Cracking the GRE with 4 Practice Tests, 2016 Edition

Part III. How to Crack the Math Section

Chapter 11. Algebra (And When to Use It)

The basics for math on the GRE are often used in the context of algebra. While comfort with algebraic operations is a good skill to have, Plugging In numbers in lieu of doing the algebra is often a much faster way of getting the correct answer. This chapter provides an introduction to Plugging In instead of doing algebra, as well as strategies for Plugging In The Answers, how to deal with exponents and square roots, and how to manipulate equations, inequalities, quadratic equations, and simultaneous equations.


Many of the hardest questions you might encounter on the GRE involve algebra. Algebra questions are generally difficult for two reasons. First, they are often complicated, multistep problems. Second, ETS studies the types of mistakes that people make when they solve questions using algebra. They generate wrong answers for the questions based on these common algebraic errors. So, if you aren’t careful, you can make an algebraic mistake and still find your answer among the choices.

If you are one of the many students who take the GRE and struggle with Algebra, you’re in luck. Plugging In is a strategy that will make even the hardest, messiest GRE algebra problem and turn it into an arithmetic problem.

Let’s look at an example of how Plugging In can make a seemingly messy algebra problem much easier to work with.

Why Plug In?

Plugging In is a powerful tool that can greatly enhance your math score, but you may be wondering why you should plug in when algebra works just fine. Here’s why:

Plugging In converts algebra problems into arith-metic problems. No matter how good you are at algebra, you’re better at arithmetic. Why? Because you use arithmetic every day, every time you go to a store, balance your checkbook, or tip a waiter. Chances are you rarely use algebra in your day-to-day activities.

Plugging In is oftentimes more accurate than algebra. When you plug in real numbers, you make the problems concrete rather than abstract. Once you’re working with real numbers, it’s easier to notice when and where you’ve messed up a calculation. It’s much harder to see where you went wrong (or to even know you’ve done something wrong) when you’re staring at a bunch of x’s and y’s.

The GRE allows the use of a calculator. A calculator can do arithmetic but it can’t do algebra, so Plugging In allows you to take advantage of the calculator function.

ETS expects its students to attack the problems algebraically and many of the tricks and the traps built into the problem are designed to catch students who do the problems with algebra. By Plugging In, you’ll avoid these pitfalls.

As you can see, there are a number of excellent reasons for Plugging In. Mastering this technique can have a significant impact on your score.

Dale gives Miranda x bottles of water. He gives Marcella two fewer bottles of water than he gives to Miranda and he gives Mary three more bottles of water than he gives to Marcella. How many bottles of water did Dale give to Miranda, Marcella and Mary, in terms of x ?

    3x − 1


    3x + 1

    3x + 2

    x − 2

Here’s How to Crack It

This problem can definitely be solved using algebra. However, the use of Plugging In makes this problem much easier to solve. The problem has one variable in it, x, so start plugging in by picking a number for x. An easy number to use would be 10, so use your scratch paper and write down x = 10. Now read the problem again and follow the directions, only this time do the arithmetic instead of the algebra on the scratch paper. So, Miranda gets 10 bottles of water. The problem then states that Marcella gets two fewer bottles of water than Miranda, so Marcella gets 8 bottles. Next, Mary gets three more bottles than Marcella, so Mary gets 11 bottles. That’s a total of 10 + 8 + 11 = 29 bottles of water. The problem asks for how many bottles of water did Dale give to Miranda, Marcella, and Mary, so the answer to the question is 29 bottles of water. This is the target answer, which should always be circled on the scratch paper so you don’t forget it. Now plug in 10 for the variable x in all the answer choices and see which answer choice equals 29. Be sure to check all five answer choices.


3(10) – 1 = 29

Looks good!


3(10) = 30



3(10) + 1 = 31



3(10) + 2 = 32



10 – 2 = 8


The correct answer to this question is (A), and if you successfully completed the algebra you would have gotten the same answer. Pretty easy compared to the algebra, huh?

As you can see, Plugging In turned this algebra problem into an arithmetic problem. The best news is that you can solve any problem with variables by using Plugging In.

Here are the steps:

Step 1:

Recognize the opportunity. See variables in the problem and answer choices? Get ready to Plug In. The minute you see variables in a question or answer choices, you should start thinking about opportunities to Plug In.

Step 2:

Set up the scratch paper. Plugging In is designed to make your life easier. Why make it harder again by trying to solve problems in your head? You are not saving any notable amount of time by trying to work out all the math without writing it down, so use the scratch paper. Even if it seems like an easy question of translating a word problem into an algebraic equation, remember that there are trap answer choices. Whenever you recognize the opportunity to Plug In, set up the scratch paper by writing answer choices A-E down before you start to solve.

Step 3:

Plug In. If the question asks for “x apples,” come up with a number for x. The goal here is to make your life easier, so plugging in numbers such as 2, 3, 5, 10, 100 are all good strategies. However, for the first attempt at Plugging In on any given problem, avoid the numbers 1 or 0. These numbers can oftentimes create a situation where more than one answer choice produces the target answer. If you Plug In a number and the math starts getting difficult (for example, you start getting fractions or negative numbers), don’t be afraid to just change the number you plug in.

Step 4:

Solve for the Target. The Target is the value the problem asks you to solve for. Remember to always circle the Target so you don’t forget what it is you are solving for.

Step 5:

Check All The Answer Choices. Anywhere you see a variable, plug in the number you have written down for that variable and do the arithmetic. The correct answer is the one that matches the Target. If more than one answer matches the Target, just plug in a different number for the variables and test the answer choice you were unable to eliminate with the original number.

Can I Just Plug In Anything?

You can plug in any numbers you like, as long as they’re consistent with any restrictions stated in the problem, but it’s more effective if you use easy numbers. What makes a number easy? That depends on the problem, but in most cases, lesser numbers are easier to work with than greater numbers. Usually, it’s best to start with a lesser number, such as 2 for example. Avoid the numbers 0 and 1; both 0 and 1 have special properties, which you’ll hear more about later. You want to avoid these numbers because they will often make more than one answer choice match the target. For example, if we plug in 0 for a variable such as x, then the answers 2x, 3x, and 5x would all equal 0. Also, try to avoid plugging in any numbers that are repeats of numbers that show up a lot in the question or answer choices. If you can avoid plugging in 0, 1, or repeat numbers, you can oftentimes avoid situations that may make you have to Plug In again.

Plug in numbers that make
the calculations EASY.

