you only have time to tackle a few articles, take a look at two of the most important math strategies for improving your math score--plugging in answers and plugging in numbers. Knowing these strategies will help you take on some of the more challenging questions on the ACT in no time. 

Looking to get a perfect score? Check out our guide to getting a 36 on the ACT math section, written by a perfect-scorer. 



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Courtney Montgomery
About the Author

Courtney scored in the 99th percentile on the SAT in high school and went on to graduate from Stanford University with a degree in Cultural and Social Anthropology. She is passionate about bringing education and the tools to succeed to students from all backgrounds and walks of life, as she believes open education is one of the great societal equalizers. She has years of tutoring experience and writes creative works in her free time.



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