## Cracking the SAT

# Part III

# How to Crack the Math Section

# Chapter 12

# Algebra: Cracking the System

In the last chapter we reviewed some fundamental math concepts featured on the SAT. Many questions on the SAT combine simple arithmetic concepts with more abstract algebraic concepts. This is one way the test writers raise the difficulty level of a question—they replace numbers with variables, or letters that stand for unknown quantities. In this chapter you will learn multiple ways to answer these algebraic questions, including avoiding algebra altogether.

## SAT ALGEBRA: CRACKING THE SYSTEM

The SAT generally tests algebra concepts that you probably learned in the eighth or ninth grade. So unless you suffer from arithmophobia (fear of numbers), you are probably pretty familiar with the level of math on the test. In fact, many people who take the SAT are currently taking math classes, such as calculus or trigonometry, that cover topics far more advanced than those on the SAT.

The SAT Math section tests not only your math skills, but also your reading skills. It is important that you read the questions carefully and translate the words in the problem into mathematical symbols.

Here are some words and their equivalent symbols:

Here are two examples:

**Words:** 14 is 5 more than some number.

**Equation:** 14 = 5 + *x*

**Words:** If one-eighth of a number is 3, what is one-half of the same number?

**Equation:** *n* = 3, *n* = ?

Later in the chapter we’re going to show you some extremely important techniques for bypassing a great majority of the algebra on the test, but first we need to go over some of the basics. If you feel comfortable with your algebra skills, feel free to skip ahead.

## BASIC PRINCIPLES: FUNDAMENTALS OF SAT ALGEBRA

Many problems on the SAT require you to work with variables and equations. In algebra class you learned to solve equations by “solving for *x*” or “solving for *y*.” To do this, you isolate *x* or *y* on one side of the equal sign and put everything else on the other side. The good thing about equations is that, to isolate the variable, you can do anything you want to them—add, subtract, multiply, divide, square—provided you perform the same operation to all the numbers in the equation.

Hence, the golden rule of equations:

Whatever you do to the items on one side of the equal sign, you must do to the items on the other side of it as well.

Let’s look at a simple example of this rule:

**1.** If 2*x* – 15 = 35, what is the value of *x* ?

**A Little Terminology**

Here are some words that you will need to know to follow this chapter. The words themselves won’t show up on the SAT, so after you finish the chapter you can forget about them.

*Term:* An equation is like a sentence, and a term is the equivalent of a word. For example, 9 × 2 is a term in the equation 9 × 2 + 0*x* = 5*y*.

*Expression:* If an equation is like a sentence, then an expression is like a phrase or a clause. An expression is a combination of terms and mathematical operations with no equal or inequality sign. For example, 9 × 2 + 3*x* is an expression.

*Polynomial:* A polynomial is any expression containing two or more terms. Binomials and trinomials are both polynomials.

Here’s How to Crack It

You want to isolate the variable. First, add 15 to each side of the equation. Now you have the following:

2*x* = 50

Divide each side of the equation by 2. Thus, *x* equals 25.

### Solving for Expressions

Some SAT algebra problems ask you to find the value of an expression rather than the value of a variable. In most cases, you can find the value of the expression without finding the value of the variable.

**5.** If 4*x* + 2 = 4, what is the value 4*x* − 6 ?

(A) −6

(B) −4

(C) 0

(D) 4

(E) 8

#### The Princeton Review Solution

Since ETS is asking for the value of an expression (4*x* − 6) rather than the value of *x*, you correctly suspect that there might be a shortcut. So, you look for a way to turn 4*x* + 2 into 4*x* − 6 and realize that subtracting 8 from both sides of the equation will do just that.

So, you just do (4*x* + 2) − 8 = 4 − 8 = −4 and you’ve got answer B.

The Princeton Review solution will save you time—provided that you see it quickly. So, while you practice, you should train yourself to look for these sorts of direct solutions whenever you are asked to solve for the value of an expression.

However, don’t worry too much if you don’t always see the faster way to solve a problem like this one. The math class way will certainly get you the right answer.

Here’s another example:

**9.** If = *x* − 2, what is the value of (*x* − 2)^{2} ?

(A)

(B)

(C) 5

(D) 9

(E) 25

Here’s How to Crack It

If you were to attempt the math class way, you’d find that *x* = + 2 and then you’d have to substitute that into the provided expression. There’s got to be an easier way!

The problem is much easier if you look for some sort of direct solution. Then, you notice that all the problem wants you to do is to square the expression on the right of the equal sign. Well, if you square the expression on the right, then you’d better square the expression on the left, too. So ()^{2} = 5 = (*x* − 2)^{2}, and the answer is C. That was pretty painless by comparison.

### Solving Simultaneous Equations

Some SAT problems will give you two equations involving two variables and ask for the value of an expression. These problems are very similar to the problems on the previous page. ETS would like you to solve for the value of each variable. But, all you really need to do is add or subtract the equations.

Here’s an example:

If 4*x* + *y* = 14 and 3*x* + 2*y* = 13, then *x* – *y* = ?

Here’s How to Crack It

You’ve been given two equations here. But instead of being asked to solve for a variable (*x* or *y*), you’ve been asked to solve for an expression (*x* – *y*). Why? Because there must be a direct solution.

