SAT Math 1 & 2 Subject Tests

Chapter 4


Algebra questions ask you to solve for an unknown amount. In this chapter, we’ll show you how ETS uses algebra (and often tries to trick you with it). You’ll learn some great techniques to help you avoid ETS traps. We’ll also review concepts, such as solving for x, inequalities, factoring, simultaneous equations, and quadratic equations.


Algebra questions will make up about 30 percent of the questions on the Math Level 1 Subject Test and about 20 percent of the questions on the Math Level 2 Subject Test. Many of these questions are best answered by using the simple algebra rules outlined in this chapter. Others can be shortcut with The Princeton Review techniques, which you’ll also find in the following pages.


Here are some algebraic terms that will appear on the Math Subject Tests. Make sure you’re familiar with them. If the meaning of any of these vocabulary words keeps slipping your mind, add those words to your flash cards.


An unknown quantity in an equation represented by a letter (usually from the end of the alphabet), for example, x, y, or z.


An unchanging numerical quantity—either a number or a letter that represents a number (usually from the beginning of the alphabet), for example, 5, 7.31, a, b, or k.


An algebraic unit consisting of constants and variables multiplied together, such as 5x or 9x2.


In a term, the constant before the variable. In ax2a is the coefficient. In 7x, 7 is the coefficient.


An algebraic expression consisting of more than one term joined by addition or subtraction. For example, x2 − 3x2 + 4x − 5 is a polynomial with four terms.


A polynomial with exactly two terms, such as (x − 5).


A quadratic expression is a polynomial with one variable whose largest exponent is a 2, for example, x2 − 5x + 6 or y = x2 + 4.


A root of a polynomial is a value of the variable that makes the polynomial equal to zero. More generally, the roots of an equation are the values that make the equation true. Roots are also known as zeros, solutions, and x-intercepts.


Many questions on the Math Subject Tests will require you to solve simple algebraic equations. Often these algebraic questions are in the form of word problems. Setting up an equation from the information contained in a word problem is the first step to finding the solution, and is the step at which many careless mistakes are made. The translation chart on this page is very useful for setting up equations from information given in English.

An algebraic equation is an equation that contains at least one unknown—a variable. “Solving” for an unknown means figuring out its value. Generally, the way to solve for an unknown is to isolate the variable—that is, manipulate the equation until the unknown is alone on one side of the equal sign. Whatever’s on the other side of the equal sign is the value of the unknown. Take a look at this example.

5(3x3 − 16) − 22 = 18

In this equation, x is the unknown. To solve for x, you need to get x alone. You isolate x by undoing everything that’s being done to x in the equation. If x is being squared, you need to take a square root; if x is being multiplied by 3, you need to divide by 3; if x is being decreased by 4, you need to add 4, and so on. The trick is to do these things in the right order. Basically, you should follow PEMDAS in reverse. Start by undoing addition and subtraction, then multiplication and division, then exponents and roots, and, last, what’s in parentheses.

The other thing to remember is that any time you do something to one side of an equation, you’ve got to do it to the other side also. Otherwise you’d be changing the equation, and you’re trying to rearrange it, not change it. In this example, you’d start by undoing the subtraction.

5(3x3 − 16) − 22 = 18

                    + 22 + 22

        5(3x3 − 16) = 40

Then undo the multiplication by 5, saving what’s in the parentheses for last.

Once you’ve gotten down to what’s in the parentheses, follow PEMDAS in reverse again—first the subtraction, then the multiplication, and the exponent last.

At this point, you’ve solved the equation. You have found that the value of x must be 2. Another way of saying this is that 2 is the root of the equation 5(3x3 − 16) − 22 = 18. Equations containing exponents may have more than one root (see “Exponents,” in the last chapter).

Solving Equations with Absolute Value

The rules for solving equations with absolute value are the same. The only difference is that, because what’s inside the absolute value signs can be positive or negative, you’re solving for two different results.

Let’s look at an example:

20. |x − 2| = 17

Now, we know that either (x − 2) is a negative number or a non-negative number. When a number is negative, the absolute value makes it the inverse, or multiplies it by −1 to yield a positive result. If the number is positive, it remains the same after being sent through the absolute value machine. So when we remove the absolute value bars, we’re left with two different equations:

Vocab Review

Remember that a non-negative
number can be
either a positive number or
zero. Since zero is neither
positive nor negative, if
we said “positive number”
that wouldn’t include zero.

Now simply solve both equations:

And that’s all there is to it!


Practice solving equations in the following examples. Remember that some equations may have more than one root. The answers to these drills can be found in Chapter 12.

  1.  If  = 4, then x =

  2.  If n2 = 5n, then n =

  3.  If , then a =

  4.  If  = 21, then s =

  5.  If  = 4, then x =

  6.  If |2m + 5| = 23, then m =

  7.  If  = 4, then r =


When manipulating algebraic equations, you’ll need to use the tools of factoring and distributing. These are simply ways of rearranging equations to make them easier to work with.


Factoring simply means finding some factor that is in every term of an expression and “pulling it out.” By “pulling it out,” we mean dividing each individual term by that factor, and then placing the whole expression in parentheses with that factor on the outside. Here’s an example:

x3 − 5x2 + 6x = 0

On the left side of this equation, every term contains at least one x—that is, x is a factor of every term in the expression. That means you can factor out an x:

x3 − 5x2 + 6x = 0

x(x2 − 5x + 6) = 0

The new expression has exactly the same value as the old one; it’s just written differently, in a way that might make your calculations easier. Numbers as well as variables can be factored out, as seen in the example below.

17c − 51 = 0

On the left side of this equation, every term is a multiple of 17. Because 17 is a factor of each term, you can pull it out.

17c − 51 = 0

17(c − 3) = 0

      c − 3 = 0

             c = 3

As you can see, factoring can make equations easier to solve.