Good Numbers Make Life Easier

However, numbers of lesser value aren’t always the best choices for Plugging In. What makes a number good to work with depends on the context of the problem, so be on the lookout for clues to help choose the numbers you are going to use to Plug In. For instance, in a problem involved percentages the numbers 10 and 100 are good numbers to use. In a problem that involves minutes or seconds any multiple or factor of 60, such as 30 or 120 are often good choices.

Plug in real numbers for
variables to turn algebra
into arithmetic!

Let’s use the Plugging Insteps from above to work through the following problem.

Mara has six more than twice as many apples as Robert and half as many apples as Sheila. If Robert has x apples, then, in terms of x, how many apples do Mara, Robert, and Sheila have combined?

    2x + 6

    2x + 9

    3x + 12

    4x + 9

    7x + 18

On the GRE, Plugging In
is often more accurate,
and easier, than doing the

Here’s How to Crack It

Step 1:

Identify the Opportunity. Look at the question. There is the variable x in the question stem and the answer choices. This is a clear indication to start thinking about Plugging In.

Step 2:

Set up the scratch paper. Keep yourself organized by listing out answer choices (A) through (E) on the scratch paper. Leave some space to work the problem.

Step 3:

Plug In. Plug in a good number. The problem states that Robert has x apples, and doesn’t indicate that the number of apples needs to be anything specific for choose an easy number so as x = 4.

Step 4:

Solve for the Target. Now use x = 4 to read the problem again and solve for the target. The problem states that “Mara has six more than twice as many apples as Robert.” If Robert has 4 apples, then Mara must have 14. Next, the problem states that Mara has “half as many apples as Sheila.” That means that Sheila must have 28 apples. The question asks for the number of apples that Robert, Sheila, and Mara have combined so add 4 + 14 + 28 = 46 apples. This is the target number, so circle it.

Step 5:

Check All The Answer Choices. Plug in x = 4 for all of the variables in the answer choices and use the scratch paper to solve them, eliminating any answer choice that does not equal 46.

(A) 2(4) + 6 = 14—This is not 46, so eliminate it.

(B) 2(4) + 9 = 17—Eliminate this too.

(C) 3(4) + 12 = 24—Also not 46, so eliminate this.

(D) 4(4) + 9 = 25—This is still not 46, so eliminate this as well.

(E) 7(4) + 18 = 46—Bingo! This is the correct answer.

On the GRE, Plug In for variables in the question stem and answer choices. Remember to plug in numbers that will be easy to work with based on the problem, as some numbers can end up causing more trouble than they are worth.

Always be on the lookout
for variables and if you see
them, get ready to Plug In!

When Plugging In, follow these rules:

1.    1. Avoid plugging in 0 or 1. These numbers, while easy to work with, have special properties.

2.    2. Avoid plugging in numbers that are already in the problem; this often leads to more than one answer matching your target.

3.    3. Avoid plugging in the same number for multiple variables. For example, if a problem has xy, and z in it, pick three different numbers to plug in for the three variables.

4.    4. Avoid plugging in conversion numbers. For example, don’t use 60 for a problem involving hours, minutes, or seconds.

Finally, Plugging In is a powerful tool, but you must remember to always check all five answer choices when you Plug In. In certain cases, two answer choices can yield the same target. This doesn’t necessarily mean you did anything wrong; you just hit some bad luck. When this happens just plug in different numbers, solve for a new target, and recheck the answer choices that worked the first time.


Some questions may not have variables in them but will try to tempt you into using algebra to solve them. We call these Plugging In The Answers questions, or PITA for short. These are almost always difficult problems but once you recognize the opportunity to PITA, these questions turn into simple arithmetic questions. In fact, the hardest part of these problems is often identifying them as opportunities for PITA. The beauty of these questions is that they take advantage of one of the inherent limitations of a multiple-choice test: the answers are given to you. ETS has actually given you the answers, and only one of them is correct. The essence of this technique is to systematically Plug In The Answers to see which answer choice works given the information in the problem.


Above is a tried-and
-true Princeton Review
strategy, PITA (which has
nothing to do with the
delicious type of bread).

Let’s look at an example of a Plugging In The Answers question.

An office supply store sells binder clips that cost 14 cents each and binder clips that cost 16 cents each. If a customer purchases 85 binder clips from this store at a total cost of $13.10, how many 14-cent binder clips does the customer purchase?






Are you tempted to try to
set up an algebraic equation?
Are there no quickly
identifiable variables? Are
the answer choices real
number? Try Plugging In
The Answers!

Here’s How to Crack It

ETS would like you to solve this problem using algebra, and there is a good chance that you started to think about the variables you could use to set up some equations to solve this problem. That urge to do algebra is actually the first sign that you can solve this problem using Plugging In The Answers. Other signs that you can Plug In The Answers to solve this problem are that the question asks for a specific amount and that the numbers in the answer choices reflect that specific amount. With all these signs, it’s definitely time to Plug In The Answers!

Start by setting up your scratch paper. To do so, just list the five answer choices in a column, with the actual numbers included. Since the problem is asking for the number of 14-cent binder clips purchased, these answer choices have to represent the number of 14-cent binder clips purchased. Label this column 14¢.

The answer choices will always be listed in either ascending or descending numerical order, so when you Plug In the Answers, start with (C). By determining whether or not (C) works, you can eliminate the other answer choices that are either greater or less than (C), based on the result of this answer choice. This effectively cuts the amount of work you need to do in half. So, start with the idea that the customer purchased 30 binder clips that cost 14 cents each. What can you figure out with this information? You’d know that the total spent on these binder clips is 30 × $0.14 = $4.20. So, make a column with the heading “amount spent” and write $4.20 next to (C). Now, look for the next thing you’d know from this problem. If the customer purchased a total of 85 binder clips and 30 of them cost 14 cents each, that means that the customer purchased 55 16-cent binder clips. Make another column with the heading “16¢” and write 55 in the row for (C). Next, make another column for the amount spent on 16-cent binder clips, label it “amount spent,” and write 55 × $0.16 = $8.80 under this column in the row for (C). The next piece of information in the problem is that the customer spends a total of $13.10 on the binder clips. This information allows you to determine if (C) is correct. All Plugging In the Answers questions contain a condition like this that lets you decide if the answer is correct. In this case, $4.20 + $8.80 = $13.00, which is less than $13.10, so eliminate (C). Since the total was not great enough, you can determine that to increase the total, the customer must have purchased more 16-cent binder clips. Since (D) and (E) would increase the number of 14-cent binder clips purchased, they cannot be correct. Eliminate (D) and (E) as well.

