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Inequalities on ACT Math: Strategies and Practice


Inequality questions come in a variety of shapes and forms on the ACT, but, no matter their form, you will see approximately three inequality questions on any given test. This means that inequality questions make up 5% of your overall ACT math test. Now, 5% of your test might not sound like a lot, but with only a quick brush-up on inequalities, that's an additional 5% of your questions that you're bound to rock!

This will be your complete guide to inequalities on the ACT: what they are, the different types of ACT math problems on inequalities, and how to solve them.


What Are Inequalities?

An inequality is a representation that two values are not equal or that two values are possibly not equal. There are different types of inequalities and different symbols to denote these different relationships.


≠ is the "unequal" sign. Whenever you see this sign, you know that two values are not equal, but nothing more. We don't know which value is greater or less than, just that they are not the same.

If we have $y ≠ x$, we do not know if $y$ is greater or less than $x$, just that they do not equal one another.


> is the "greater than" sign. Whichever number or variable is facing the opening of the sign is always the larger of the two values. (Some of you may have learned that the sign is a "crocodile" and that the crocodile always wants to eat the larger value).

For instance, $x > 14$ means that $x$ can be anything larger than 14 (it can even be 14.00000000001), but it cannot be 14 and it cannot be less than 14.


< is the "less than" sign. Whichever number is facing away from the opening of the sign is the lesser of the two values. This is just the greater than sign in reverse.

So $14 < x$ is the exact same equation we had earlier. $x$ must be larger than 14, 14 must be smaller than $x$.


≥ is the "greater than or equal to" sign. This acts exactly the same as the greater sign except for the fact that our values can also be equal.

Whereas $x > 14$ meant that $x$ could only be any number larger than 14, $x ≥ 14$ means that $x$ could be equal to 14 or could be any number larger than 14.


≤ is the "less than or equal to" sign. Just as the less than sign acted as a counter to the greater than sign, the less than or equal to sign acts counter to the greater than or equal to sign.

So $x ≥ 14$ is the exact same thing as saying $14 ≤ x$. Either way, we are saying that 14 is less than or equal to $x$, $x$ is greater than or equal to 14.


Each symbol describes the relationship between two values, but we can also link multiple values in a string. For instance, we can say:

$5 < x < 15$

This gives us both an upper and a lower limit on our $x$ value, because we know it must be both larger than five and less than 15. If we only had $5 < x$, the upper limit of $x$ would stretch into infinity, and the same with the lower limit if we only had $x < 15$.

For tips on how to keep track of which signs mean which, check out this article.



The inequality crocodile is always hungry for the most it can get, om nom nom.


How to Represent Inequalities

We can represent inequalities in one of three different ways:

  • A written expression

  • A number line

  • A graph

Let's look at all three.


Inequalities as written expressions use only mathematical symbols and no diagrams. They are exactly what we have been working with above (e.g., $y > 37$).


An inequality number line allows us to visualize the set of numbers that represents our inequality. We use a dark line to show all the numbers that match our inequality, and we mark where the inequality begins and/or ends in two different ways. To mark the beginning of an inequality that is "greater than" or "less than," we use an open circle. This shows that the starting number is NOT included.



To mark the beginning of an inequality that is "greater than or equal to" or "less than or equal to," we use a closed circle. This shows that the starting number IS included.



We can also combine these symbols if our inequality equation requires us to use two different symbols. For instance, if we have $-3 < x ≤ 3$, our number line would look like:




And finally, we can have inequalities in graphs for any and all types of graphs on the coordinate plane (more on the coordinate plane here). "Greater than" will be above the line of the graph, while "less than" will be below the line of the graph.




This is true no matter which direction the line of the graph extends.





In terms of markings, inequality graphs follow the same rules as inequalities on number lines. Just as we use an open circle for "greater than" or "less than" inequalities, we use a dashed line for inequality graphs that are "greater than" or "less than."



And just how we use a closed circle for "greater than or equal to" or "less than or equal to" inequalities, we use a solid line for our graphs that are greater or less than or equal to.