Distributing is factoring in reverse. When an entire expression in parentheses is being multiplied by some factor, you can “distribute” the factor into each term, and get rid of the parentheses. For example:

3x(4 + 2x) = 6x2 + 36

On the left side of this equation the parentheses make it difficult to combine terms and simplify the equation. You can get rid of the parentheses by distributing.

3x(4 + 2x) = 6x2 + 36

12x + 6x2 = 6x2 + 36

And suddenly, the equation is much easier to solve.

12x + 6x2 = 6x2 + 36

−6x2 −6x2

12x = 36

x = 3


Practice a little factoring and distributing in the following examples, and keep an eye out for equations that could be simplified by this kind of rearrangement. The answers to these drills can be found in Chapter 12.

  3.  If (11x)(50) + (50x)(29) = 4,000, then x =

(A)   2,000

(B)   200

(C)   20

(D)   2

(E)   0.2

17.  If ab ≠ 0,  =

(A)   –3

(B)   –2

(C)   0

(D)   1

(E)   3

36.  If x ≠ −1,  =

(A)   4x2

(B)   x2

(C)   4x

(D)   x

(E)   4


Plugging In is a technique for short-cutting algebra questions. It works on a certain class of algebra questions in which relationships are defined, but no real numbers are introduced. For example:

To Number or Not to Number?

Let’s say you walk into a candy store. The store is selling certain pieces of candy for 5 cents and 10 cents each. You want to get 3 pieces of the 5 cent candy and 6 pieces of the 10 cent candy. You give the cashier a $5 bill. What’s your change?

Ok, now let’s say you walk into a candy store run by ETS. This store is selling certain pieces of candy for x cents and y cents each. You want to get m pieces of the x cent candy and n pieces of the y cent candy. You give the cashier a $z bill. What’s your change?

Which problem would be easier to solve? The one with the numbers! Numbers make everything easier. So why bother with variables when you don’t have to?

11.  The use of a neighborhood car wash costs n dollars for a membership and p cents for each wash. If a membership includes a bonus of 4 free washes, which of the following reflects the cost, in dollars, of getting a membership at the car wash and washing a car q times, if q is greater than 4 ?

(A)   100n + pq − 4p

(B)   n + 100pq − 25p

(C)   n + pq − 

(D)   n +  − 

(E)   n +  − 

Here’s How to Crack It

In this problem you see a lot of variables in the question and in the answer choices. That’s a big clue!

When you see variables in the answer choices, PLUG IN!

Let’s try Plugging In with this problem. We’ll start with n, the membership fee.

Plug In an easy number like 3, so that a membership costs $3.00.

Then, Plug In a number for p, the charge per wash. Since this number is in cents, and we’ll need to convert it to dollars in the answers, choose a number that can be converted easily to dollars, like 200. Let’s make p = 200, so a wash costs $2.00.

Last, let’s say that q, the number of washes, is 5. That’s as easy as it gets. With 4 free washes, you’re paying for only 1.

Then, just work out the answer to the question using your numbers. How much does it cost for a membership and 5 washes? Well, that’s $3.00 for a membership, 4 washes free, and 1 wash for $2.00. The total is $5.00. That means that if you plug your numbers into the answer choices, the right answer should give you 5. We call that your target number—the number you are looking for in the answer choices. Put a double circle around your target number, so that it stands out from all the other numbers you’ve written down. It looks like a bull’s-eye that you’re trying to hit:

When you plug n = 3, p = 200, and q = 5 into the answer choices, the only answer choice that gives you 5 is (D). That means you’ve hit your target number, and you’re done.

A Big Clue

There will be times when
ETS will give you questions
that include variables
and the phrase “in terms
of” (for example, “in terms
of x”). This is a big clue
that you can Plug In. Cross
off the phrase “in terms of
x,” because you don’t need
it to solve the problem.

Take a look at one more:

13.  The size of an art collection is tripled, and then 70 percent of the collection is sold. Acquisitions then increase the size of the collection by 10 percent. The size of the art collection is then what percent of its size before these three changes?

(A)      240%

(B)      210%

(C)      111%

(D)      99%

(E)      21%

Here’s How to Crack It

Here’s another question in which you aren’t given numbers. In this case, you don’t know the original size of the art collection. Instead of variables, though, the question and answers contain percents. This is another sign that you can Plug In whatever numbers you like. Because you’re working with percentages, 100 is a good number to Plug In—it’ll make your math easier.

You start with a collection of 100 items. It’s tripled, meaning it increases to 300. Then it’s decreased by 70%. That’s a decrease of 210, so the collection’s size decreases to 90. Then, finally, it increases by 10%. That’s an increase of 9, for a final collection size of 99. Since the collection began at 100, it’s now at 99% of its original size. The answer is (D). It doesn’t matter what number you choose for the original size of the collection—you’ll always get the right answer. The trick to choosing numbers is picking ones that make your math easier.

The idea behind Plugging In is that if these relationships are true, then it doesn’t matter what numbers you put into the question; you’ll always arrive at the same answer choice. So the easiest way to get through the question is to Plug In easy numbers, follow them through the question, and see which answer choice they lead you to.

Not Sure When to Plug In? Here Are Some Hints

·        The answer choices contain variables, percentages, fractions, or ratios.

·        There are unknown quantities or variables in the question.

·        The question seems to call for an algebraic equation.

·        You see the phrase “in terms of” followed by a variable (for example “in terms of p”). Cross off the phrase “in terms of p,” because you don’t need it to solve the problem.

Occasionally, more than one answer choice will produce the correct answer. This often occurs when the question asks for something that “must be true.” When that happens, eliminate the answer choices that didn’t work out, and Plug In some different kinds of numbers. Some numbers you might try are odd and even integers, positive and negative numbers, fractions, zero, positive or negative one, and really big or really small numbers, like 1,000 or −1,000. The new numbers will produce a new target number. Use this new target number to eliminate the remaining incorrect answer choices. You will rarely have to Plug In more than two sets of numbers.