Now, do the same steps starting with (B). If the customer purchased 25 of the 14-cent binder clips, they cost $3.50. The customer also purchased 60 of the 16-cent binder clips at a cost of $9.60. The total amount spent is $3.50 + $9.60 = $13.10. Since this matches the amount spent in the problem, (B) is correct.

Here’s what your scratch paper should look like after this problem:

When you want to Plug In the Answers, here are the steps that you should follow.

Step 1:

Recognize the Opportunity. There are three ways to do this. The first triggers are the phrases “how much…,” “how many…,” or “what is the value of….” When you see one of these phrases in a question, its a good indicator that you may be able to Plug In The Answers. The second tip-off is specific numbers in the answer choices in ascending or descending order. The last tip-off is your own inclination. If you find yourself tempted to write your own algebraic formulas and to invent your own variables to solve the problem, it’s a good sign that you can Plug In The Answer choices.

Step 2:

Set up the Scratch Paper. The minute you recognize the opportunity, list the numbers in the answer choices in a column on the scratch paper.

Step 3:

Label the First Column. The question asks you to find a specific number of something so the answer choices must be options for that number. At the top of the column above the answer choices, write down what the numbers represent.

Step 4:

Start with (C). Choice (C) will always be the number in the middle. This the most efficient place to start because it will allow you to eliminate as many as three answer choices if it is wrong.

Step 5:

Create Your Spreadsheet. Use (C) to work through the problem. It is always easier to understand the problem using a specific number. Work through the problem one step at a time, and every time you have to do something with the number, make a new column. Each column is a step in solving the problem that you may need to use again with a different answer choice, so don’t leave anything out.

Step 6:

Repeat with the Other Answer Choices. On single-answer multiple-choice questions, only one answer choice can work. If (C) is correct, you are finished with the problem. If it is not correct, you may be able to determine if the value of the number is too great or too less. If it is too great, you can eliminate it and every answer choice that it is greater than. The same thing can be done if the value of the resulting answer is lesser than the value indicated by the problem. At this point, you have basically created your own little spreadsheet that is specfically designed to calculate the correct answer. Check the remaining answer choices by using the spreadsheet. As soon as you find an answer choice that works, you’re finished.

On PITA questions, you don’t need to check all five answer choices because only one of them can be correct. Once you have found an answer that works with the problem, select it and move on to the next problem. PITA is a great tool but it requires a high level of organization, so make sure to keep track of everything that you do on the scratch paper.


Quantitative Comparison questions with variables can be extremely tricky because the obvious answer is often wrong, whereas finding the correct answer may involve a scenario that is not commonly thought of. On the other hand, there is a simple set-up and approach that you can use to help find the correct answers. As always, whenever you see variables, replace them with real numbers. On quant comp questions, however, it is crucial that you Plug In more than once and specifically that you plug in different kinds of numbers that may not occur to you to think of initially. A good way to help you think of this is to always keep the nature of the answer choices in mind. Picking (A) means that you believe that the quantity in column A will always be greater than Quantity B—no matter what number you plug in. Choice (B) means that the quantity in column B will always be greater than Quantity A—no matter what number you plug in, and so forth. To prove that one of these statements is true you have to plug in every possible number that could change the outcome. Don’t worry. We have a simple process to help figure out what to plug in and how to track your progress as you do.

Quantitative Comparison
questions often test your
knowledge of the
properties of fractions,
zero, one, negatives, and
other weird numbers.

Here are the steps:

Step 1:

Recognize the Opportunity. The first seven or eight questions of any math section will be quant comp. When a quant comp question appears and you see variables, you know that you can Plug In.

Step 2:

Set up the Scratch Paper. The minute you see quant comp and variables set up the scratch paper. The recommended set up should look something like the diagram below. Place Quantity A and B on either side. Quant Comp questions only have 4 potential answer choices, so write a, b, c, and d down as well, so you can eliminate answers as you go. Finally, leave space to write down the numbers that you plug in for the variables in between the Quantities, so you can stay organized.

Step 3:

Plug In and Eliminate. Start with an easy number, just like outlined in the earlier Plugging In section, but make sure that you also follow any conditions in the problem. With the number you plugged in for the variable, solve for Quantity A and Quantity B and write the solutions down. If Quantity A is greater than Quantity B, eliminate (B) and (C). If Quantity B is greater than Quantity A, eliminate (A) and (C). If both quantities are the same, eliminate (A) and (B).

Step 4:

Plug In Again Using FROZEN Numbers. On Quant Comp questions with variables, you always need to Plug In more than once and the second time you do it, you need to use FROZEN numbers. FROZEN is an acronym that will be explained in the next section, as well the entire concept behind why to Plug In more than once, so keep reading!

Always Plug In More Than Once on Quant Comp Questions

Quant Comp questions only have four options for answer choices but one of those options, (D), can be selected if the relationship between Quantity A and Quantity B cannot be determined based on the information given. After you Plug In the first time, you need to Plug In again but this time you need to try and choose a number that will produce a different outcome for the question. While the first time you Plug In you can usually reliably eliminate two of answer choices (A), (B), or (C), you Plug In again to try and make sure that you can eliminate (D). Choice (D) can be eliminated only when you have a high level of confidence that no matter what number you plug in, the answer will always remain the same. If even one of the numbers you choose creates a different answer, (D) should be selected.

So, to eliminate (D) you need to choose a different number. But what makes a number different and what makes for a good number to choose that might create a different outcome for the problem? When you Plug In for the second (or sometimes third or fourth) time in a Quant Comp question, you should pick a FROZEN number. FROZEN is an acronym that highlights different types of numbers and it stands for:







On quant comp, plug in
easy numbers such as 2
or 5, and eliminate two
choices. Then plug in
FROZEN numbers (Fractions,
Repeats, One, Zero,
Extremes, Negatives)
to try to disprove your
first answer. If different
numbers give you different
answers, you’ve proved
that the answer is (D).

Fractions are numbers such as  or  that are great to use if the problem contains exponents or roots, as fractions respond to these two stimuli in a different way from whole numbers. Repeats are numbers that are found in the question stem, can be used in both Quantities, or numbers that are implied by the question stem (such as using the number 60 if the question is about seconds, minutes, or hours). One and zero are special numbers because they can oftentimes result in two quantities being equal to each other and number react to them in different ways that are unique only to them. Extreme numbers are numbers such as 10 or 100 that should be used to see if the relationship between the quantities changes for numbers that are greater than the one that was initially chosen. Negative numbers, such as –2 or –3, are numbers that create different outcomes when plugged in for variables, as they can make Quantities negative or positive, which can alter the outcome.