In math class, you’re taught to solve one equation for one variable in terms of a second variable and to substitute that value into the second equation to solve for the first variable.

Forget it. These methods are far too time consuming to use on the SAT, and they put you at risk of making mistakes. There’s a better way. Just stack them on top of each other, then add or subtract the two equations; either addition or subtraction will produce an easy answer. Let’s try it.

Adding the two equations gives you this:

Unfortunately, that doesn’t get us anywhere. So try subtracting:

4*x* + *y* = 14

(3*x* + 2*y* = 13)

When you subtract equations, just change the signs of the second equation and add. So the equation above becomes

The value of (*x* – *y*) is precisely what you are looking for, so this is the answer.

### Solving Inequalities

In an equation, one side equals the other. In an inequality, one side does not equal the other. The following symbols are used in inequalities:

≠ |
is not equal to |

> |
is greater than |

< |
is less than |

≥ |
is greater than or equal to |

≤ |
is less than or equal to |

Solving inequalities is pretty much like solving equations. You can collect similar terms, and you can simplify by doing the same thing to both sides. All you have to remember is that if you multiply or divide both sides of an inequality by a negative number, the direction of the inequality symbol changes. For example, here’s a simple inequality:

*x* > *y*

Now, just as you can with an equation, you can multiply both sides of this inequality by the same number. But if the number you multiply by is negative, you have to change the direction of the symbol in the result. For example, if you multiply both sides of the inequality above by –2, you end up with the following:

–2*x* < –2*y*

Remember: When you multiply or divide an inequality by a negative number, you must reverse the inequality sign.

### Simplifying Expressions

If a problem contains an expression that can be factored, it is very likely that you will need to factor it to solve the problem. So, you should always be on the lookout for opportunities to factor. For example, if a problem contains the expression 2*x* + 2*y*, you should see if factoring it to produce the expression 2(*x + y*) will help you to solve the problem.

If a problem contains an expression that is already factored, you should consider using the distributive law to expand it. For example, if a problem contains the expression 2(*x + y*), you should see if expanding it to 2*x* + 2*y* will help.

Here are five examples that we’ve worked out:

1. 4*x* + 24 = 4(*x*) + 4(6) = 4(*x* + 6)

2. = 5(*x* − 6) = 5*x* − 30

3.

4. 2(*x* + *y*) + 3(*x* + *y*) = (2 + 3)(*x* + *y*) = 5(*x* + *y*)

5. *p*(*r* + *s*) + *q*(*r* + *s*) = (*p* + *q*)(*r* + *s*)

### Multiplying Binomials

Multiplying binomials is easy. Just be sure to use FOIL (first, outer, inner, last):

### Combine Similar Terms First

In manipulating long, complicated algebraic expressions, combine all similar terms before doing anything else. In other words, if one of the terms is 5*x* and another is –3*x*, simply combine them into 2*x*. Then you won’t have as many terms to work with. Here’s an example:

### Evaluating Expressions

Sometimes ETS will give you the value of one of the variables in an algebraic expression and ask you to find the value of the entire expression. All you have to do is plug in the given value and see what you come up with.

Here is an example:

**Problem:**

If 2*x* = –1, then (2*x* – 3)^{2} = ?

**Solution:**

Don’t solve for *x*; simply plug in –1 for 2*x*, like this:

### Solving Quadratic Equations

To solve quadratic equations, remember everything you’ve learned so far: Look for direct solutions and either factor or expand when possible.

Here’s an example:

If (*x* + 3)^{2} = (*x* – 2)^{2}, then *x* =

Here’s How to Crack It

Expand both sides of the equation using FOIL:

Now you can simplify. Eliminate the *x*^{2}’s, because they are on both sides of the equal sign. Now you have 6*x* + 9 = – 4*x* + 4, which simplifies to

#### Factoring Quadratics

To solve a quadratic, you might also have to factor the equation. Factoring a quadratic basically involves doing a reverse form of FOIL.

For example, suppose you needed to know the factors of *x ^{2}* + 7

*x*+ 12. Here’s what you would do:

1. Write down 2 sets of parentheses and put an *x* in each one because the product of the first terms is *x*^{2}.

*x*^{2} + 7*x* + 12 = (*x* )(*x* )

2. Look at the number at the end of the expression you are trying to factor. Write down its factors. In this case, the factors of 12 are 1 and 12, 2 and 6 and 3 and 4.

3. To determine which set of factors to put in the parentheses, look at the coefficient of the middle term of the quadratic expression. In this case, the coefficient is 7. So, the correct factors will also either add or subtract to get 7. Write the correct factors in the parentheses.

*x*^{2} + 7*x* + 12 = (*x* __ 3)(*x* __ 4)

4. Finally, determine the signs for the factors. To get a positive 12, the 3 and the 4 are either both positive or both negative. But, since 7 is also positive, the signs must both be positive.

*x*^{2} + 7*x* + 12 = (*x* + 3)(*x* + 4)

You can always check that you have factored correctly by FOILing the factors to see if you get the original quadratic expression.

Now, try this one:

**16.** In the expression *x*^{2} + *kx* + 12, *k* is a negative integer. Which of the following is a possible value of *k*?