When using Plugging In, keep a few simple rules in mind:

·        Avoid Plugging In 1 or 0, which often makes more than one answer choice produce the same number. For the same reason, avoid Plugging In numbers that appear in the answer choices—they’re more likely to cause several answer choices to produce your target number.

·        Plug In numbers that make your math easy—2, 3, and 5 are good choices in ordinary algebra. Multiples of 100 are good in percentage questions, and multiples of 60 are good in questions dealing with seconds, minutes, and hours.

Plugging In can be an incredibly useful technique. By Plugging In numbers, you’re checking your math as you do the problem. When you use algebra, it takes an extra step to check your work with numbers. Also, there are fewer chances to mess up when you Plug In. And you can Plug In even when you don’t know how to set up an algebraic equation.

Plugging In is often safer because ETS designs the answer choices so that, if you mess up the algebra, your result will be one of the wrong answers. When your answer matches one of the choices, you think it must be right. Very tempting. Furthermore, all of the answer choices look very similar, algebraically. This is how ETS camouflages correct answers. But when you Plug In, the answers often look very different. Often you’ll be able to approximate to eliminate numbers that are obviously too big or too small, without doing a lot of calculation, and that will save you lots of time!


Try solving the following practice questions by Plugging In. Remember to check all your answer choices, and Plug In a second set of numbers if more than one answer choice produces your target number. The answers to these drills can be found in Chapter 12.

  5.  The price of an item in a store is p dollars. If the tax on the item is t%, what is the total cost in dollars of n such items, including tax?

(A)   npt

(B)   npt + 1


(D)   100n(p + pt)


  8.  Vehicle A travels at x miles per hour for x hours. Vehicle B travels a miles per hour faster than Vehicle A, and travels b hours longer than Vehicle A. Vehicle B travels how much farther than Vehicle A, in miles?

(A)   x2 − ab

(B)   a2 + b2

(C)   ax + bx + ab

(D)   x2 + abx + ab

(E)   2x2 + (a + b)x + ab

17.  For any real number n, |5 − n| − |n − 5| =

(A)   −2

(B)   −1

(C)   0

(D)   1

(E)   2

20.  If Company A builds a skateboards per week, and Company B builds b skateboards per day, then in m weeks, Company A builds how many more skateboards than Company B ?

(A)   7bm

(B)   m(a − 7b)

(C)   7(ma − mb)

(D)   7m(a − b)


23.  If a > 3 and b < 3, then which of the following could be true?

    I.   a − b > 3

   II.   a + b < 3

  III.   |a + b| < 3

(A)   I only

(B)   III only

(C)   I and II only

(D)   II and III only

(E)   I, II, and III

30.  For all real numbers, x3 < y3. Which of the following must be true?

    I.   x < y

   II.   x2 < y2

  III.   |x| < |y|

(A)   I only

(B)   III only

(C)   I and II only

(D)   II and III only

(E)   I, II, and III


Plugging In the Answers (PITA) is another approach to solving algebra questions. It uses numbers instead of algebra to find the answer. As you’ve just seen, Plugging In is useful on questions whose answer choices contain variables, percentages, fractions, or ratios—not actual numbers. PITA, on the other hand, is useful on questions whose answer choices do contain actual numbers.

ETS always organizes answers in numerical order—usually from least to greatest. You can use this to your advantage by combining PITA and POE.

To use PITA on an algebra question, take (C), the middle answer choice, and stick it back into the problem. If it makes all of the statements in the question true, then it’s the right answer. If it doesn’t, eliminate (C) and try another answer choice. Usually, you’ll know from your experience with (C) whether you want to try a smaller or larger answer choice. If (C) is too small, you can eliminate the smaller two choices and try again with the remaining two.

Like Plugging In, PITA can open doors for you when you’re unsure how to approach a question with algebra. Also, like Plugging In, it checks your answers as you pick them, eliminating careless errors. Particularly at the tough end of a Math Subject Test, where you’re getting into hard material, Plugging In and PITA can enable you to solve problems that would otherwise stump you.

Let’s take a look at a PITA example.

10.   A duck travels from point A to point B. If the duck flies  of the way, walks  of the way, and swims the remaining 10 kilometers of her trip, what is the total distance in kilometers traveled by the duck?

(A)   36

(B)   45

(C)   56

(D)   72

(E)   108

Here’s How to Crack It

To use PITA on this question, you’d start with answer choice (C). The answer choices represent the quantity asked for in the question—in this case, the total distance traveled by the duck. Always know what question you’re answering. Answer choice (C), therefore, means that the duck traveled a total distance of 56 kilometers. Follow this information through the problem.

The duck flies  of the way.  of 56 is 42 kilometers.

The duck walks  of the way.  of 56 is 6.22 kilometers.

That makes 48.22 kilometers, which leaves 7.78 kilometers in the trip.

BUT the duck swims 10 kilometers!

That means that (C) isn’t the right answer. It also tells you that you need a longer trip to fit in all that flying, walking, and swimming; move down to (D), the next largest answer, and try again. At this point, you can also eliminate (A), (B), and (C) because they are too small.

The duck flies  of the way.  of 72 is 54 kilometers.

The duck walks  of the way.  of 72 is 8 kilometers.

That makes 62 kilometers, which leaves 10 kilometers in the trip.

And THAT’S exactly how far the duck swims, 10 kilometers. Right answer.

Generally, you’ll never have to try more than two answer choices when using PITA thanks to POE—and sometimes, the first answer you try, (C), will be correct. Keep your eyes open for PITA opportunities on the Math Subject Tests, particularly when you run into an algebra question that you’re not sure how to solve.

You Should Try PITA Whenever

·        there is an unknown or variable in the question, the question asks for an actual value, and the answer choices are all numbers arranged in increasing or decreasing order

·        you have the bizarre urge to translate a word problem into a complicated algebraic equation

·        you find yourself reading a long, convoluted story about some number, and you don’t even know what the number is

·        you have no idea how to solve the problem

If, after you Plug In answer choice (C), you’re not sure which way to go in the answer choices, don’t haggle for too long. Just eliminate (C), pick a direction, and go! If you go the wrong way, you’ll know pretty quickly, and then you can head the other way.