FROZEN numbers can also be combined to create different numbers, such as –100, −, or −1. Often ETS will create a Quant Comp question that has a correct answer that depends on using these types of numbers. They do that because they know that most people will not think of these numbers, which is why it is important to Plug In more than once and, when you do, use FROZEN numbers.

Let’s look at an example problem:

Quantity A

Quantity B



    Quantity A is greater.

    Quantity B is greater.

    The two quantities are equal.

    The relationship cannot be determined from the information given.

Here’s How to Crack It

Step 1:

Recognize the Opportunity. This is a quant comp question and there are variables in the quantities, so this is a Plug In problem.

Step 2:

Set up the Scratch Paper. Get yourself organized and ready to answer the problem by setting up the scratch paper.

Step 3:

Plug In. Let’s start with an easy number such as 2. Plug in 2 for x. When x = 2, the quantity in column A is 2 × 23 = 16, and the quantity in column B is 4 × 22 = 16 as well. Since both quantities are the equal, neither (A) nor (B) is always greater than the other, so eliminate them both.

Step 4:

Plug In Again Using FROZEN Numbers. Now look at the problem and try to decide on a FROZEN number for x that may create a different answer. Try One. If x = 1, then Quantity A is 2 × 13 = 2 and Quantity B is 4 × 12 = 4. Quantity B is now greater than Quantity A, which means that (C) is incorrect. Eliminate (C) and select (D), which is the correct answer.

If you chose to follow the recommended set up for the scratch paper, it should look like this:

You might also have noticed that choosing different FROZEN numbers, such as Fractions or Zero, would also yield a different result that would have allowed you to eliminate (C). This is not uncommon as ETS is hoping you forget to use these FROZEN numbers when Plugging In. Make sure you use these numbers aggressively on Quant Comp problems because they can radically affect the relationship between the two Quantities.


Now that you’ve become familiar with fractions and percents, we’ll show you a great method for solving many of these problems. When you come to regular multiple-choice questions, or multiple-choice, multiple-answer questions, that involve fractions or percents, you can simply plug in a number and work through the problem using that number. This approach works even when the problem doesn’t have variables in it. Why? Because, as you know, fractions and percents express only a relationship between numbers—the actual numbers don’t matter. For example, look at the following problem:

Plugging In on fraction
and percent problems is a
great way to make these
problems much easier.

A recent survey of registered voters in City X found that  of the respondents support the mayor’s property tax plan. Of those who did not support the mayor’s plan,  indicated they would not vote to reelect the mayor if the plan were implemented. Of all the respondents, what fraction indicated that they would not vote for the mayor if the plan were enacted?






What important information
is missing from the

Here’s How to Crack It

Even though there are no variables in this problem, we can still Plug In. On fraction and percent problems, ETS will often leave out one key piece of information: the total. Plugging In for that missing value will make your life much easier. What crucial information did ETS leave out of this problem? The total number of respondents. So let’s plug in a value for it. Let’s say that there were 24 respondents to the survey. 24 is a good number to use because we’ll have to work with  and , so we want a number that’s divisible by both those fractions. Working through the problem with our number, we see that  of the respondents support the plan.  of 24 is 8, so that means 16 people do not support the plan. Next, the problem says that  of those who do not support the plan will not vote for the mayor.  of 16 is 2, so 2 people won’t vote for the mayor. Now we just have to answer the question: Of all respondents, how many will not vote for the mayor? Well, there were 24 total respondents and we figured out that 2 aren’t voting. So that’s , or . Choice (B) is the one we want.


While Plugging In is a great strategy to make Algebra problems easy on the GRE by turning them into Arithmetic, in many cases being comfortable manipulating variables in an equation is necessary to answering a question. Plugging In will help you solve for a variable in a question but sometimes the question only requires you to manipulate an equation to get the correct answer.

Dealing with Variables

The previous chapter familiarized you with number concepts and the previous section showed you how to turn Algebra into Arithmetic by using Plugging In. However, its time to learn the basics of dealing with variables and manipulating equations to help make problems easier to work with and give you the best opportunity to optimize your score.

Manipulating Equations

When working with equations, you can do pretty much anything you want to them as long as you follow the golden rule:

Whatever you do on one side of the equals sign you must also do on the other side.

Solving for One Variable

Let’s begin the discussion of manipulating equations with one variable by solving for one variable. When presented with an equation with one variable, start by isolating the variable on one side of the equation and the numbers on the other side. You can do this by adding, subtracting, multiplying, or dividing both sides of the equation by the same number. Just remember that anything you do to one side of an equation, you must do to the other side. Let’s look at a simple example:

3x – 4 = 5

Don’t assume you’ll
always need to solve
for the variable on the
GRE; sometimes you’ll
simply have to manipulate
the equation to get the

Here’s How to Crack It

When presented with a problem like this, your goal is to isolate the variable on one side of the equation with all the real numbers, or constants, on the other. In the example above, begin manipulating this question by adding 4 to both sides of the equation. In general, you can eliminate negative numbers by adding them to both sides of the equation, just as you can eliminate positives by subtracting them from both sides of the equation.

The variable is not quite isolated yet, as it is still being multiplied by 3. In the same way that you manipulated the equation earlier move the 4 to the other side of the equation, you must move the 3. Since the 3 is being multiplied to the variable, move it by doing the opposite operation, in this case division. This allows you to solve for x and finish the problem.

x = 3

Let’s try another one:

5x – 13 = 12 – 20x

Here’s How to Crack It

Again, we want to get all the x values on the same side of the equation. This time, however, there is more than one instance of x so begin the question by combining the x values.

As the problems get more
involved, make sure to
keep yourself organized by
utilizing the scratch paper
given to you.

Now that the values of x are combined, isolate the x by moving the negative 13 to the other sides of the question.

Solve for x by finishing the isolation by moving the 25 that it is being multiplied by.

x = 1

Let’s try one more that is slightly more complicated.

Here’s How to Crack It

The first thing you probably notice here is the fraction. Whenever you see an equation like this that contains a fraction, begin by “clearing” the fraction. To clear the fraction, multiply all the terms in the equation by the denominator of the fraction. In this case, multiply all the terms by 2.