(A) −13

(B) −12

(C) −6

(D) 7

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

Here’s How to Crack It

Of course, you’re going to eliminate E because that’s the Joe Bloggs answer. While you are eliminating answers, you can also get rid of D because the question says that *k* is negative.

To solve the question, you need to factor. This question is just a twist on the example used above. Don’t worry that we don’t know the value of *k*. The question said that *k* was an integer so you need to consider only the integer factors of 12. The possible factors of 12 are 1 and 12, 2 and 6, and 3 and 4. Since 12 is positive and *k* is negative, then you’ll need subtraction signs in both factors.

The possibilities are as follows:

*x*^{2} + *kx* + 12 = (*x* – 1)(*x* – 12)

*x ^{2}* +

*kx*+ 12 = (

*x*– 2)(

*x*– 6)

*x ^{2}* +

*kx*+ 12 = (

*x*– 3)(

*x*– 4)

If you FOIL each of these sets of factors, you’ll get the following expressions:

(*x* – 1)(*x* – 12) = *x ^{2}* – 13

*x*+ 12

(*x* – 2)(*x* – 6) = *x ^{2}* – 8

*x*+ 12

(*x* – 3)(*x* – 4) = *x ^{2}* – 7

*x*+ 12

The correct answer is A, as −13 is the only value from above included in the answers. Of course, you didn’t need to write them all out if you started with 1 and 12 as your factors.

## ETS FAVORITES

ETS plays favorites when it comes to quadratic equations. There are three equations that they use all the time. You should memorize these and be on the lookout for them. Whenever you see a quadratic that contains two variables, it is almost certain to be one of these three.

(*x* + *y*) (*x* – *y*) = *x*^{2} – *y*^{2}

(*x* + *y*)^{2} = *x*^{2} + 2*xy* + *y*^{2}

(*x* – *y*)^{2} = *x*^{2} – 2*xy* + *y*^{2}

Here’s an example of how ETS is likely to test these equations. Try it:

**11.** If 2*x* − 3*y* = 5, what is the value of 4*x*^{2} – 12*xy* + 9*y*^{2} ?

(A)

(B) 5

(C) 12

(D) 25

(E) 100

Here’s How to Crack It

Look! A quadratic equation with two variables! What do you think ETS is up to? Yes, that’s right, it’s one of their favorites.

In this case, work with 2*x* − 3*y* = 5. If you square the left side of the equations, you get

(2*x* – 3*y*)^{2} = 4*x*^{2} – 12*xy* + 9*y*^{2}

That’s precisely the expression for which you need to value. It’s also the third of the equations from the box. Now, since you squared the left side, all you need to do is square the 5 on the right side of the equation to discover that the expression equals 25, answer D.

Did you notice that this question was just another version of ETS asking you to solve for the value of an expression rather than for a variable? Quadratics are one of their favorite ways to do that.

## SOLVING QUADRATICS SET TO ZERO

Before factoring most quadratics, you need to set the equation equal to zero. Why? Well, if *ab* = 0, what do you know about *a* and *b*? At least one of them must equal 0, right? That’s the key fact you need to solve most quadratics.

Here’s an example:

**9.** If 3 − = *x* + 7, and *x* ≠ 0 , which of the following is a possible value for *x* ?

(A) −7

(B) −1

(C) 1

(D) 3

(E) 7

Here’s How to Crack It

ETS has tried to hide that the equation is actually a quadratic. Start by multiplying both sides of the equation by *x* to get rid of the fraction.

3*x* – 3 = *x*^{2} + 7*x*

Now, just rearrange the terms to set the quadratic equal to 0. You’ll get *x ^{2}* + 4

*x*+ 3 = 0. Now, it’s time to factor:

*x*^{2} + 4*x* + 3 = (*x* + 1)(*x* + 3) = 0

So, at least one of the factors must equal 0. If *x* + 1 = 0, then *x* = −1. If *x* + 3 = 0, then *x* = −3. Since −1 is answer B, that’s the one you want.

## WHEN VALUES ARE ABSOLUTE

Absolute value is just a measure of the distance between a number and 0. Since distances are always positive, the absolute value of a number is also always positive. The absolute value of a number is written as |*x*|.

When solving for the value of a variable inside the absolute value bars, it is important to remember that variable could be either positive or negative. For example, if |*x*| = 2, then *x* = 2 or *x* = −2 since both 2 and −2 are a distance of 2 from 0.

Here’s an example:

|*x* + 3| = 6

|*y* – 2| = 7

**9.** For the equations shown above, which of the following is a possible value of *x* − *y* ?

(A) −14

(B) −4

(C) −2

(D) 1

(E) 14

Here’s How to Crack It

To solve the first equation, set *x* + 3 = 6 and set *x* + 3 = −6. If *x* + 3 = −6, then the absolute value would still be 6. So, *x* can be either 3 or −9. Now, do the same thing to solve for *y*. Either *y* = 9 or *y* = −5.

To get the credited answer, you need to try the different combinations. One combination is *x* = −9 and *y* = −5. So, *x* − *y* = −9 − (−5) = −4, which is answer B.