Solve the following questions by using PITA. Remember to start with (C), the middle answer choice. The answers to these drills can be found in Chapter 12.

11.  Matt has 4 more hats than Aaron and half as many hats as Michael. If the three together have 24 hats, how many hats does Michael have?

(A)   7

(B)   9

(C)   12

(D)   14

(E)   18

17.  A shipment of 3,200 items is divided into 2 portions so that the difference between the portions is one-half of their average. What is the ratio of the smaller to the larger portion?

(A)   1 : 2

(B)   1 : 3

(C)   2 : 5

(D)   3 : 5

(E)   5 : 8

27.  Three distinct positive integers have a sum of 15 and a product of 45. What is the largest of these integers?

(A)   1

(B)   3

(C)   5

(D)   9

(E)   15


Inequalities can be treated just like equations. You can add, subtract, multiply, and divide on both sides of the inequality sign. And you still solve by isolating the variable. There is one major difference between solving an equation and solving an inequality.

Reading Inequality

Here’s how you should
read the four basic
inequality signs:

a < b    a is less than b

a > b    a is greater than b

a ≤ b    a is less than or
equal to b

a ≥ b    a is greater than
or equal to b

Whenever you multiply or divide both sides of an inequality by a negative, flip the inequality sign.

Multiplying across an inequality by a negative flips the signs of all of the terms in the inequality. The inequality sign itself must also flip.

−1(4n − 20 > −3n + 15)

−1(x ≥ 5)

−4n + 20 < 3n − 15

x ≤ −5

As long as you remember this rule, you can treat inequalities just like equations and use all of your algebra tools to solve them.


Practice solving inequalities in the following exercises. The answers to these drills can be found in Chapter 12.

  1.  If ≤3, then

  2.  If < 5, then

  3.  If ≤ 5, then

  4.  If 8(3x + 1) + 4 < 15, then

  5.  If 23 − 4t ≥11, then

  6.  If 4n − 25 ≤ 19 − 7n, then

  7.  If −5(p + 2) < 10p − 13, then

  8.  If  ≥ 2s + 1, then

  9.  If −3x − 16 ≤ 2x + 19, then

10.  If  ≥ s − 1, then


Inequalities are also used when discussing the range of possible values a variable could equal. Sometimes you’ll see an algebraic phrase in which there are two inequality signs. These are called ranges. Your variable can be any number within a range of numbers. For example: 2 < x < 10. This means that x can be any number between, but not equal to, 2 and 10. Let’s look at this next example:

At a certain amusement park, anyone under 12 years of age is not permitted to ride the Stupendous Hurlcoaster, because the person could easily lose his or her mind due to the ride’s extreme funkiness. Anyone over 60 years of age is also prohibited from the ride, as the incredible velocity of the Hurlcoaster may cause spontaneous coronary explosion. If x is the age of a rider of the Stupendous Hurlcoaster, what is the range of possible values of x ?

The end values of the range are obviously 12 and 60. But are 12 and 60 included in the range themselves, or not? If you read carefully, you’ll see that only those under 12 or over 60 are barred from riding the Hurlcoaster. If you’re 12 or 60, you’re perfectly legal. The range of possible values of x is therefore given by 12 ≤ x ≤ 60. Noticing the difference between “greater than” and “greater than or equal to” is crucial to many range questions.

You can manipulate ranges in a couple of ways. You can add and subtract ranges, as long as their inequality signs point the same way. You can also multiply or divide across a range to produce new information, as long as you obey that basic rule of inequalities—flip the sign if you multiply or divide by a negative number.


If the range of possible values for x is given by −5 < x < 8, find the range of possible values for each of the following:

1.   −x:

2.   4x:

3.   x + 6:

4.   (2 − x):

5.   :

Adding Ranges

Occasionally, a question on the Math Subject Tests will require you to add, subtract, or multiply ranges. Take a look at this example:

If 3 < a < 10 and −6 < b < 3, what is the range of possible values of a + b ?

Here, the range of (a + b) will be the sum of the range of a and the range of b. The easy way to do this is to list out the four ways you can combine the endpoints of the two ranges. To do this, take the smallest a and add it to the smallest b. Then, add the smallest a to the biggest b. Then add the biggest a to the smallest b. Finally, take the biggest a and add it to the biggest b. The biggest and smallest results you get will be the endpoints of the range of (a + b). Watch!

3 + −6 = −3

3 + 3 = 6

10 + −6 = 4

10 + 3 = 13

The smallest number you found is −3, and the biggest is 13, so the range of possible values looks like the following:

−3 < a + b < 13

Subtracting Ranges

To subtract one range from another, combine the endpoints just as you did when adding ranges, but in this case, subtract the four combinations of endpoints. Make sure you’re subtracting in the order the question asks you to. Let’s look at this example.

If −4 < a < 5 and 2 < b < 12, then what is the range of possible values of a − b ?

This time, take the smallest a and subtract the smallest b. Then, find the smallest a minus the biggest b, and so on.

−4 − 2 = −6

−4 − 12 = −16

5 − 2 = 3

5 − 12 = −7

So the range you’re looking for is:

−16 < a − b < 3

Multiplying Ranges

To multiply ranges, follow the same steps, but multiply the endpoints. Let’s try one.

If −3 < f < 4 and −7 < g < 2, then what is the range of possible values of fg ?

These are the four possible products of the bounds of f and g.

(−3)(−7) = 21

(−3)(2) = −6

    (4)(−7) = −28

(4)(2) = 8  

The greatest of these values is 21 and the least is −28. So the range of possible values of fg is:

−28 < fg < 21

And that’s all there is to dealing with ranges.


Try the following range questions. The answers to these drills can be found in Chapter 12.