And now to dive right in to ACT inequality problems! (Awkward flailing optional).


Typical ACT Inequality Problems

There are three different types of inequality questions you'll see on the ACT, in the order from most to least common:

#1: Solve an inequality equation (find the solution set)

#2: Identify or answer questions about an inequality graph or number line

#3: Find alternate inequalities that fulfill given information

Let's look at each type—what they mean and how you'll see them on the ACT.


#1: Solving an Inequality Equation

This is by far the most common type of inequality question you'll see on the ACT. You will be given one or two inequality equations and must solve for the solution set of your variable.

Inequality problems work exactly the same way as a single variable equation and can be solved in the same way. Just think of the inequality sign as being the same as the equals sign. So you will perform the same actions (adding, subtracting, multiplying, and dividing) on each side.

For instance:

$9 + 12x > 45$

$12x > 36$

$x > 3$

The only difference between equations and inequalities is that the inequality sign flips if you multiply or divide each side by a negative.

For instance,

$10 - 4x < 50$

$-4x < 40$

$x > -10$

Because we had to divide each side by -4, we had to reverse the sign of the inequality.

Alternatively, we can also use the strategy of plugging in answers (PIA) or plugging in numbers (PIN) to solve our inequality problems. Because all ACT math problems are multiple choice, we can simply test out which answers match our equation (and which do not) or we can choose our own values for x based on the information we know, depending on the problem.

Let's look at an example of how this looks on the ACT, whether we solve the problem algebraically or by PIA.

The inequality $3(x+2)>4(x-3)$ is equivalent to which of the following inequalities?

F. $x<-6$
G. $x<5$
H. $x<9$
J. $x<14$
K. $x<18$


Solving Method 1: Algebra

First, distribute out the variable on each side.

$3(x + 2) > 4(x - 3)$

$3x + 6 > 4x - 12$

Now, we must isolate our variable just as we would with a single variable equation.

$6 > x - 12$

$18 > x$

Just as we saw back in our definitions, we know that we can also flip the inequality sign if we also switch the sides of our values. So $18 > x$ is the same as saying $x < 18$.

Our final answer is K, $x < 18$


Solving Method 2: Plugging in Answers

Though it will often take a little longer, we can also solve our inequality problems by testing out the values in our answer choices. Let's, as usual when using PIA, start with answer choice C.

Answer choice C says $x$ is less than 9, so let us see if this is true by saying that $x = 8$. If we plug in 8 for $x$ in the equation, we'll get:

$3(x + 2) > 4(x - 3)$

$3(8 + 2) > 4(8 - 3)$

$3(10) > 4(5)$

$30 > 20$

This is true, but that doesn't necessarily mean that it is the correct answer. Just because we know that $x$ can be equal to 8 or less doesn't mean it can't also be greater than 8. All we know for sure is that we can eliminate answer choices F and G, since we've problem that $x$ can be larger than each of them. So let us now go the opposite route and look at the highest value $x$ can be, given our answer choices.

Answer choice J gives us $x < 14$ and answer choice K says that $x < 18$, so what would happen is we gave $x$ a value between the two? Let us say that $x = 16$

$3(x + 2) > 4(x - 3)$

$3(16 + 2) > 4(16 - 3)$

$3(18) > 4(13)$

$54 > 52$

Because our inequality works for $x = 16$, we know that $x$ can be greater than $x < 14$ and can, therefore, be greater than all the answer choices except for answer choice K (the answer choice that gives us our largest possible value for $x$). This is enough to tell us that our final answer is K.

Our final answer is, again, K, $x < 18$


#2: Inequality Graph and Number Line Questions

For these types of questions, you will be asked to identify a graph or a number line from a given equation. Alternatively, you may be asked to infer information from a given inequality graph. Either way, you will always be given the graph on the coordinate plane.


We know that the sum of $x$ and $y$ must be greater than 1, so let us imagine that one of those two variables is equal to 0.

If we say that $x = 0$, then y alone has to be greater than 1 to make the sum of $x$ and $y$ still be greater than 1. We also know that we indicate that a value is "greater than" on a graph with a dashed line at the value in question and a filled in area above the value.