## PRINCETON REVIEW ALGEBRA, OR HOW TO AVOID ALGEBRA ON THE SAT

Now that you’ve reviewed some basic algebra, it’s time for some Princeton Review algebra. At The Princeton Review, we like to avoid algebra whenever possible. You read that correctly: We’re going to show you how to avoid doing algebra on the SAT. Now, before you start crying and complaining that you love algebra and couldn’t possibly give it up, just take a second to hear us out. We have nothing against algebra—it’s very helpful when solving problems, and it impresses your friends—but on the SAT, using algebra can actually hurt your score. And we don’t want that.

### But Algebra Would Never Hurt Me!

We know it’s difficult to come to terms with this. But if you use algebra on the SAT, you’re doing exactly what ETS wants you to do. You see, when the test writers design the problems on the SAT, they expect the students to use algebra to solve them. ETS builds little traps into the problems to take advantage of that fact. But if you don’t use algebra, there’s no way you can fall into those traps.

Plus, when you avoid algebra, you have one other powerful tool at your disposal: your calculator! Even if you have a super-fancy calculator that plays games and doubles as a global positioning system, chances are it doesn’t do algebra. Arithmetic, on the other hand, is easy for your calculator. It’s what calculators were invented for.

Our goal, then, is to turn all the algebra on the SAT into arithmetic. We do that using something we call *Plugging In.*

## PLUGGING IN THE ANSWERS (PITA)

Algebra uses letters to stand for numbers, but very few other things do. You don’t go to the grocery store to buy *x* eggs or *y* gallons of milk. Most people think in terms of numbers, not letters that stand for numbers.

You should think in terms of numbers on the SAT as much as possible. On many SAT algebra problems, even very difficult ones, you will be able to find ETS’s answer without using any algebra at all. You will do this by working backward from the answer choices instead of trying to solve the problem using math-class algebra.

*Plugging In The Answers* is a technique for solving word problems in which the answer choices are all numbers. Many algebra problems on the SAT can be solved simply and quickly by using this powerful technique.

In algebra class at school, you solve word problems by using equations. Then, if you’re careful, you check your solution by plugging in your answer to see if it works. Why not skip the equations entirely by simply checking the five solutions ETS offers on the multiple-choice questions? One of these has to be correct. You don’t have to do any algebra, you will seldom have to try more than two choices, and you will never have to try all five. Note that you can use this technique only for questions that ask for a specific amount.

Here’s an example:

**9.** Zoë won the raffle at a fair. She will receive the prize money in 5 monthly payments. If each payment is half as much as the previous month’s payment and the total of the payments is $496, what is the amount of the first payment?

(A) $256

(B) $96

(C) $84

(D) $16

(E) $4

Here’s How to Crack It

ETS would like you to go through all the effort of setting up this equation:

Then, of course, they want you to solve the equation. But, look at all those fractions! There are lots of opportunities to make a mistake and you can bet that ETS has figured most of them out so they can have a trap answer waiting. So, let’s work with the answers instead.

To work with the answer choices, first you need to know what they represent so that you can label them. In this case, the question asks for the first payment, so write something like first payment over the answers.

Now, it’s time to start working the steps of the problem. But first, notice that the answer choices are in numerically descending order. ETS likes to keep their problems organized so they will always put the answers in order. You can use that to your advantage by starting with answer choice C.

Grab answer choice C and ask yourself “If the first payment is $84, what’s the next thing I can figure out?” In this case, you could figure out the second payment. So, make a chart and write down 42 (half of 84) next to the 84. Keep doing that to find the values of the third, fourth, and fifth payments.

When you have worked all the steps, your problem should look like this:

**9.** Zoë won the raffle at a fair. She will receive the prize money in 5 monthly payments. If each payment is half as much as the previous month’s payment and the total of the payments is $496, what is the amount of the first payment?

You need to determine if that was the correct answer. The problem says that the total is supposed to be $496, so add up the payments: 84 + 42 + 21 + 10.50 + 5.25 = 162.75, which is much smaller than 496. So, cross off choices C, D and E.

Now, all you need to do is try answer B. If B works, then you’re done. And, if B doesn’t work, you’re still done because the answer must be A. That’s putting your POE to good use! If the first payment is $96, then the payments are 96 + 48 + 24 + 12 + 6 = 186, which is still too small. That means the answer must be A and you don’t really need to check it.

Here are the steps for solving a problem using the PITA approach:

To solve a problem by Plugging In The Answers

1. Label the answer choices.

2. Starting with answer C, work the steps of the problem. Be sure to write down a label for each new step.

3. Look for something in the problem that tells you what must happen for the answer to be correct.

4. When you find the correct answer, STOP.

Here’s another problem that PITA makes easy:

**14.** The units digit of a two-digit number is 3 times the tens digit. If the digits are reversed, the resulting number is 36 more than the original number. What is the original number?

(A) 26

(B) 36

(C) 39

(D) 54

(E) 93

Here’s How to Crack It

It’s pretty hard to come up with the algebraic equation to solve this one, so try Plugging In The Answers. First, label the answers as the original number.

Start at choice C. If the original number is 39, what must be true? The units digit is supposed to be 3 times the tens digit and 9 is three times 3, so don’t rule this answer choice out yet. Next, the problem says to reverse the digits, so the new number would be 93. Finally, the difference is supposed to be 36 but 93 − 39 = 54, so this answer gets eliminated.