  1.  If −2 ≤ a ≤ 7 and 3 ≤ b ≤ 9, then what is the range of possible values of b − a ?

  2.  If 2 ≤ x ≤ 11 and 6 ≥ y ≥ −4, then what is the range of possible values of x + y ?

  3.  If −3 ≤ n ≤ 8, then what is the range of possible values of n2 ?

  4.  If 0 < x < 5 and −9 < y < −3, then what is the range of possible values of x − y ?

  5.  If −3 ≤ r ≤ 10 and −10 ≤ s ≤ 3, then what is the range of possible values of r + s ?

  6.  If −6 < c < 0 and 13 < d < 21, then what is the range of possible values of cd ?

  7.  If |3 −x| ≤ 4, then what is the range of possible values of x ?

  8.  If |2a + 7| ≥13, then what is the range of possible values of a ?


Trouble with absolute
value? Pay special attention
to the explanations
of questions 7 and 8 from
this drill.


Direct and indirect variation are specific relationships between quantities. Quantities that vary directly are said to be in proportion or proportional. Quantities that vary indirectly are said to be inversely proportional.

Direct Variation

If x and y are in direct variation, that can be said in several ways: x and y are in proportion; x and y change proportionally; or x varies directly as y. All of these descriptions come down to the same thing: x and y increase and decrease together. Specifically, they mean that the quantity  will always have the same numerical value. That’s all there is to it. Take a look at a question based on this idea.

A Great Way to Remember

To remember direct variation, think “direct means divide.” So in order to solve, you set up a proportion with a fraction on each side of the equation. Just solve for the one number you don’t know. There are two formulas associated with direct variation that may be familiar to you. They are:  or y = kx, where k is a constant.

17.  If n varies directly as m, and n is 3 when m is 24, then what is the value of n when m is 11 ?

(A)   1.375

(B)   1.775

(C)   1.95

(D)   2.0

(E)   2.125

Here’s How to Crack It

To solve the problem, use the definition of direct variation:  must always have the same numerical value. Set up a proportion.

Solve by cross-multiplying and isolating n.

24n = 33

            n = 33 ÷ 24

        n = 1.375

And that’s all there is to it. The correct answer is (A). All direct variation questions can be answered this way.

Indirect Variation

If x and y are in inverse variation, this can be said in several ways as well: x and y are in inverse proportion; x and y are inversely proportional; or x varies indirectly as y. All of these descriptions come down to the same thing: x increases when y decreases, and decreases when y increases. Specifically, they mean that the quantity xy will always have the same numerical value.

Opposites Attract

A great way to remember indirect or inverse variation is that direct and indirect are opposites. What’s the opposite of division? Multiplication! So set up an inverse variation as two multiplication problems on either side of an equation. There are two formulas associated with indirect variation that may be familiar to you. They are: x1y1 = x2y2 or y = , where k is a constant.

Take a look at this question based on inverse variation:

15.  If a varies inversely as b, and a = 3 when b = 5, then what is the value of a when b = 7 ?

(A)   2.14

(B)   2.76

(C)   3.28

(D)   4.2

(E)   11.67

Here’s How to Crack It

To answer the question, use the definition of inverse variation. That is, the quantity ab must always have the same value. Therefore, you can set up this simple equation.

3 × 5 = a × 7

7a = 15

         a = 15 ÷ 7

            a = 2.142857

So the correct answer is (A). All inverse variation questions on the Math Subject Tests can be handled this way.


Try these practice exercises using the definitions of direct and inverse variation. The answers to these drills can be found in Chapter 12.

15.  If a varies inversely as b, and a = 3 when b = 5, then what is the value of a when b = x ?




(D)   3x

(E)   3x2

18.  If n varies directly as m, and n = 5 when m = 4, then what is the value of n when m = 5 ?

(A)   4.0

(B)   4.75

(C)   5.5

(D)   6.25

(E)   7.75

24.  If p varies directly as q, and p = 3 when q = 10, then what is the value of p when q = 1 ?

(A)   0.3

(B)   0.43

(C)   0.5

(D)   4.3

(E)   4.33

26.  If y varies directly as x2, and y = 24 when x = 3.7, what is the value of y when x = 8.3 ?

(A)   170.67

(B)   120.77

(C)   83.23

(D)   64.00

(E)   53.83


Word problems dealing with work and travel tend to cause a lot of careless mistakes, because the relationships among distance, time, and speed—or among work-rate, work, and time—sometimes confuse test takers. When working with questions about travel, just remember this:

distance = rate × time

When working with questions about work being done, remember this:

work done = rate of work × time

Look familiar?

If these two formulas
seem the same, it’s
because they are. After all,
what is work done if not
the distance from the start
of work to the end? Don’t
worry about learning too
many equations; generally
speaking, the few you’ll
need are more useful than
their wording indicates.


Answer the following practice questions using these formulas. The answers to these drills can be found in Chapter 12.

11.  A factory contains a series of water tanks, all of the same size. If Pump 1 can fill 12 of these tanks in a 12-hour shift, and Pump 2 can fill 11 tanks in the same time, then how many tanks can the two pumps fill, working together, in 1 hour?

(A)   0.13

(B)   0.35

(C)   1.92

(D)   2.88

(E)   3.33

12.  A projectile travels 227 feet in one second. If there are 5,280 feet in 1 mile, then which of the following best approximates the projectile’s speed in miles per hour?

(A)   155

(B)   170

(C)   194

(D)   252

(E)   333

18.  A train travels from Langston to Hughesville and back in 5.5 hours. If the two towns are 200 miles apart, what is the average speed of the train in miles per hour?

(A)   36.36

(B)   72.73

(C)   109.09

(D)   110.10

(E)   120.21

25.  Jules can make m muffins in s minutes. Alice can make n muffins in t minutes. Which of the following gives the number of muffins that Jules and Alice can make together in 30 minutes?