The only graph with a dashed line at $y = 1$ and that has a shaded area above this value is graph J. This means graph J is more than likely our answer, but let's confirm it just to be safe.

Because the sum of $x$ and $y$ must be greater than 1, the alternative possibility to $x = 0$ and $y > 1$ is that $y$ equals zero, so $x$ must be greater than 1. To show this, we would need a dashed line at $(1, 0)$ and a shaded area above it, all of which graph J has.

Now, to finish confirming that graph J is indeed our answer, we would simply do what we did to locate the lower limit of our graph in reverse so that we can find the upper limit. If $x + y < 2$, then, when $x = 0$, $y$ must be less than 2, and when $y = 0$, $x$ must be less than 2. This would give us dashed lines at $(0, 2)$ and $(2, 0)$, both of which are on graph J.

Our final answer is J.


#3: Finding Alternate Inequality Expressions

The rarest form of inequality questions on the ACT will ask you to use given inequalities and find alternate inequalities that must be true based off this given information.

Let's look at one of these in action, to better see how this type of question works.

If $x$ and $y$ are real numbers, such that $x>1$ and $y<-1$, then which of the following inequalities must be true?

A. $x/y>1$

B. $|x|^2>|y|$

C. $x/3-5>y/3-5$

D. $x^2+1>y^2+1$

E. $x^{-2}>y^{-2}$


There are two different ways we can solve this problem, by plugging in our own numbers or by working through it based on our logic and knowledge of algebra. We'll go through both methods here.


Solving Method 1: Plugging in Numbers (PIN)

Because we have a problem with multiple variables in both the problem and in the answer choices, we can make life a little easier and give our variables numerical values. Now, we do have to be careful when using this method, however, because there are infinite variables to choose from for both $x$ and $y$ and so more than one answer choice might work for any given variables we give to $x$ and $y$. If two or more answer choices work, we must simply pick new variables—eventually only the correct answer will be left, as it must work for ALL values of $x$ and $y$.

When it comes to picking our values for $x$ and $y$, we can also make life easy by picking values that are easy to work with. We know that we must divide both $x$ and $y$ by 3 in answer choice C, so let us pick values that are divisible by 3, and we know we must square our values in several answer choices, so let us pick numbers that are fairly small.

Now let's just say that $x = 6$ and $y = -9$ (Why those numbers? So long as they fulfill the given information—and they do—then why not!)

And let us plug these values into our answer choices.

Answer choice A gives us:

$x/y > 1$

If we plug in our values, we get:



This is NOT greater than 1, so we can eliminate answer choice A.


Answer choice B gives us:

$|x|^2 > |y|$

If we plug in our values, we get:

$|6|^2 > |-9|$

$36 > 9$

This is correct, so we will keep answer option B in the running for right now.


Answer choice C gives us:

$x/3 - 5 > y/3 - 5$

If we plug in our values, we get:

$6/3 - 6 > {-9}/3 - 5$

$2 - 6 > -3 - 5$

$-4 > -8$

This is correct, so we will keep answer option C in the running for now as well. Because B and C are both correct, we will need to come back and test them both again with different values later.


Answer choice D gives us:

$x^2 + 1 > y^2 + 1$

$6^2 + 1 > -9^2 + 1$

$36 + 1 > 81 + 1$

$37 > 82$

This is NOT true, so we can eliminate answer choice D.


Answer choice E gives us:

$x^{-2} > y^{-2}$

$6^{-2} > -9^{-2}$

$1/{6^2} > 1/{-9^2}$

$1/36 > 1/81$

Now this is indeed true, but what if we had chosen different values for x and y? Let's say that we said $x = 9$ and $y = -6$ instead (remember—so long as the numbers fit with the given information, we can use any values we like).

$x^{-2} > y^{-2}$

$9^{-2} > -6^{-2}$

$1/{9^2} > 1/{-6^2}$

$1/81 > 1/36$

Whoops! Answer choice E is no longer correct, which means we can eliminate it. We are looking for the answer choice that is always true, so it cannot possibly be answer E.