It’s hard to tell if you need a bigger number or smaller number, so just pick a direction and try it. Let’s try A. If the original number is 26, then the units digit is three times the tens digit. So far, so good. Next, if the digits are reversed, the new number is 62. Finally, 62 − 26 = 36. Bingo! A is the answer.

Always make sure that you follow *all* of the steps of a problem. If you had just checked to be sure that the units digit was three times the tens digit, you might have chosen answer C and been wrong!

One last thing about PITA. Here’s how you spot that you can use this approach to solve the problem.

Three ways to know that it’s time for PITA:

1. There are numbers in the answer choices.

2. The question asks for a specific amount such as “what was the first payment.”

3. You have the urge to write an algebra equation to solve the problem.

### Plugging In: Advanced Principles

Plugging In is the same on difficult problems as it is on easy and medium ones. You just have to watch your step and make certain you don’t make any careless mistakes or fall for Joe Bloggs answers.

Here’s one of our examples:

**15.** A baseball team won 54 more games than it lost. If the team played a total of 154 games and there were no ties, how many games did the team win?

(A) 50

(B) 98

(C) 100

(D) 102

(E) 104

Here’s How to Crack It

Doesn’t seem so bad, right? Well, that’s what Joe thinks, too. What would Joe choose? If you thought that Joe might choose C, you’ve tapped into your inner Joe and you can use that knowledge to eliminate wrong answers. Joe likes answer C because he thinks all he needs to do is 154 – 54 = 100. But, that’s just too easy to be the answer to a hard question like number 15. Cross off C!

If you’re itching to write an equation, that’s a sure sign that it’s time for PITA. So, start by labeling the answer choices as *games won*. Since C has been eliminated, start at choice D. If the team won 102 games, what else can you figure out? The team lost 102 – 54 = 48 games. How will you know if that’s the right answer? The team is supposed to have played 154 games but with choice D the team has played only 102 + 48 = 150 games. That’s not enough, so cross off D. The team needs to have won more games, so the answer must be E. (It is. Really. You can check it if you’d like.)

Here’s another example:

**17.** Committee *A* has 18 members and Committee *B* has 3 members. How many members from Committee *A* must switch to Committee *B* so that Committee *A* will have twice as many members as Committee *B* ?

(A) 4

(B) 6

(C) 7

(D) 11

(E) 15

Here’s How to Crack It

Because it’s fairly hard to write the correct equation to solve this problem, only a very small percentage of students get it right. But, if you Plug In The Answers, you won’t have any trouble.

This problem is about two committees, so the first thing you should do is quickly draw a picture in your test booklet to keep from getting confused:

Now plug in the answer choices, starting with answer choice C. If you move 7 members out of Committee *A,* there will be 11 members left in *A* and 10 members in *B.* Is 11 twice as many as 10? No, eliminate.

As you work through the choices, keep track of them, like this:

Choice C didn’t work. To make the question work out right, you need more members in Committee *A* and fewer in Committee *B.* In other words, you need to try a smaller number. Try the smallest one, choice A. Moving 4 members from Committee *A* will leave 14 in *A* and 7 in *B.* Is 14 twice as many as 7? Yes, of course. This is the answer.

## PLUGGING IN YOUR OWN NUMBERS

Plugging In The Answers enables you to find the answer on problems when the answer choices are all numbers. What about problems that have answer choices containing variables? On these problems, you will usually be able to find the answer by Plugging In your own numbers.

Plugging In is easy. It has three steps:

1. Pick numbers for the variables in the problem.

2. Use *your* numbers to find an answer to the problem. Circle your answer.

3. Plug your numbers from step 1 into the answer choices to see which choice equals the answer you found in step 2.

### The Basics of Plugging In Your Own Numbers

This sort of Plugging In is simple to understand. Here’s an example:

**3.** If Jayme will be *j* years old in 3 years, then in terms of *j*, how old was Jayme 5 years ago?

(A) *j* – 8

(B) *j* – 5

(C) *j* – 3

(D) *j* + 5

(E) *j* + 8

Here’s How to Crack It

First, pick a number for *j*. Pick something easy to work with, like 10. In your test booklet, write 10 directly above the letter *j* in the problem, so you won’t forget.

If *j* = 10, then Jayme will be 10 years old in 3 years. That means that she is 7 right now. Because the problem asked you how old she was 5 years ago, just calculate 7 – 5 = 2. She was 2! Write a nice big 2 in you test booklet and circle it. That’s your target. The correct answer will be the choice that, when you plug in 10 for *j,* equals 2.

Now plug in!

Plug in 10 for *j* in A and you get 10 – 8, or 2. This is the number that you are looking for, so this must be the right answer! Go ahead and try the other choices just to make sure that you’re right and to practice Plugging In.

Here’s another example:

**12.** The sum of four consecutive positive even integers is *x*. In terms of *x,* what is the sum of the second and third integers?