10x + 3 = 14x

You must always do the
same thing to both sides
of an equation.

Notice how all the terms have by multiplied by 2! This is very important, so don’t forget to do it! Now, manipulate the equation to collect all the x’s on the same side of the equation.

Now finish isolating the x by moving the 4.


Many times, however, an equation on the GRE will involve two variables. An example of an equation with two variables looks like this:

3x + 10y = 64

Here’s How to Crack It

The important thing to note about this situation is that we cannot solve this equation. Why, you ask? The problem is that since there are two variables, there are many possible solutions to this equation all of which are equally valid. For example, plugging in the values x = 8 and y = 4 would satisfy the equation. But the equation would also be satisfied if you plugged in the values x = 10 and y = 3.4. Therefore, the GRE cannot test an equation with two variables without either providing a definitive way to solve for one of the variables, or providing a second equation. By giving two equations, you are able to find definitive values for the variables. So a more likely problem would look something like this:

3x + 10y = 64

6x – 10y = 8    

You can’t solve an
equation with two
variables unless you
have a second equation.

Now there are 2 variables and 2 equations, which means we can solve for the variables. When two equations are given, look to combine them by adding or subtracting the entire equations. We do this so that we can cancel out one of the variables, leaving us with a simple equation with one variable. In this case, it’s easier to add the two equations together, which will eliminate the y variable as seen below.

Add these two equations to get 9x = 72. This is a simple equation, just like the ones discussed in the previous section, which we can solve to find x = 8. Once we’ve done that, we can solve for the other variable by inserting the value of x into one of the original equations. For example, if we substitute x = 8 into the first equation, we get 3(8) + 10y = 64, we can solve to find that y = 4.

The GRE will rarely give you two equations that line up as nicely as the above example does, though. You are more likely to find two equations with two variables and, while the variables match, the numbers associated with the variables are not equal. In this case, you will need to manipulate one equation so the numbers associated with a variable are equal. Doing this will allow the eliminate of a variable when the two equations are added or subtracted. Try the next problem as an example.

4x + 7y = 41

2x + 3y = 19

Here’s How to Crack It

Notice here that the numbers associated with the variables are not equal, which means that you cannot eliminate a variable. Adding the two equations yields 6x + 10y = 60. That doesn’t help; it’s a single equation with two variables, which is impossible to solve. Subtracting the equations leaves 2x + 4y = 22, which is also a single equation with two variables. To solve this question, you need to make one of the variables equivalent to the same variable in the other equation. In this case, try multiplying the second equation by 2.

2(2x + 3y) = 2(19)

This gives us the following:

4x + 6y = 38

Now we can subtract this equation from the first equation. Doing this operation yields y = 3. Now you can substitute y = 3 into either one of the original equations to find that x = 5.

Simultaneous Equations

Thus far, we have learned how to manipulate equations with one variable and two equations with two variables in order to solve for the variables. However, it is not uncommon for ETS to give you two equations and ask you to use them to find the value of a given expression. Much like manipulating two equations with two variables, all you need to do is add or subtract the two equations! The only difference is this time you won’t end up solving for an individual variable.

Here’s an example:

If 5x + 4y = 6 and 4x + 3y = 5, then what does x + y equal?

Here’s How to Crack It

Remember that the problem is asking you to solve for x + y. This may appear like you need to solve for the variables individually, but try to add or subtract the equations first to see what they yield. First, try adding the two equations together.

Since the problem wants the value of x + y and this gives us the value of 9x + 7y this is not useful. So try subtracting the two equations.

Bingo. The value of the expression (x + y) is exactly what we’re looking for. You could have tried to solve for each variable individually and solved the problem that way, but since the question is asking for the value of an expression, it was easier to manipulate the equations like this. So remember, on the GRE you may see two equations written horizontally. Now you know that you don’t need complicated math to solve them! Just rewrite the two equations, putting one on top of the other, and simply add or subtract them.


The difference between an equation and an inequality is that in an equation one side always equals the other and in an inequality on side does not equal the other. Equations contain equal signs, while inequalities contain one of the following symbols:

≠        is not equal to

>       is greater than

<        is less than

≥        is greater than or equal to

≤        is less than or equal to

The point of the inequality
sign always points to the
lesser value.

The good news is that inequalities are manipulated in the same way that you manipulated any of the equations in the previous sections of this chapter. However, there is one critical difference. When you multiply or divide both sides of an inequality by a negative number, the direction of the inequality symbol must change. So, if the inequality x > y is multiplied by –1, the resulting inequality is –x < –y.

To see this rule in action, take a look at the following inequality:

12 – 6x > 0

Here’s How to Crack It

There are two ways to solve this inequality. You could manipulate this inequality without ever multiplying or dividing by a negative number by just adding 6x to both sides and then dividing both sides of the inequality by the positive 6. In this case, the sign would not change, as seen below.

The other way to solve this inequality is to subtract 12 from both sides first. This will create a situation where you need to divide both sides of the equation by –6, as shown below.

Notice that the sign flipped because you divided both sides by a negative number, but the answer for both methods of solving this inequality is the same thing. The first answer says that the number 2 is greater than x, and the second says that x is less than the number 2!

Flip the sign! When you
multiply or divide both
sides of an inequality by
a negative number, the
greater than/less
than sign points the
opposite way.

Inequalities show up on the GRE in a variety of ways. For instance, ETS may give you a range for two variables and then ask you to combine them in some way. This type of problem looks like the following question.

If 0 ≤ x ≤ 10, and –10 ≤ y ≤ –1, then what is the range for x – y?

Here’s How to Crack It

First, determine what the question is asking you to do. The question is asking you to solve for the range for the expressions x – y. To determine this you need to consider all possible combinations of x – y. Since the inequalities are ranges themselves, find the greatest and least possible values of x – y by calculating the largest x minus the largest y, the largest x minus the least y, the least x minus the largest y, and the least x minus the least y. The greatest value of x is 10 and the least value of x is 0. The greatest value of y is –1 and the least value is –10. Calculate these values and keep yourself organized by writing this all down on the scratch paper.

The calculations look as follows:

10 – (–1) = 11

10 – (–10) = 20

0 – (–1) = 1

0 – (–10) = 10

Since the least possible value of x – y is 0 – (–1) = 0 and the greatest possible value of x – y is 10 – (–10) = 20, the range is 1 ≤ x – y ≤ 20.