(C)   30(mt + ns)



Average Speed

The “average speed” question is a specialized breed of travel question. Here’s what a basic “average speed” question might look like.

15.  Roberto travels from his home to the beach, driving at 30 miles per hour. He returns along the same route at 50 miles per hour. If the distance from Roberto’s house to the beach is 10 miles, then what is Roberto’s average speed for the round-trip in miles per hour?

(A)   32.5

(B)   37.5

(C)   40.0

(D)   42.5

(E)   45.0

The easy mistake to make on this question is to simply choose answer choice (C), the average of the two speeds. Average speed isn’t found by averaging speeds, however. Instead, you have to use this formula:

Don’t Be Joe!

Remember, Joe takes the
easy way out. He thinks
that if you need the average
of two averages, you
should just average them.
No! But knowing what
Joe would do helps you.
Now you know you can
eliminate (C), because it’s
what Joe would pick.

average speed = 

The total distance is easy to figure out—10 miles each way is a total of 20 miles. Total time is a little trickier. For that, you have to use the “distance = rate × time” formula. Here, it’s useful to rearrange the eqution to read as follows:

time = 

On the way to the beach, Roberto traveled 10 miles at 30 mph, which took 0.333 hours, according to the formula. On the way home, he traveled 10 miles at 50 mph, which took 0.2 hours. That makes 20 miles in a total of .533 hours. Plug those numbers into the average-speed formula, and you get an average speed of 37.5 mph. The answer is (B).

Look Familiar?

This formula may look familiar to you. That’s because it’s taken from our average pie. Another way to work with average speed questions is to use the average pie where the total is the total distance and the number of things is the time. So it would look like this:

Here’s a general tip for “average speed“ questions: On any round-trip in which the traveler moves at one speed heading out and another speed returning, the traveler’s average speed will be a little lower than the average of the two speeds.


Try these “average speed” questions. The answers to these drills can be found in Chapter 12.

19.  Alexandra jogs from her house to the lake at 12 miles per hour and jogs back by the same route at 9 miles per hour. If the path from her house to the lake is 6 miles long, what is her average speed in miles per hour for the round-trip?

(A)   11.3

(B)   11.0

(C)   10.5

(D)   10.3

(E)   10.1

24.  A truck travels 50 miles from Town S to Town T in 50 minutes, and then immediately drives 40 miles from Town T to Town U in 40 minutes. What is the truck’s average speed in miles per hour, from Town S to Town U ?

(A)   1

(B)   10

(C)   45

(D)   60

(E)   90

33.  Ben travels a certain distance at 25 miles per hour and returns across the same distance at 50 miles per hour. What is his average speed in miles per hour for the round-trip?

(A)   37.5

(B)   33.3

(C)   32.0

(D)   29.5

(E)   It cannot be determined from the information given.


It’s possible to have a set of equations that can’t be solved individually but can be solved in combination. A good example of such a set of equations would be:

4x + 2y = 18

x + y = 5

You can’t solve either equation by itself. But you can if you put them together. It’s called simultaneous equations. All you do is stack them and then add or subtract them to find what you’re looking for. Often, what you’re looking for is another equation. For example, the question that contains the two equations you were given wants to know what the value of 10x + 6y is. Do you need to know x or y? No! You just need to know 10x + 6y. Let’s try adding the two equations:

Did adding help? It did! Even though we didn’t get what they were asking for, we did get half of what they were asking for. So just multiply the entire equation by 2 and you have your answer: 46.

That Nasty Phrase
“In Terms Of”

Remember how we had
you cross off the phrase
“in terms of” when you
plugged in because it
doesn’t help you at all?
Well, solving x “in terms
of” y for simultaneous
equations doesn’t help
either. It takes too much
time and there is too much
room for error to solve in
terms of one variable and
then put that whole thing
into the other equation.
And much of the time,
that’s unnecessary
because we don’t care
what the values of the
individual variables are!

Here’s another example of a system of simultaneous equations as they might appear on a Math Subject Test question. Try it.

  7.  If x and y are real numbers such that 3x + 4y = 10 and 2x − 4y = 5, then what is the value of x ?

Add It Up

Do you notice how adding
brings you close to what
the question is asking for?

In the question above, instead of solving to find a third equation, you need to find one of the variables. Your job doesn’t change: Stack ’em; then add or subtract. This will be the case with every simultaneous equations question. Every once in a while you may want to multiply or divide one equation by a number before you add or subtract.

Try another one. Solve it yourself before checking the explanation.

  8.  If 12a − 3b = 131 and 5a − 10b = 61, then what is the value of a + b ?

This time adding didn’t work, did it? Let’s go through and see what subtraction does:

Avoid Subtraction

If adding doesn’t work and
you want to try subtracting,
wait! Multiply one of
the equations by −1 and
add instead. That way
you ensure that you don’t
make any calculation
errors along the way.

A little practice will enable you to see quickly whether adding or subtracting will be more helpful. Sometimes it may be necessary to multiply one of the equations by a convenient factor to make terms that will cancel out properly. For example:

  6. If 4n − 8m = 6, and −5n + 4m = 3, then n =

4n − 8m = 6

−5n + 4m = 3

Here, it quickly becomes apparent that neither adding nor subtracting will combine these two equations very usefully. However, things look a little brighter when the second equation is multiplied by 2.

Occasionally, a simultaneous equation can be solved only by multiplying all of the pieces together. This will generally be the case only when the equations themselves involve multiplication alone, not the kind of addition and subtraction that the previous equations contained. Take a look at this example:

# of Equations = # of

We’ve been talking about
two equations, two
variables. But ETS doesn’t
stop there. A good rule of
thumb is, if the number of
equations is equal to the
number of variables, you
can solve the equations.
So count ’em and don’t
get discouraged! They’re
always easier than they

ab = 3

bc = 

ac = 15

34.  If the above statements are true, what is one possible value of abc ?

(A)   5.0

(B)   8.33

(C)   9.28

(D)   18.54

(E)   25.0

Here’s How to Crack It

This is a tough one. No single one of the three small equations can be solved by itself. In fact, no two of them together can be solved. It takes all three to solve the system, and here’s how it’s done:

ab × bc × ac = 3 ×  × 15

aabbcc = 25

a2b2c2 = 25

Where’s the Trap?