(A)

(B)

(C) 2*x* + 6

(D)

(E)

Here’s How to Crack It

Let’s pick four numbers: 2, 4, 6, and 8. The sum of these four is 20, so *x* = 20; write that in your test booklet. The second and third numbers, 4 and 6, add up to 10; this is your target number, circle it. You are looking for the choice that will equal 10 when you plug in 20. Try each choice:

### Which Numbers?

Although you can plug in any number, you can make your life much easier by Plugging In “good” numbers—numbers that are simple to work with or that make the problem easier to manipulate. Picking a small number, such as 2, will usually make finding the answer easier. If the problem asks for a percentage, plug in 10 or 100. If the problem has to do with minutes, try 30 or 120.

*Except in special cases, you should avoid plugging in 0 and 1*; these numbers have weird properties. Using them may allow you to eliminate only one or two choices at a time. You should also avoid plugging in any number that appears in the question or in any of the answer choices. These could make more than one answer match your target.

Many times you’ll find that there is an advantage to picking a particular number, even a very large one, because it makes solving the problem easier.

Here’s an example:

**14.** If 60 equally priced downloads cost *x* dollars, then how much do 9 downloads cost?

(A)

(B) 9*x* – 60

(C)

(D) 60*x* + 9

(E)

Here’s How to Crack It

Should you plug in 2 for *x*? You could, but plugging in 120 would make the problem easier. After all, if 60 downloads cost a total of $120, then each download costs $2. Write *x* = 120 in your test booklet.

If each download costs $2, then 9 downloads cost $18. Write an 18 in your test booklet and circle it. You are looking for the answer choice that works out to 18 when you plug in $120 for *x*. Let’s try each choice:

(A)

(B) 9(120) – 60 ≠ 18

(C)

(D) 60(120) + 9 ≠ 18

(E) Here’s your answer.

Let’s try another example:

**20.** A watch loses *x* minutes every *y* hours. At this rate, how many hours will the watch lose in one week?

(A) 7*xy*

(B)

(C)

(D)

(E)

Here’s How to Crack It

This is an extremely difficult problem for students who try to solve it the math-class way. You’ll be able to find the answer easily, though, if you plug in carefully.

What numbers should you plug in? As always, you can plug in anything. However, if you think just a little bit before choosing the numbers, you can make the problem easier to understand. There are three units of time—minutes, hours and weeks—and that’s a big part of the reason this problem is hard to understand. If you choose units of time that are easy to think about, you’ll make the problem easier to handle.

Start by choosing a value for *x*, which represents the number of minutes that the watch loses. You might be tempted to choose *x* = 60 and that would make the math pretty easy. However, it’s usually not a good idea to choose a conversion factor such as 60, the conversion factor between minutes and hours, when plugging in. It would actually turn out okay on this problem but when dealing with time, 30 is usually a safer choice. So, write down *x* = 30.

Next, you need a number for *y*, which represents the number of hours. Again, you might be tempted to use *y* = 24 but that’s the conversion factor between hours and days. So, how about *y* = 12 as a safer choice? Write down *y* = 12.

Now, it’s time to solve the problem to come up with a target. If the watch loses 30 minutes every 12 hours, then it loses 60 minutes every 24 hours. Put another way, the watch loses an hour each day. In one week, the watch will lose 7 hours. That’s your target so be sure to circle it.

Now, you just need to check the answer choices to see which one gives you 7 when *x* = 30 and *y* = 12.

### Inequalities

Plugging In works on problems containing inequalities, but you will have to follow some different rules. Plugging in one number is often not enough; to find ETS’s answer, you may have to plug in several numbers, including weird numbers like –1, 0, 1, , and −.

The five numbers just mentioned all have special properties. Negatives, fractions, 0, and 1 all behave in peculiar ways when, for example, they are squared. Don’t forget about them!

Sometimes it can actually be easier or faster to simplify. Here’s an example:

**8.** If –3*x* + 6 ≥ 18, which of the following must be true?

(A) *x* ≤ –4

(B) *x* ≤ 6

(C) *x* ≥ –4

(D) *x* ≥ –6

(E) *x* = 2

Here’s How to Crack It

The inequality in the problem can be simplified quite a bit:

We’re close to one of the answer choices, but not quite there yet. Multiply both sides by –1 to make *x* positive. *Remember to change the direction of the inequality sign!*

*x* ≤ –4

So choice A is the answer.

### Other Special Cases

Sometimes SAT algebra problems will require you to determine certain characteristics of a number or numbers. Is *x* odd or even? Is it small or large? Is it positive or negative?

On questions like this, you will probably have to plug in more than one number and/or plug in weird numbers, just as you do on problems containing inequalities. Sometimes ETS’s wording will tip you off. If the problem states only that *x* > 0, you know for certain that *x* is positive but you don’t know that *x* is an integer. See what happens when you plug in a fraction.

Here are some other tip-offs you should be aware of when you’re trying to eliminate answers.

### Must Be True

*Must* is a very strong word in a math problem. It means that whatever condition you are given needs to work for every number that you are allowed to try. When you plug in on a question that uses *must*, it is very likely that you will need to plug in at least twice to find the answer. If you choose the first answer that meets the condition, you could wind up choosing something that * could be true* rather than something that

*must be true.*

Try the following problem:

**14.** If *a* – *b* is a multiple of 7, which of the following must also be a multiple of 7 ?

(A) *ab*

(B) *a* + *b*

(C)

(D)

(E) *b* – *a*

Here’s How to Crack It

Because there are variables in the answer choices, you can plug in. First, plug in easy numbers that make the given statement (*a* – *b* is a multiple of 7) true. Make *a* = 14 and *b* = 7. So the whole thing would be 7, which is indeed a multiple of 7. The question asks which of the following must also be a multiple of 7. Plug in the number and cross off any answer choices that are not true.