Quadratic Equations

Quadratic equations are special types of equations that involve four terms. Here is an example of a quadratic:

(x + 4)(x – 7)

In order to work with quadratics on the GRE, you must be familiar with two concepts: FOIL and factoring.


When you see a quadractic, you need to multiply every term in the first set of parentheses by every term in the second set of parentheses. Use the acronym FOIL to remember this method. FOIL stands for firstouterinnerlast. For example, if you see (x + 4) (x + 3), you would multiply the first terms (x × x), the outer terms (x × 3), the inner terms (4 × x), and the last terms (4 × 3), as follows:

(x × x) + (x × 3) + (4 × x) + (4 × 3) =

x2 + 3x + 4x + 12 =

x2 + 7x + 12

We know to use plus signs inside the parentheses because both the 7 and the 12 are positive. Now we have to think of two numbers that, when added together, give us 7, and when multiplied together, give us 12. Find these numbers by listing the factor pairs of 12. Those pairs are 1 and 12, 2 and 6, and 3 and 4. The only pair that equals 7 when they are added together is 4 and 3, so insert those into the equation.

Quadratic Equations

There are three quadratic equations that frequently appear on the GRE. Knowing these equations both in their factored and unfactored forms, can drastically improve your time on these questions. Here they are:

1.    Factored form: x2 – y2 (the difference between two squares)
Unfactored form: (x + y)(x – y)

2.    Factored form: (x + y)2
Unfactored form: x2 + 2xy + y2

3.    Factored form: (x – y)2
Unfactored form: x2 – 2xy + y2

(x + 4) (x + 3) = 0

To find the solutions, set each factor equal to 0 and solve. So, x + 4 = 0 and x + 3 = 0. So x can either be –4 or –3.

Let’s look at a question that uses quadratics that you may see on the GRE.

Quantity A

Quantity B


     Quantity A is greater.

     Quantity B is greater.

     The two quantities are equal.

     The relationship cannot be determined from the information given.

Here’s How to Crack It

First, eliminate (D) because there are only numbers in this question, so the answer can be determined. Now, Quantity A looks like a job for FOIL! Multiply the first terms, and you get 16. Multiply the outer terms and you get – 4. Multiply the inner terms to get 4. Multiply the last terms to get –6. So, Quantity A is now 16 – 4 + 4 –6. The two inner terms cancel each other out and all the remains is 16 – 6, or 10. Since Quantity B is also equal to 10, the two answer choices are equal and the correct answer is (C). You might also notice that Quantity A is one of the common quadratics: (x + y)(x – y) = x2 – y2. Therefore, (4 + ) (4 - )= 42 – 2 = 16 – 6 = 10.


The process of factoring “undoes” the FOIL process. Factoring is commonly tested on the GRE so you should be very familiar with the process. Think of factoring as taking a quadratic in the opposite direction of FOIL. Here is a quadratic in its unfactored, or expanded, form:

x2 – 10x + 24

We are going to factor this quadratic using the following steps.

1.    Separate the x2 into (x )(x ).

2.    Find the factors of the third term that, in this case the number 24, when added or subtracted, yield the second term, the number 10. Note here that we are not concerned with the variable x.

3.    Figure out the signs (+/–) for the terms. The signs have to yield the middle number when added and the last term when multiplied.

If we apply these steps to the quadratic given earlier, we begin the problem by splitting x2 into

(x  )(x  )

Next, write down the factors of the third term, 24. The factors are 1 and 24, 2 and 12, 3 and 8, and 4 and 6. Of these pairs of factors, which contains two numbers that we can add or subtract to get the second term, 10? 4 and 6 is the only pair that works, so put those into the parentheses.

(x   4)(x   6)

The final step is to figure out the signs. We need to end up with a negative 10 and a positive 24. If we add –6 and –4, we’ll get –10. Similarly, if we multiply –6 and –4, we’ll end up with 24. So the factored form of the quadratic is

(x – 4)(x – 6)

Solving Quadratic Equations

ETS likes to use quadratic equations because they have an interesting quirk; when you solve a quadratic equation, you usually get two possible answers as opposed to one. For this reason, quadratic equations are perfect ways for ETS to try to trick you.

Here’s an example:

x2 + 2x – 15 = 0

Quantity A

Quantity B



    Quantity A is greater.

    Quantity B is greater.

    The two quantities are equal.

    The relationship cannot be determined from the information given.

Quadratic equations
usually have two solutions.

Here’s How to Crack It

In order to solve a quadratic equation, the equation must be set equal to zero. Normally, this will already be the case on the GRE, as it is in this example. But if you encounter a quadratic equation that isn’t set equal to zero, you must first manipulate the equation so that it is. Next you must factor the equation; otherwise you cannot solve it. So let’s factor the quadratic equation in this example. We need to figure out the factors of 15 that we can add or subtract to give us 2. The only possible factors are 3 and 5. In order to get a negative 15 and a positive 2, we need to use 5 and –3. So that leaves us

(x – 3)(x + 5) = 0

Next, we’re going to solve each of the two factors within parentheses separately:

x – 3 = 0 and x + 5 = 0

Thus, x = 3 and x = –5. If x = 3, then Quantity B is greater, but if x = –5 then Quantity A is greater. This means that the correct answer is (D).

Let’s try another one:

If x2 + 8x + 16 = 0, then what is the value of x ?

Here’s How to Crack It

Let’s factor the equation. Start with (x   ) (x   ). Next, find the factors of 16 that add or subtract to 8. The factors of 16 are 1 and 16, 2 and 8, and 4 and 4. Of these pairs, only 4 and 4 have a sum of 8. Since we have a positive 8 and a positive 16, the signs for both numbers must be positive. Thus, we end up with (x + 4) (x + 4) = 0. Now, we need to solve the equation. If x + 4 = 0, then x = –4. This is the number we’d enter into the text box on the GRE.

Let’s look at one more example.

If x and y are positive integers, and if x2 + 2xy + y2 = 25, then what is the value of (x + y)3 ?






Here’s How to Crack It

While this problem may look like a lot of work, if you have committed the common quadratic equations from earlier in this section to memory the answer is easier to come by. The equation in this question is reflective of the common quadratic: x2 + 2xy + y2 = (x + y)2. The question tells us that x2 + 2xy + y2 is equal to 25, which means that (x + y)2 is also equal to 25. Think of x + y as one unit that, when squared, is equal to 25. Since this question specifies that x and y are positive integers, what positive integer squared equals 25? The number 5 does, so x + y = 5. The question is asking for (x + y)3. In other words, what’s 5 cubed, or 5 × 5 × 5? The answer is (E), 125.