Now we are left with answer choices B and C. Let's look at them each again.

While we saw that our values for $x$ and $y$ meant that answer choice B was indeed true, let's see what would happen if we choose a much smaller value for $y$. Nothing is stopping us from choosing -6,000 for $y$—remember, all that we are told is that $y < -1$. So let us use $y = -6,000$ instead.

$|x|^2 > |y|$

$|6|^2 > |-6,000|$

$36 > 6,000$

This inequality is NOT true anymore, which means we can eliminate answer choice B.


This means that answer choice C must be the right answer by default, but let's test it to make absolutely sure.

Let us try what we did with answer option E and reverse the absolute values of our $x$ and $y$. So instead of $x = 6$ and $y = -9$, we will say that $x = 9$ and $y = -6$.

$x/3 - 5 > y/3 - 5$

$9/3 - 5 > {-6}/3 - 5$

$3 - 5 > -2 - 5$

$-2 > -7$

No matter how many numbers we choose for $x$ and $y$, answer choice C will always be correct.

Our final answer is C, $x/3 - 5 > y/3 - 5$


Solving Method 2: Algebraic Logic

As we can see, using PIN was successful, but required a good deal of time and trial and error. The alternative way to solve the problem is by thinking of how negatives and positives work and how exponents and absolute values alter these rules.

We know that $x$ must be positive and $y$ must be negative to fulfill the requirements $x > 1$ and $y < -1$. Now let us look through our answer choices to see how these expressions are affected by the idea that $x$ must always be positive and $y$ must always be negative.


Answer choice A gives us:

$x/y > 1$

We know that any fraction with a positive numerator and a negative denominator will be negative. And any negative number is less than 1.

Answer choice A can never be correct.


Answer choice B gives us:

$|x|^2 > |y|$

An absolute value means that the negative sign on $y$ has been negated, so this might be correct. But y can be any number less than -1, which means its absolute value could potentially be astronomically large, and $x$ can be any number greater than 1, which means its absolute value might be comparatively tiny.

This means that answer choice B is not always correct, which is enough to eliminate it from the running.


Answer choice C gives us:

$x/3 - 5 > y/3 - 5$

Now let's look at each side of the inequality.

We know that any fraction with a positive number in both the numerator and in the denominator will give us a positive value. This means we will have some positive value minus 5 on the left side.

We also know that any time we have a negative value in the numerator and a positive value in the denominator, we will have a negative fraction. This means we will have some negative value minus 5 on the right side. We also know that a negative plus a negative will give us an even greater negative (a smaller value).

If we put this information together, we know that the left side may or may not be a negative value, depending on the value of $x$, but the right side will only get more and more negative. In other words, no matter what values we give to $x$ and $y$, the left side will always be greater than the right side, which means the expression is always true.

Now this should be enough for us to select our right answer as C, but we should give a look to the other answer choices just in case.


Answer choice D gives us:

$x^2 + 1 > y^2 + 1$

We know that if we square both a positive number and a negative number, we will get a positive result, so the negative value for $y$ is no longer in play. This inequality will therefore be true if the absolute value of $x$ is greater than the absolute value of $y$ (e.g., $x = 10$ and $y = -9$), but it won't be true if the absolute value of $y$ is greater than the absolute value of $x$ (e.g., $x = 9$, $y = -10$).

This means that the inequality will sometimes be true, but not always, which is enough to eliminate it.


Finally, answer choice E gives us:

$x^{-2} > y^{-2}$

We know that a number to a negative exponent is equal to 1 over that number to the positive exponent (e.g., $5^{-3} = 1/{5^3}$). This means that each value will be a fraction of 1 over the square of our $x$ and $y$ values.

This will give us two positive fractions and $1/{x^2}$ will only be larger if the absolute value of $x$ is smaller than the absolute value of $y$. But, because our $x$ and $y$ values can be anything so long as $y$ is negative and $x$ is positive, this will only sometimes be true. We can therefore eliminate answer choice E.

This leaves us with only answer choice C that is always true.