(A) 14(7) = 98 is a multiple of 7, so keep it.

(B) 14 + 7 = 21 is a multiple of 7, so keep it.

(C) is not a multiple of 7. Cross it off.

(D) is not a multiple of 7. Cross it off.

(E) 7 – 14 = –7 is a multiple of 7, so keep it.

Since the question asks for something that *must* be true and you are left with three answer choices, you must plug in again. The first time you plugged in, you used two other multiples of 7 (*a* = 14 and *b* = 7) to satisfy the condition. But it doesn’t say that *a* and *b* must be multiples of 7, just that the *expression* must be. So now, plug in two numbers that are *not* multiples of 7, but still make the statement true. Plug in 10 for *a* and 3 for *b*. The statement becomes 10 – 3 = 7, which works. Now check the answers that you didn’t eliminate the first time:

(A) (10)(3) = –30. Cross it off.

(B) 10 + 3 = 13. Cross it off.

(E) 3 – 10 = –7. It still works

(and is the only one left), so keep it.

The answer is E.

### Plugging In: Advanced Principles

If there are variables in the answer choices, you should definitely Plug In. However, sometimes a question will be a plug in question that doesn’t have variables in the answer choices. It is, instead, a hidden Plug In question. it will refer to some unknown amount, but never actually give you a number. So, you’re going to have to make up your own number.

Here’s an example:

**16.** If the remainder is 3 when positive integer *a* is divided by 8, then what is the remainder when 7*a* is divided by 8?

(A) 1

(B) 3

(C) 5

(D) 6

(E) 7

Here’s How to Crack It

The question here asks about the variable *a*, but never gives you a value for *a* or a way to solve for *a*. In that case, Plug In!

You need a number that you can divide by 8 and get a remainder of 3. If you’re not sure what number to pick, start with something easy and adjust it until you get a number that works. You can’t have *a* = 8, because 8 ÷ 8 = 1 with remainder of 0. Adjust your number a bit: *a* = 9 gives 9 ÷ 8 = 1 with a remainder of 1. *a* = 10 gives a remainder of 2, so *a* = 11 works. (You may have noticed that other numbers, such as 19 or 83, also work.)

Now work through the problem. What is the remainder when 7*a* is divided by 8? Since 7*a* = 7(11) = 77, find the remainder for 77 ÷ 8: 8 goes into 77 a total of 9 times (8 × 9 = 72), with 5 left over. The answer is C.

## ADVANCED PRINCIPLES OF SAT ALGEBRA

### Solving Rational Equations

A rational equation is basically a fraction with a polynomial in the numerator and a polynomial in the denominator. Rational equations look scary, but there are very simple ways of solving them. One way is to factor out like terms and then cancel. All in all, ETS can’t get too messy here, so they will keep the math nice and tidy.

Try one:

**18.** If , then which of the following could be a value of *x* ?

(A) –7

(B) –5

(C) 0

(D) 6

(E) 16

Here’s How to Crack It

Hate factoring? PITA! Start with answer choice C and plug in 0 for *x*. Does everything work out? In this case, it doesn’t. Keep trying other answer choices until you find one that works. Choice A does, so that’s the correct answer choice. See? These are all bark and no bite.

### Solving Radical Equations

Radical equations are just what the name suggests: an equation with a radical (√) in it. Not to worry, just remember to get rid of the radical first by raising both sides to that power.

Here’s an example:

**7.** If 7 − 24 = 11, what is the value of *x* ?

(A)

(B)

(C) 5

(D) 25

(E) 35

Here’s How to Crack It

Start by adding 24 to both sides to get 7 = 35. Now, divide both sides by 7 to find that = 5. Finally, square both sides to find that *x* = 25, which is answer D.

Don’t forget that you can also get the answer by using PITA. If you are ever unsure about how to solve an equation, PITA is a great approach.

### Functions

Think of a function as just a machine for producing ordered pairs. You put in one number and the machine spits out another. The most common function is an *f*(*x*) function. You’ve probably dealt with it in your algebra class.

Let’s look at a problem:

**9.** If *f*(*x*) = *x*^{3} – 4*x* + 8, then *f*(5) =

(A) 67

(B) 97

(C) 113

(D) 147

(E) 153

Here’s How to Crack It

Any time you see a number inside the parentheses, such as *f*(5), plug in that number for *x*. The question is actually telling you to plug in! Let’s do it:

That’s choice C.

Sometimes you’ll get more complicated questions. As long as you know that when you put in *x*, your function will spit out another number, you’ll be fine. Try this next one:

**20.** Let the function *g* be defined by *g*(*x*) = 5*x* + 2. If = 6, what is the value of *a* ?

(A)

(B)

(C)

(D)

(E)

Here’s How to Crack It

This may look complicated, but just follow the directions. You know that *g*(*x*) = 5*x* + 2. You also know that = 6. First, get rid of the square root by squaring both sides. Now you have *g* = 36. Usually there’s an *x* inside the parentheses. Treat this the same. This statement says that *g*of some number equals 6. We know that *g* of some number is the same as 5*x* + 2. So 5*x* + 2 = 6. Simplify and you get . Careful, you’re not done. You now know that , so *a* = , or D.