Finally, the last section of this chapter is going to deal with exponents and square roots. Questions with exponents and square roots are common on the GRE and solving these questions often requires manipulating the exponents or roots. Here’s the information you need to know in order to work with them.

What Are Exponents?

Exponents are a sort of mathematical shorthand for repeated multiplication. Instead of writing (2)(2)(2)(2), you can use an exponent and write 24. The little 4 is the power and the 2 is called the base. The power tells you how many times to multiply the base by itself. Knowing this terminology will be helpful in following the discussion in this section.

The Five Rules of Working with Exponents

For the GRE there are five major rules that apply when you work with exponents. The more comfortable you are with these rules, the more likely you will be to approach an exponent question with confidence and get the answer correct!

The first three rules deal with the combination and manipulation of exponents. Those three rules are represented by the acronym MADSPM, which stands for:







These three rules will be explained in more detail shortly, but for now just remember:

·        when you see exponents with equal bases which are being multiplied, you add the powers;

·        when equal bases are divided you subtract the exponents; and

·        when an exponent is raised to a power, you multiply the powers.

The fourth rule is the definition of a negative exponent. The fifth and final rule is the definition of a zero exponent.

The Multiply-Add Rule of Exponents

When two exponents with equal bases are multiplied, you must add the exponents. Consider the following example:

32 × 33

As defined earlier, a power just tells you how many times to multiply a base by itself. So another way to write this expression is:

32 × 33 = (3 × 3)(3 × 3 × 3) = 35

As you can see, the number of bases, which in this case is the integer 3, that are actually being multiplied together is five, as there are two 3’s that are represented by 32 and three, 3’s that are represented by 33.

Now solve this question more quickly by using the multiply-add rule.

32 × 33 = 32+3 = 35

The Divide-Subtract Rule of Exponents

When two exponents with equal bases are divided, you must subtract the exponents. Consider the following example and expand the exponents to make it clear.

Now, instead of expanding the exponents just apply the divide-subtract rule for the same problem.

The Power-Multiply Rule of Exponents

When an expression with an exponent is raised to another power, multiply the powers together. Consider the following example and expand the exponents to make it clear.

(62)3 = (62) (62) (62) = (6 × 6) (6 × 6) (6 × 6) = 66

Now, apply the power-multiply rule to solve the same problem.

(62)3 = 62×3 = 66

For all of these rules, the bases must be the same. So, for example, you could not divide-subtract the expression  because the bases are not the same.

Negative Exponents

A negative exponent is another way ETS uses exponents on the GRE. Consider the following example.

So when you have a negative exponent, all that needs to be done is to put the entire expression in a fraction, with 1 in the numerator and the exponent in the denominator, and change the negative exponent to a positive. A term raised to a negative power is the reciprocal of that term raised to the positive power.

Zero Exponents

Sometimes ETS will give you an exponent question that, after you have successfully manipulated it, results in a base number raised to a power of 0. Any nonzero number raised to a power of 0 is equal to 1. Consider the following example.

Exponent Tips Beyond the Five Rules

Sometimes you will be presented with, or will be working on, an exponent problem and none of the five rules discussed apply. If you reach this point there are two tips to keep you moving forward.

Tip 1: Rewrite Terms Using Common Bases

ETS will always write questions that work out nicely, so if none of the bases in an exponent question seem to match up, see if you can find a way to rewrite the bases so that they match, and you will be able to use one of the five rules.

Tip 2: Look for a Way to Factor the Expression

Factoring the expression is often a way to reveal something about the exponent expression that you may not have noticed before. If you get stuck with an exponent question try to factor the expression and see if there is a way to use one of the five rules.

It will be uncommon for ETS to just test one or two of these concepts on a GRE problem. Most times, two or more of these concepts will be combined to create a problem. Let’s look at a couple of examples.

If y ≠ 0, which of the following is equivalent to 






Here’s How to Crack It

Begin by simplifying the denominator of the fraction. Use the power-multiply rule to combine (y2)3 into y6. Since a number, or in this case a variable, by itself is the same thing as having that number or variable raised to a power of 1, use the add-multiply rule to combine y(y6) into y7. Now use the divide-subtract rule to solve the question, so and the correct answer is choice (B).

Let’s look at another problem.

Quantity A

Quantity B

1515 – 1514

1514(14) – 1

    Quantity A is greater.

    Quantity B is greater.

    The two quantities are equal.

    The relationship cannot be determined from the information given.

Here’s How to Crack It

The question wants you to compare the two quantities but since none of the rules for exponents apply here, see if there is something else you can do to this problem. The expression in Quantity A can be factored so begin there. Quantity A is now 1515 – 1514 = 1514(15 – 1), which is the same thing as 1514(14). Notice how this is the same as Quantity B, expect Quantity B is 1 less than 1514(14). Therefore, Quantity A is greater and the correct answer is (A).

Take a look at one more exponent problem.

If x ≠ 0 and 643 = 8x then what is the value of x2 ?






Here’s How to Crack It

Solve this question by determining the value for x. To compare these exponents in the equation, begin by making similar bases. Since 64 can be written as 82, the equation can be rewritten as (82)3 = 8x. Since the bases are the same, for the equation to be equal the powers have to be the same as well. (82)3 can be rewritten as 86 because of the power-multiply rule, so if 86 = 8x then x = 6. Now plug that number into the value for x2. This is now 62 which equals 36, so the correct answer is (D).

The Peculiar Behavior of Exponents

·        Raising a number greater than 1 to a power greater than 1 results in a greater number. For example, 22 = 4.

·        Raising a fraction that’s between 0 and 1 to a power greater than 1 results in a lesser number. For example,  = .

·        A negative number raised to an even power results in a positive number. For example, (–2)2 = 4, because (–2)(–2) = 4.

·        A negative number raised to an odd power results in a negative number. For example, (–2)3 = –8, because (–2)(–2)(–2) = –8.

·        A number raised to the first power ALWAYS results in the number itself. For example, 1,0001 = 1,000.

What Is a Square Root?

The radical sign  indicates the square root of a number. For example,  means that some number times itself is equal to 4. In this case, since 22 = 4 it can be determined that  = 2. Think of square roots as the opposite of exponents. If you want to eliminate a square root in an equation, all you need to do is raise that square root to a power of 2. Just remember to do that for all of the elements in the equation!