Our final answer is C, $x/3 - 5 > y/3 - 5$

body_strategy-8"Win a war," "Rock the ACT"—we'd say the two are basically one and the same.


ACT Math Strategies for Inequality Problems

Though there are a few different types of inequality problems, there are a few strategies you can follow to help you solve them most effectively.


#1: Write Your Information Down and Draw It Out

Many problems on the ACT, inequalities included, appear easier or less complex than they actually are and can lead you to fall for bait answers. This illusion of ease may tempt you to try to solve inequality questions in your head, but this is NOT the way to go.

Take the extra moment to work your equations out on the paper or even draw your own diagrams (or draw on top of the diagrams you're given). The extra few seconds it will take you to write out your problems are well worth the points you'll gain by taking the time to find the right answer.


#2: Use PIN (or PIA) When Necessary

If all you know about $x$ is that it must be more than 7, go ahead and pick a value for $\bi x$. This will help you more easily visualize and work through the rest of the problem, since it is generally always easier to work with numbers than it is to work with variables.

As you use this strategy, the safest bet is to choose two values for your variable—one that is close to the definition value and one that is very far away. This will allow you to see whether the values you chose work in all instances.

For instance, if all you know is $x > 7$, it's a good idea to work through the problem once under the assumption that $x = 8$ and another time under the assumption that $x = 400$. If the problem must be true for all values $x > 7$, then it should work for all numbers of $x$ greater than 7.


#3: Keep Very Careful Track of Your Negatives

One of the key differences between inequalities and single variable equations is in the fact that the inequality sign is reversed whenever you multiply or divide both sides by a negative. And you can bet the house that this is what the ACT will try to test you on again and again.

Though the ACT is not engineered to trick you, the test-makers are still trying to challenge you and test whether or not you know how to apply key mathematical concepts. If you lose track of your negatives (an easy thing to do, especially if you're working in your head), you will fall for one of the bait answers. Keep a keen eye.


#4: Double-Check Your Answer by Working Backwards (Optional)

If you feel unsure about your answer for any reason (because so many of the answer choices look the same, because you're not sure if you handled the issue of negative numbers correctly, etc.), you can work backwards to see if your expression is indeed correct.

For instance, let us look at the inequality we had earlier, when talking about the function of negatives on inequalities:

$10 - 4x < 50$

Again, we would go through this just as we would a single variable equation.

$-4x < 40$

$x > -10$

But now maybe that answer doesn't feel right to you or you just want to double-check to be sure. Well, if we're told that $x$ must be greater than -10 to fulfill the inequality, let's make sure that this is true.

Let us solve the expression with $x = -9$ and see if we are correct.

$10 - 4x < 50$

$10 - 4(-9) < 50$

$10 + 36 < 50$

$46 < 50$

This is correct, so that's promising. But we found that $x$ needed to be greater than -10, so our expression should also be INCORRECT if $x$ were equal to -10 or if $x$ were less than -10. So let us see what happens if we have $x = -10$.

$10 - 4x < 50$

$10 - 4(-10) < 50$

$10 + 40 < 50$

$50 < 50$

The inequality is no longer correct. This means that we know for certain that the solution set we found, $x > -10$ is true.

You will always be able to work backwards in this way to double-check your inequality questions. Though this can take a little extra time, it might be worth your peace of mind to do this whenever you feel unsure about your answers.



Ready, set? It's test time!


Test Your Knowledge

Now let's put all that inequality knowledge to the test on some real ACT math problems.


The inequality $6(x+2)>7(x-5)$ is equivalent to which of the following inequalities?
A. $x<-23$
B. $x<7$
C. $x<17$
D. $x<37$
E. $x<47$





If $r$ and $s$ can be any integers such that $s>10$ and $2r+s=15$, which of the following is the solution set for $r$?

A. $r≥3$
B. $r≥0$
C. $r≥2$
D. $r≤0$
E. $r≤2$



Which of the following is the solution statement for the inequality shown below?