Remember that a number
34 is a difficult question.
What do you notice about
choice (E)?

Once you’ve multiplied all three equations together, all you have to do is take the square roots of both sides, and you’ve got a value for abc.

a2b2c2 = 25

        abc = 5, −5

And so (A) is the correct answer.


Try answering the following practice questions by solving equations simultaneously. The answers to these drills can be found in Chapter 12.

26.  If a + 3b = 6, and 4a − 3b = 14, a =

(A)   −4

(B)   2

(C)   4

(D)   10

(E)   20

31.  If 2x − 7y = 12 and −8x + 3y = 2, which of the following is the value of x − y ?

(A)   12.0

(B)   8.0

(C)   5.5

(D)   1.0

(E)   0.8

ab = bc = 6, ac = 3

34.  If all of the above statements are true, what is one possible value of abc ?

(A)   3.75

(B)   2.25

(C)   2.0

(D)   1.5

(E)   0.25

37.  If xyz = 4 and y2z = 5, what is the value of  ?

(A)   20.0

(B)   10.0

(C)   1.25

(D)   1.0

(E)   0.8


A binomial is an algebraic expression that has two terms (pieces connected by addition or subtraction). FOIL is how to multiply two binomials together.

The letters of FOIL stand for:

Suppose you wanted to do the following multiplication:

(x + 5)(x − 2)

You would multiply the two first terms together, (x)(x) = x2.

And then the outside terms, (x)(−2) = −2x.

And then the inside terms, (5)(x) = 5x.

And finally the two last terms, (5)(−2) = −10.

String the four products together and simplify them to produce an answer.

x2 − 2x + 5x − 10

x2 + 3x − 10

And that’s the product of (x + 5) and (x − 2).


Practice using FOIL on the following binomial multiplications. The answers to these drills can be found in Chapter 12.

  1.  (x − 2)(x + 11) =

  2.  (b + 5)(b + 7) =

  3.  (x − 3)(x − 9) =

  4.  (2x − 5)(x + 1) =

  5.  (n2 + 5)(n − 3) =

  6.  (3a + 5)(2a − 7) =

  7.  (x − 3)(x − 6) =

  8.  (c − 2)(c + 9) =

  9.  (d + 5)(d − 1) =


An expression like x2 + 3x + 10 is a quadratic polynomial. A quadratic is an expression that fits into the general form ax2 + bx + c, with a, b, and c as constants. An equation in general quadratic form looks like this:

General Form of a Quadratic Equation

ax2 + bx + c = 0

Often, the best way to solve a quadratic equation is to factor it into two binomials—basically FOIL in reverse. Let’s take a look at the quadratic you worked with in the previous section, and the binomials that are its factors.

x2 + 3x − 10 = (x + 5)(x − 2)

Notice that the coefficient of the quadratic’s middle term (3) is the sum of the constants in the binomials (5 and −2), and that the third term of the quadratic (−10) is the product of those constants. That relationship between a quadratic expression and its factors will always be true. To factor a quadratic, look for a pair of constants whose sum equals the coefficient of the middle term, and whose product equals the last term of the quadratic. Suppose you had to solve this equation:

x2 − 6x + 8 = 0

Your first step would be to factor the quadratic polynomial. That means looking for a pair of numbers that add up to −6 and multiply to 8. Because their sum is negative but their product is positive, you know that the numbers are both negative. And as always, there’s only one pair of numbers that fits the bill—in this case, −2 and −4.

x2 − 6x + 8 = 0

(x − 2)(x − 4) = 0

Since zero multiplied by anything is equal to zero, this equation will be true if (x − 2) = 0 or if (x − 4) = 0. Therefore,

x = {2, 4}

Two and four are therefore called the zeros of the equation. They are also known as the roots or solutions of the equation.

Once a quadratic is factored, it’s easy to solve for x. The product of the binomials can be zero only if one of the binomials is equal to zero—and there are only two values of x that will make one of the binomials equal to zero (2 and 4). The equation is solved.


Solve the following equations by factoring the quadratic polynomials. Write down all of the roots of each equation (values of the variable that make the equations true). The answers to these drills can be found in Chapter 12.

1.   a2 − 3a + 2 = 0

2.   d2 + 8d + 7 = 0

3.   x2 + 4x− 21 = 0

4.   3x2 + 9x − 30 = 0

5.   2x2 + 40x + 198 = 0

6.   p2 + 10p = 39

7.   c2 + 9c + 20 = 0

8.   s2 + 4s − 12 = 0

9.   x2 − 3x − 4 = 0

10.  n4 − 3n2 − 10 = 0

Special Quadratic Identities

There are a few quadratic expressions that you should be able to factor at a glance. Because they are useful mathematically, and above all, because ETS likes to put them on the Math Subject Tests, you should memorize the following identities:

(x + y)2 = x2 + 2xy + y2

(x − y)2 = x2 − 2xy + y2

(x + y)(x − y) = x2 − y2              

Here are some examples of these quadratic identities in action.

1.  n2 + 10n + 25 = (n + 5)(n + 5) = (n + 5)2

2.  r2 − 16 = (r + 4)(r − 4)

3.  n2 − 4n + 4 = (n − 2)(n − 2) = (n − 2)2

But knowing the quadratic identities will do more for you than just allow you to factor some expressions quickly. ETS writes questions based specifically on these identities. Such questions are easy to solve if you remember these equations and use them, and quite tricky (or even impossible) if you don’t. Here’s an example.