Unlike exponents, however, square roots can only exist on the GRE with non-negative numbers. If the problem states that x2 = 16, then x = ±4 as both a positive and a negative 4, when multiplied by itself, yields 16. However, when ETS asks you for the square root of any number, the result will always be positive.

Rules for Square Roots

There are rules that dictate what you can and cannot do with square roots, just like there are rules about exponents.

Adding and Subtracting Square Roots

You can only add or subtract square roots if the values under the radical sign are equal. So, for example, the expression  can be simplified to  because the value under the radical sign is equal. Conversely, the expression  cannot be reduced any further because the values of the roots are not the same.

You can multiply and
divide any square roots,
but you can add or
subtract roots only when
the number under the
radical sign is the same.

Rules for Adding and Subtracting Square Roots

Multiplying and Dividing Square Roots

Any square roots can be multiplied or divided. There aren’t any restrictions on this so keep an eye out for opportunities to combine roots by multiplying or dividing that could make a root easier to work with. For example, . Roots can be divided as well; for example, .

Rules for Multiplying and Dividing Square Roots

Simplifying Square Roots

Often times when you multiply square roots on the GRE, you will not get numbers under the radical sign that work out perfectly. When this happens, you will need to simplify the square root. You simplify a square root by look for ways to factor the number under the root that results in a least one perfect square. A perfect square is an integer that when the square root of that integer is taken, the result is another integer. For example, 4 is a perfect square because = 2. Similarly, 9 and 25 are perfect squares because  and , respectively. Look at the following expression and try to simplify it.

You can combine this expression to result in . However, this is not in the most simplified form. Look for ways to factor 20 in which one of the pairs of numbers is a perfect square. The factors of 20 are 1 and 20, 2 and 10, and 4 and 5. Since 4 is a perfect square, this can now be simplified even further.

Now let’s take a look at some examples of how ETS might test roots on the GRE.

What is the value of the expression 






Here’s How to Crack It

The problem is subtracting roots. Since roots cannot be subtracted unless the numbers under the radical sign are equal, look for a way to simplify the roots. Since 5 cannot be simplified any further, work with 80. The factors of 80 are 1 and 80, 2 and 40, 4 and 20, 5 and 16, and 8 and 10. Two of these pairs of factors contain a perfect square, but one contains a perfect square and a prime number. This is a good thing. This means that it could be reduced no further, so choose 5 and 16 and simplify to read . Now that the bases are equal, subtract the expression to find that , which is (C). The same answer would have been found if the numbers 4 and 20 had been chosen as the factors of 80, but there would have been another round of simplifying the root, as 20 would have needed to be reduced to 4 and 5 as factors.

Here’s another problem.

z2 = 144

Quantity A

Quantity B


    Quantity A is greater.

    Quantity B is greater.

    The two quantities are equal.

    The relationship cannot be determined from the information given.

Here’s How to Crack It

The trap answer here is (C). Remember that if z2 = 144, then the value of z is either 12 or –12. However, when a value is under a radical sign it can only be positive. Therefore, Quantity A could be 12 or –12 and Quantity B can only be 12. Since there is no way to ensure that one is always greater than the other, the correct answer is (D).

Try one more problem:

What is the value of  ?






To Simplify Roots:

1. Rewrite the number as the
product of two factors,
one of which is a perfect square.
Use the multiplication
rule for roots.

Here’s How to Crack It

First, simplify each of these roots.  has a factor that is a perfect square—25, so it can be rewritten as  and simplified to  has the perfect square 9 as a factor, so it can be written as  and then simplified to . This means that  is equal to ; the  in the numerator and denominator cancel, leaving . The correct answer is (A).

Algebra (And When To Use It) Drill

Now it’s time to try out what you have learned on some practice questions. Try the following problems and then check your answers in Part V.

1 of 10

The original selling price of an item at a store is 40 percent more than the cost of the item to the retailer. If the retailer reduces the price of the item by 15 percent of the original selling price, then the difference between the reduced price and the cost of the item to the retailer is what percent of the cost of the item to the retailer?

2 of 10

x2 + 8x = –7

Quantity A

Quantity B



    Quantity A is greater.

    Quantity B is greater.

    The two quantities are equal.

    The relationship cannot be determined from the information given.

3 of 10

If 33 × 912 = 3x, what is the value of x ?

4 of 10

If A = 2x – (y – 2c) and B = (2x – y) – 2c, then A – B =






5 of 10

A merchant sells three different sizes of canned tomatoes. A large can costs the same as 5 medium cans or 7 small cans. If a customer purchases an equal number of small and large cans of tomatoes for the same amount of money needed to buy 200 medium cans, how many small cans does she purchase?






6 of 10

If 6k – 5l > 27 and 3l – 2k < –13 and 5k – 5l > j, what is the value of j ?

7 of 10

If the integer a is multiplied by 3 and the result is 4 less than 6 times the integer b, what is the value of a – 2b ?






8 of 10

Quantity A

Quantity B

    Quantity A is greater.

    Quantity B is greater.

    The two quantities are equal.

    The relationship cannot be determined from the information given.

9 of 10

11x + 14y = 30 and 3x + 4y = 12

Quantity A

Quantity B

x + y

(x + y)–2

    Quantity A is greater.

    Quantity B is greater.

    The two quantities are equal.

    The relationship cannot be determined from the information given.

10 of 10

If x = 3a and y = 9b, then all of the following are equal to 2(x + y) EXCEPT

    3(2a + 6b)

    6(a + 3b)

    24(a + b)

    (18a + 54b)

    12(a + b)


·        Plugging In converts algebra problems to arithmetic problems. Plug In by replacing variables in the question with real numbers or by working backwards from the answer choices provided.

·        Use easy numbers first when Plugging In for variables. If the need arises to Plug In again, use the FROZEN numbers to help eliminate tricky answer choices on math problems.

·        The golden rule of equations: Whatever you do to one side of the equation, you must do to the other. This applies to all equations, whether they are equal or not.

·        In order to solve an equation with two variables, you need two equations. Stack them up and add or subtract to cancel out one of the variables.

·        Inequalities are manipulated the same way that another other equations are, with one notable difference: Always remember to flip the sign when you multiply or divide by a negative number.

·        Use the FOIL process to expand quadratics. To solve a quadratic equation, set it equal to zero, and factor.

·        An exponent is shorthand for repeated multiplication. When in doubt on exponent problems, look to find common bases or ways to factor the expressions.

·        Think of a square root as the opposite of an exponent. Square roots are always positive.