F. $-5<x<10$
G. $-3<x$
H. $-3<x<2$
J. $-2<x<3$
K. $x<-3$ or $x>2$




Answers: E, E, E, H, D


Answer Explanations

1. This is a standard inequality equation, so let us go through our solve accordingly. First, let's begin by distributing out our equation.

$6(x + 2) > 7(x - 5)$

$6x + 12 > 7x - 35$

$12 > x - 35$

$47 > x$

Because we did not have to divide or multiply by a negative, we were able to keep the inequality sign intact. And because the expression $47 > x$ and $x < 47$ mean the same thing, we can see that this matches one of our answer choices.

Our final answer is E, $x < 47$


2. We are given two graphs with equations attached and we must identify when one equation/graph is less than the other. We don't even have to know anything about what these equations means and we do not have to fuss with solving the equations—we can simply look at the diagram.

The only place on the diagram where the graph of $y = (x - 1)^4$ is less than (aka lower than) the graph of $y = x - 1$ is between $x = 1$ and $x = 2$ on the coordinate plane.


In other words, this inequality is true when $x > 1$ and when $x < 2$, or $1 < x < 2$.

Our final answer is E, $1 < x < 2$.


3. We know that $s > 10$ and it must be an integer, so let us make life easy and just say that $s = 11$. Now we can use this number to plug into the equation.

$2r + s = 15$

$2r + 11 = 15$

$2r = 4$

$r = 2$

We know that $r$ can be equal to 2 and that it is the nearest integer to our definition. This means that our answer will either be C or E. So let us now find which direction our inequality sign must face.

Let's now try one integer larger than 11 to see whether our solution set must be less or equal to 2 or greater than or equal to 2. If we say that $s = 12$, then our equation becomes:

$2r + s = 15$

$2r + 12 = 15$

$2r = 3$

$r = 1.5$

We can see now that, as $s$ increases, $r$ will decrease. This means our solution set will be that $r$ is equal to or less than 2.

Our final answer is E, $r≤ 2$


4. Though this problem is made slightly more complex due to the fact that it is a double inequality expression, we still solve the inequality the same way we normally would.

$-5 < 1 - 3x < 10$

If we think of this expression as two different inequality equations, we would say:

$-5 < 1 - 3x$


$1 - 3x < 10$

So let us solve each of them.

$-5 < 1 - 3x$

$-6 < -3x$

Because we now must divide by a negative, we must reverse the inequality sign.

$2 > x$

And now let's solve our second expression:

$1 - 3x < 10$

$-3x < 9$

Again, we must reverse our inequality sign, since we need to divide each side by a negative.

$x > -3$

Now, if we put the two results together, our expression will be:

$-3 < x < 2$

Our final answer is H, $-3 < x < 2$


5. Because we have a number line with two closed circles, we know that must use less than or equal to and greater than or equal to signs.

We can see that the right side of the graph gives us a set of numbers equal to or greater than 3, which means:

$x ≥ 3$

The left side of the graph gives us a set of numbers less than or equal to -1, which means:

$x ≤ -1$

Our final answer is, therefore, D, $-1 ≥ x$ and $x ≤ 3$.


body_cute-1And now, your reward for solving your inequality problems is oodles of Cuteness.


The Take-Aways

Inequalities are so similar to single variable equations that it can be easy to treat the two as the same. The test-makers know this, so it pays to be careful when it comes to your inequality questions. Remember the key differences (multiplying or dividing by a negative reverses the sign, and you can flip your inequality signs so long as you flip both sides of the expression) and keep careful track of the details to avoid all the common pitfalls and bait answers.

After you've mastered the art of answering your inequality questions, that's another 5% of the test that you've dominated. You're well on your way to that score goal of yours now!


What's Next?

Want to brush up on any of your other math topics? Check out our individual math guides to get the walk-through on each and every topic on the ACT math test.

Been procrastinating on your ACT studying? Learn how to get over your desire to procrastinate and make a well-balanced study plan.

Running out of time on the ACT math section? We'll teach you how to beat the clock and maximize your ACT math score.

Looking to get a perfect score? Check out our guide to getting a perfect 36 on ACT math, 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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