Even though these questions may have looked like one you might get on an algebra test, remember you’re still dealing with ETS. Sometimes they’ll throw you a curveball by inserting a function question that has crazy symbols instead of *f*(*x*). Just follow the directions of the function and you’ll do fine.

Here’s an example:

**14.** If *x* # *y* = for all values of *x* and *y* such that *x* ≠ *y*, then, what is the value of ?

(A) 6

(B)

(C)

(D) –1

(E) –6

Here’s How to Crack It

Finding ETS’s answers is just a matter of simple substitution. Just substitute and for *x* and *y* in the function.

ETS’s answer, therefore, is choice A.

## WHAT’S THE POINT?

Why did math folks come up with functions? To graph them of course! When you put in a value for *x*, and your machine (or function) spits out another number, that’s your *y*. You now have an ordered pair. Functions are just another way to express graphs. Knowing the connection between functions and graphs is useful, because ETS will show you a graph and ask you questions about it.

## Drill 1

Work these problems using the Plugging In techniques and algebra tips we’ve covered in this chapter. Answers can be found on this page.

**4.** If 4 more than twice a number is 6 less than that number, what is the number?

(A) –10

(B) 2

(C) 10

(D) 16

(E) 24

**6.** If *n* is a negative integer, which of the following must be a positive integer?

(A) *n* + 2

(B) 2*n*

(C) 2*n* + 4

(D) *n*^{2} – 5

(E) *n*^{2} + 1

**7.** Ashley is 7 years younger than Sarah and three times as old as Cindy. If Cindy is *c* years old, how old is Sarah in terms of *c*?

(A) *c* – 7

(B)

(C) + 7

(D) 3*c*

(E) 3*c* + 7

**8.** If *r* and *s* are both odd integers, which of the following must be an even integer?

(A)

(B)

(C) *rs*

(D) 2*rs*

(E) *3rs*

**9.** John buys 40 sacks of flour for *y* dollars. At the same rate, how much would it cost to buy 12 sacks of flour?

(A)

(B)

(C) 3*y*

(D)

(E)

**11.** If *x* percent of 50 is 20 percent of *y*, what is the value of *x* in terms of *y*?

(A)

(B)

(C) 2*y*

(D) 5*y*

(E) 10*y*

**12.** If Alex can fold 12 napkins in *x* minutes, how many napkins can he fold in *y* hours?

(A)

(B)

(C)

(D)

(E) 720*xy*

**13.** Alice had to read 350 pages of a book over the weekend. If on Sunday, she read 50 pages more than half the amount she read on Saturday, how many pages did she read on Saturday?

(A) 150

(B) 175

(C) 200

(D) 225

(E) 250

**14.** If *xy* < 0, which of the following must be true?

I. *x* + *y* = 0

II. 2*y* – 2*x* < 0

III. *x*^{2} + *y*^{2} > 0

(A) I only

(B) III only

(C) I and III

(D) II and III

(E) I, II, and III

**15.** Alan, Fred, and Mark are going to buy a computer that costs $540. If Alan pays $40 more than Fred and Fred pays twice as much as Mark, then how much does Mark pay?

(A) $100

(B) $140

(C) $160

(D) $200

(E) $240

**16.** Allie has three fewer than twice the number of coins that Jonathan has. If Jonathan gave 2 coins to Allie, she would have three times as many coins as he would. How many coins does Allie have?

(A) 2

(B) 3

(C) 5

(D) 7

(E) 9

**18.** If 2*t* + = *r*, what is *s* in terms of *r* and *t*?

(A) 2*r* – 3*t*

(B) 2(*r* – 3*t*)

(C) 2(2*r* – 3*t*)

(D) 2(2*r* + 3*t*)

(E) 3(2*r* – *t*)

## Summary

· Don’t “solve for *x*” or “solve for *y*” unless you absolutely have to. (Don’t worry; your math teacher won’t find out.) Instead, look for direct solutions to SAT problems. ETS never uses problems that *necessarily* require time consuming computations or endless fiddling with big numbers. There’s almost always a trick—if you can spot it.

· If a problem contains an expression that can be factored, factor it. If it contains an expression that already has been factored, unfactor it.

· To solve simultaneous equations, simply add or subtract the equations.

· When an algebra question has numbers in the answer choices, plug each of the numbers in the answer choices into the problem until you find one that works.

· When you plug in, use “good” numbers—ones that are simple to work with and that make the problem easier to manipulate: 2, 5, 10, or 100 are generally easy numbers to use.

· Plugging In works on problems containing inequalities, but you will have to be careful and follow some different rules. Plugging in one number is often not enough; to find ETS’s answer, you may have to plug in several numbers.

· Not every Plug In question has variables in the answer choices. For some problems, there will be some value that can’t be calculated: in that case, try making up a number.

· Learn to recognize SAT function problems. Sometimes they have funny symbols. Solve them like playing “Simon Says”—do what you are told.