36.  If a + b = 7, and a2 + b2 = 37, then what is the value of ab ?

(A)   6

(B)   12

(C)   15

(D)   22

(E)   30

Here’s How to Crack It

Algebraically, this is a tough problem to crack. You can’t divide a2 + b2 by a + b and get anything useful. In fact, most of the usual algebraic approaches to questions like these don’t work here. Even plugging the answer choices back into the question (PITA) isn’t very helpful. What you can do is recognize that the question is giving you all of the pieces you need to build the quadratic identity: (x + y)2 = x2 + 2xy + y2. To solve the problem, just rearrange the identity a little and Plug In the values given by the question.

(a + b)2 = a2 + b2 + 2ab

(7)2 = 37 + 2ab

49 = 37 + 2ab

12 = 2ab        

6 = ab         

And presto, the answer appears. It’s not easy to figure out what a or b is specifically—and you don’t need to. Just find the value asked for in the question. If you remember the quadratic identities, solving the problem is easy.


Try solving the following questions using the quadratic identities, and take note of the clues that tell you when the identities will be useful. The answers to these drills can be found in Chapter 12.

Pencil It In

Don’t forget to put pencil
to paper on these and
other algebra questions
with lots of steps. Seeing
your work will help you
better understand the
questions and solutions
and avoid careless

17.  If n − m = −3 and n2 − m2 = 24, then which of the following is the sum of n and m ?

(A)   −8

(B)   −6

(C)   −4

(D)   6

(E)   8

19.  If x + y = 3 and x2 + y2 = 8, then xy =

(A)   0.25

(B)   0.5

(C)   1.5

(D)   2.0

(E)   2.25

24.  If the sum of two nonzero integers is 9 and the sum of their squares is 36, then what is the product of the two integers?

(A)   9.0

(B)   13.5

(C)   18.0

(D)   22.5

(E)   45.0


Unfortunately, not all quadratic equations can be factored by the reverse-FOIL method. The reverse-FOIL method is practical only when the roots of the equation are integers. Sometimes, however, the roots of a quadratic equation will be non-integer decimal numbers, and sometimes a quadratic equation will have no real roots at all. Consider the following quadratic equation:

x2 − 7x + 8 = 0

There are no integers that add up to −7 and multiply to 8. This quadratic cannot be factored in the usual way. To solve this equation, it’s necessary to use the quadratic formula—a formula that produces the root or roots of any equation in the general quadratic form ax2 + bx + c = 0.

The Quadratic Formula

x = 

The a, b, and c in the formula refer to the coefficients of an expression in the form ax2 + bx + c. For the equation x2 − 7x + 8 = 0, a = 1, b = −7, and c = 8. Plug these values into the quadratic formula and you get the roots of the equation.

In Case You
Were Worried…

On Math Level 1, the
quadratic formula is
necessary only on difficult
questions. You may be
able to skip over tough
quadratic equation questions
and avoid having to
use the quadratic formula

So the equation x2 − 7x + 8 = 0 has two real roots, 5.56 and 1.44.

It’s possible to tell quickly, without going all the way through the quadratic formula, how many roots an equation has. The part of the quadratic formula under the radical, b2 − 4ac, is called the discriminant. The value of the discriminant gives you the following information about a quadratic equation:

·        If b2 − 4ac > 0, then the equation has two distinct real roots.

·        If b2 − 4ac = 0, then the equation has one distinct real root and is a perfect square. Actually, it has two identical real roots, which ETS will call a “double root.”

·        If b2 − 4ac < 0, then the equation has no real roots. Both of its roots are imaginary.


In the following exercises, use the discriminant to find out how many roots each equation has and whether the roots are real or imaginary. For equations with real roots, find the exact value of those roots using the quadratic formula. The answers to these drills can be found in Chapter 12.

  1.  x2 − 7x + 5 = 0

  2.  3a2 − 3a + 7 = 0

  3.  s2 − 6s + 4 = 0

  4.  x2 − 2 = 0

  5.  n2 + 5n + 6.25 = 0


Perhaps the easiest way to find the roots of a hard-to-factor quadratic is to graph it on your calculator and see where the quadratic intersects the x-axis. Your calculator will most likely have several functions that can help you find the answer(s) like “zeros,” “intersect,” “trace,” and/or “table.” Consult your calculator’s manual and use whatever method you’re most comfortable with.


·        Plugging In is a great way to sidestep the landmines that ETS tries to set for you.

·        You can Plug In whenever

·        you see variables, percents, or fractions (without an original amount) in the answers

·        you’re tempted to write and then solve an algebraic equation

·        you see the phrase “in terms of” in the question

·        there are unknown quantities or variables in the question

·        Plug In the answer choices when you have numbers in the answers but don’t know where to start or you are still tempted to write an algebraic equation. Don’t forget to start with choice (C)!

·        Inequalities get solved just like equations, but when you multiply or divide by a negative number, flip the sign.

·        When combining ranges, remember to write out all four possibilities.

·        Absolute value questions often have two answers. Write out and solve both equations created by the absolute value.

·        Direct and indirect variation questions ask for the relationships between numbers:

·        Direct: as x goes up, y goes up. Direct means divide. So you’ll have an equation with two fractions.

·        Indirect: as x goes up, y goes down. Indirect (also known as inverse) means multiply. So you’ll have an equation with two quantities being multiplied.

·        Work and travel questions often require either the rate equation: distance = rate × time (or work done = rate of work × time), or the average pie, which can be used to find average speed.

·        Simultaneous equation questions don’t require solving one variable in terms of another. Just stack ’em and add or subtract to find what you need. Remember that you can multiply or divide before or after you add or subtract to get to what you want.

·        The general form for a quadratic equation is ax2 + bx + c = 0. To find the factors, just reverse FOIL the equation. There are three special quadratics that you should keep an eye out for to save time and brainpower. They are:

·        (x + y)2 = x2 + 2xy + y2

·        (x − y)2 = x2 − 2xy + y2

·        (x + y)(x − y) = x2 − y2

·        If you have a quadratic equation that you can’t factor, try using the quadratic formula: