Cracking the New SAT with 4 Practice Tests, 2016 Edition (2015)
Part IV. How to Crack the Math Test
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 complex 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.
SAT ALGEBRA: CRACKING THE SYSTEM
The SAT generally tests algebra concepts that you likely learned in eighth or ninth grade. So, you are probably pretty familiar with the level of algebra on the test. However, ETS is fairly adept at wording algebra questions in a way that is confusing or distracting in order to make the questions harder than the mathematical concepts that are being tested.
In that fashion, the SAT Math sections are testing not only your math skills, but also, and possibly even more importantly to your score improvement, your reading skills. It is imperative that you read the questions carefully and translate the words in the problem into mathematical symbols.
ENGLISH |
MATH EQUIVALENTS |
is, are, were, did, does, costs |
= |
what (or any unknown value) |
any variable (x, y, k, b) |
more, sum |
+ |
less, difference |
− |
of, times, product |
× (multiply) |
ratio, quotient, out of, per |
÷ |
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.
Thus, 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, without the distraction of answer choices.
If 2x − 15 = 35, what is the value of x ?
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:
2x = 50
Divide each side of the equation by 2. Thus, x equals 25.
The skills for algebraic manipulation work just as well for more complex equations. The following question is another example of the way the SAT may ask you to manipulate equations. Don’t panic when you see a question like this; just use the skills you already have and work carefully so you don’t make an avoidable mistake in your algebra.
A Little Terminology
Here are some words that you will need to know to understand the explanations in this chapter. These words may even show up in the text of a question, so make sure you are familiar with them.
Term: An equation is like a sentence, and a term is the equivalent of a word. It can be just a number, just a variable, or a number multiplied by a variable. For example, 18, −2x, and 5y are the terms in the equation 18 − 2x = 5y.
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 + 3x is an expression.
Polynomial: A polynomial is any expression containing two or more terms. Binomials and trinomials are both examples of polynomials.
10.Suppose the evaporation rate of water in a lake is given by the equation E = , where E is the evaporation rate in gallons/day, T_{a} is the air temperature, T_{d} is the dew point temperature, V is the volume of water in the lake, T_{w} is the water temperature, and h is the number of hours the water is exposed to sunlight. Which of the following expresses T_{w} in terms of T_{a}, T_{d},V, E, and h ?
A)
B)
C)
D)
Cut the Fat
ETS will frequently add
unnecessary information
to word problems, as
they have done in this
question. If you find the
unnecessary information
distracting, you can lightly
strike through it. In this
question, the explanation
of the variables is not necessary,
so you can cross
out everything after the
equation to “…the water
is exposed to sunlight”
without removing any
information necessary to
answering the question.
Here’s How to Crack It
The goal is to get T_{w} by itself. Anything you do to one side of the equation, you must also do to the other side of the equation. Start by multiplying both sides by h^{4} to get rid of the denominator on the right side of the equation.
The equation becomes
To get the fraction with T_{w} alone on the right side, subtract the other fraction from both sides to get
Multiply both sides by T_{w} to get that variable out of the denominator.
Now divide the whole expression by the entire thing inside the parentheses.
This doesn’t match any of the answers exactly, but it looks a lot like (D). To make it match exactly, multiply the fractional part by , which won’t change the value of the fraction, to get
Then, just swap the positions of the two parts of the denominator to get (D).
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
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 (D).
Solving Rational Equations
Since you are not always allowed to use your calculator on the SAT, there will be some instances in which you will need to solve an equation algebraically. Even on the sections in which calculator use is permitted, you may find it faster and more effective to use your mathematical skills to efficiently answer a question. Another way ETS may make your calculator less effective is by asking you to solve for an expression. A lot of the time, algebraic manipulation will be the means by which you can solve that problem.
Here is an example:
5.If , what is the value of ?
A)
B)
C)2
D)3
Here’s How to Crack It
This question appears in the No Calculator section, so you must use your math skills to solve for r. You can cross multiply to get 18r = 3(r + 10) or 18r = 3r + 30. Subtracting 3r from both sides gives you 15r = 30, so r = 2. Finally , = , which is (A).
Extraneous Solutions
Sometimes solving a rational or radical expression makes funny things happen. Let’s look at an example.
Given the equation above, what is the value of z ?
Here’s How to Crack It
Use the Bowtie method to get a common denominator for the fractions on the left side of the equation. The numerator of the fraction on the left becomes (z − 2) and the numerator of the fraction on the right becomes (z + 2). The numerators are added together, and the new denominator is the product of the old ones. The equation becomes
Since the denominators are equal, the numerators are equal. This gives you
(z − 2) + (z + 2) = 4
When you simplify the left side, you get 2z = 4, so z = 2. Sounds great, right? However, you need to plug this solution back into the original equation to make sure that it works. You get
Once simplified, two of the three denominators become zero. That is not allowed, so the solution you found isn’t really a solution at all. It is referred to as an “extraneous solution.” That term refers to any answer you get to an algebraic equation that results in a false statement when plugged back in to the original equation.
Here’s another example.
29.
= t − 2
In the equation above, what is the value of the extraneous solution, if one exists?
A)0
B)4
C)5
D)There are no extraneous solutions.
Extra Answers
Any time you are solving
for a variable, make sure
your solutions actually
work. If they do not, they
are “extraneous,” or extra.
Here’s How to Crack It
Start by squaring both sides of the equation to get rid of the radical. The equation becomes
t + 4 = (t − 2)^{2}
Use FOIL to multiply the right side of the equation to get t^{2} − 2t − 2t + 4 or t^{2} − 4t + 4. Now the equation is
t + 4 = t^{2} − 4t + 4
Subtract t and 4 from both sides to get
0 = t^{2} − 5t
The right side factors to t(t − 5), so t = 0 or 5. Eliminate (B), since 4 is not a solution at all, extraneous or otherwise. Now plug 0 and 5 back into the original equation to see if they work. If both do, the answer is (D). If one of them does not, that one is the extraneous solution.
Since the equation is false when t = 0, then 0 is the extraneous solution. The correct answer is (A).
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 4x + 2 = 4, what is the value 4x − 6 ?
A)−6
B)−4
C)4
D)8
Here’s How to Crack It
Since you’re being asked for the value of an expression (4x − 6) rather than the value of x, you correctly suspect that there might be a shortcut. So, you look for a way to turn 4x + 2 into 4x − 6 and realize that subtracting 8 from both sides of the equation will do just that.
So, you just do (4x + 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.
Math Class Solution:
In math class, you would
find the value of x and
then plug that value into
the provided expression.
So, subtract 2 from both
sides to find that 4x = 2.
Then divide both sides by
4 to find that x = . Then,
4x − 6 = 4 − 6 = − 4.
So, the answer is (B).
Here’s another example:
9.If = x − 2, what is the value of (x − 2)^{2}?
A)
B)
C)5
D)25
Learn Them, Love Them
Don’t get bogged down
looking for a direct
solution. Always ask
yourself if there is a
simple way to find the
answer. If you train
yourself to think in terms
of shortcuts, you won’t
waste a lot of time.
However, if you don’t see
a quick solution, get to
work. Something
may come to you as you
labor away.
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 a 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 or more equations involving two or more variables and ask for the value of an expression or one of the variables. These problems are very similar to the problems containing one variable. ETS would like you to spend extra time trying to solve for the value of each variable, but that is not always necessary.
Here’s an example:
If 4x + y = 14 and 3x + 2y = 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, and then add or subtract the two equations; either addition or subtraction will often 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:
4x + y = 14
(3x + 2y = 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 the answer is 1.
You can also use the method above to solve problems in which ETS asks you to solve for an expression and gives you fewer equations than variables. If you have dealt with simultaneous equations in your math classes you may know that that puts you in a bind since it may be impossible to solve for each individual variable.
Here is an example:
3a − 7b = 4d − 9
−4c + 10a = 6b + 7
−2a + 3c − 4d = 10
Given the system of equations above, what is the value of −10a − 2b + 2c ?
Here’s How to Crack It
Notice that ETS has made this problem harder by mixing up the variables. Your first step is to line up the variables on the left side of the equation and arrange them in alphabetical order; move the constants to the right side of the equation. Combine like terms in each equation, and use a place holder for any variables that are missing in each equation.
Step 1:
3a − 7b + 0c − 4d = −9
10a − 6b − 4c + 0d = 7
−2a + 0b + 3c − 4d = 10
“Nothing” is Helpful
In these equations, the
place holders are 0c,
0d, and 0b respectively.
Because 0 times any
number is 0, nothing has
been added to the equation;
these 0’s are simply
acting as placeholders, so
that all of the variables
line up vertically. You may
also choose to leave these
areas blank.
Once you have the variables aligned, complete this problem just like the previous problem by adding and subtracting the equations until you get something that looks like the expression in the question.
IF YOU ADD:
IF YOU SUBTRACT:
Don’t Forget to Share
the Love
Don’t forget that when you
subtract an entire equation,
you need to subtract each
component of the equation
—in simpler terms,
change each sign to the
opposite operation, and
then add the equations.
Neither of these answers appears to be what ETS is asking for, but on closer inspection, the equation that resulted from subtraction can be multiplied by 2 to get the expression in the question.
2(−5a − 1b + 1c) = 2(−26)
−10a − 2b + 2c = −52
Solving for Variables in Simultaneous Equations
Shortcuts are awesome, so take them whenever you can on the SAT. But occasionally, you won’t have the option of using a short-cut with simultaneous equations, so knowing how to solve for a variable is imperative.
Here’s an example:
If 3x + 2y = 17 and 5x − 4y = 21, what is the value of y ?
Here’s How to Crack It
In this case, the stack and solve method doesn’t bring us an immediate answer:
Neither of these methods gives you the value of y. The best way to approach this question is to try to eliminate one variable. To do this, multiply one or both of the equations by a number that will cause the other variable to have a coefficient of 0 when the equations are added or subtracted.
In this case, the question is asking us to solve for y, so try to make the x terms disappear. You want to make the coefficient of x zero, so you can quickly find the value of y. Here’s how:
Use the coefficient of x in the second equation, 5, to multiply the first equation:
5(3x + 2y) = 5(17)
15x + 10y = 85
Then use the original coefficient of x in the first equation to multiply the second equation:
3(5x − 4y) = 3(21)
15x − 12y = 63
Now stack your equations and subtract (or flip the signs and add, which is less likely to lead to a mistake.)
Simplify your equation and you have your answer.
22y = 22
y = 1
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; at least |
≤ |
is less than or equal to; no more than |
Hungry Gator
Think of the inequality sign
as the mouth of a hungry
alligator. The alligator eats
the bigger number.
Solving inequalities is pretty similar to solving equations. You can collect like 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:
−2x < −2y
Remember: When you multiply or divide an inequality by a negative number, you must reverse the inequality sign.
Here’s an example of how an inequality question may be framed on the test:
8.If −3x + 6 ≥ 18, which of the following must be true?
A)x ≤ −4
B)x ≤ 8
C)x ≥ −4
D)x ≥ −8
Here’s How to Crack It
Simplify the inequality like any other equation:
−3x + 6 ≥ 18
−3x ≥ 12
Remember to change the direction of the inequality sign!
x ≤ −4
So (A) is the correct answer.
A Range of Values
Another skill ETS may test is solving inequalities for a range of values. In these instances, you can simplify the process by initially treating the inequality as two separate problems.
Here’s an example:
If −8 < , what is one possible value of m?
Here’s How to Crack It
First work on the left side of the equation: −8 < − m + 1.
−40 < −3m + 5
−40 − 5 < −3m + 5 − 5
−45 < −3m
15 > m
Then work on the right side of the equation:
Once you have both pieces of the inequality simplified, you just need to put them back together.
If 15 > m and m ≥ 7, then 15 > m ≥ 7, but this inequality doesn’t make logical sense. Generally, inequalities are written with the smaller number on the left and the larger number on the right, so when you solve an inequality like this, you may need to rearrange the equation. This isn’t difficult, just make sure the “arrows” are still pointing at the same numbers when you change the order.
So a correct answer to this question would be any number between 7 and 15, which includes 7, but does NOT include 15.
Writing Your Own System of Equations
Sometimes you’ll be asked to take a word problem and create a system of equations or inequalities from that information. Usually they will not ask you to solve this system of equations/inequalities, so if you are able to locate and translate the information in the problem, you have a good shot at getting the correct answer. Always start with the most straightforward piece of information. What is the most straightforward piece of information? Well, that’s up to you to decide. Consider the problem shown next.
9.Aubrie, Bera, and Kea are running a lemonade and snack stand to earn money. They are selling lemonade for $1.07 a cup and chocolate chip cookies for $0.78 each. Their customers arrive on foot or by car. During a three-hour period, they had 47 customers each buying only one item and made $45.94. Aubrie, Bera, and Kea need to determine if they have enough supplies for tomorrow. Solving which of the following system of equations will let them know how many cups of lemonade, x, and how many cookies, y, they sold today?
A)
B)
C)
D)
Here’s How to Crack It
For some people, the most straightforward piece of information deals with the number of items being sold. For others, it may be the price of the items being sold. Whichever piece of information you choose, use the math to English translations in this chapter to help you identify the mathematical operations you will need to write your equation.
Let’s work through the previous problem.
You may have noticed that at least some of this word problem is just providing background information that isn’t really necessary to solve the problem. Lightly striking through the information will make the problem look less intimidating, as shown on the next example.
9.Aubrie, Bera, and Kea are running a lemonade and snack stand to earn money. They are selling lemonade for $1.07 a cup and chocolate chip cookies for $0.78 each. Their customers arrive on foot or by car. During a three-hour period they had 47 customers each buying only one item and they made $45.94. Aubrie, Bera, and Kea need to determine if they have enough supplies for tomorrow. Solving which of the following system of equations will let them know how many cups of lemonade, x, and how many cookies, y, they sold today?
The shortened problem makes it a lot easier to recognize important information. Start by identifying a straightforward piece of information, so you can start writing your own equations.
They are selling lemonade for $1.07 a cup and chocolate chip cookies for $0.78 each.
This is a fairly straightforward piece of information. Once you identify which variable represents lemonade and which one represents cookies, you can begin to write your equation. In this problem, the very last sentence gives us the needed information.
…how many glasses of lemonade, x, and how many cookies, y…
Now you know that the number of cups of lemonade they sold, multiplied by $1.07 per cup, will give you the amount of money they made selling lemonade, and the number of cookies they sold, multiplied by $0.78, will give you the amount of money they made selling cookies. Since the problem also gives you the total amount of money they made, $45.94, and states that customers were “each buying only one item” you can use the information above to write your first equation.
Once you have your first equation, go to your answer choices to determine which answers you can eliminate. You’ll quickly see that (A), (C), and (D) are all wrong, so you can select (B) without even having to construct the second equation.
Now try one on your own!
7.To save on helium costs, a balloon is inflated with both helium and nitrogen gas. Between the two gases, the balloon can be inflated up to 8 liters in volume. The density of helium is 0.20 grams per liter, and the density of nitrogen is 1.30 grams per liter. The balloon must be filled so that the volumetric average density of the balloon is lower than that of air, which has a density of 1.20 grams per liter. Which of the following system of inequalities best describes how the balloon will be filled, if x represents the number of liters of helium and y represents the number of liters of nitrogen?
A.
B.
C.
D.
Here’s How to Crack It
Start with the most straightforward piece of information and translate that. In this case, it is probably the information about the total volume of the balloon. “Between the two gases” would indicate addition, and the gases are represented as x and y, so you need x + y in the correct answer. They all have that, so go to the next piece. Together, x and y can be “up to 8 liters.” This translates to “less than or equal to 8,” because the balloon’s volume can’t be greater than 8, but it could be 8 or less. Therefore, one equation must be x + y ≤ 8. You can eliminate (A) and (B), as those answers have different equations related to x + y. The next part may be hard to translate, so pick some numbers and try them out. Make x = 2 and y = 2. The second equation in (D) is easier to work with, so see if these numbers make that equation true. 0.2(2) + 1.30(2) = 0.4 + 2.60 = 3. This is not less than 1.20, so eliminate (D) and choose (C).
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 2x + 2y, 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 2x + 2y will help.
Something to Hide
Because factoring or
expanding is usually the
key to finding the answer
on such problems, learn to
recognize expressions that
could be either factored or
expanded. This will earn
you more points. The test
writers will try to hide the
answer by factoring or
expanding the result.
Here are five examples that we’ve worked out:
1.4x + 24 = 4(x) + 4(6) = 4(x + 6)
2. = 5(x − 6) = 5x − 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)
Here’s how this might be tested on the SAT.
10.Which of the following is equivalent to + f ?
A)
B)
C)
D)
Here’s How to Crack It
Depending on what you see when you approach this problem, you may choose to solve with distribution or factoring. You may notice that there is an f in each term of the expression. In this case, you may choose to factor the expression to , but that doesn’t give you a possible answer; however, you can eliminate (B), since is not equivalent to your expression. Now you are left with an expression you may find hard to manipulate. Let’s go back and look at the mathematics behind our initial factorization so the next step in the process will make more sense.
When you factor an f out of the expression, what you are really doing is this:
Multiply the expression by 1 so that you do not change the expression.
Distribute the .
Cancel where possible.
We can follow the exact same steps to factor an like you see in (A).
This leaves you with the same expression as (A).
If you notice that the answers are all expressions themselves, you may choose to distribute the variable in front of the parentheses instead of trying to factor the expression.
Start with (A):
Distribute to each term within the binomial:
Cancel where you can: + f
Both methods give you the same answer; however, this type of algebra leaves you open to making mistakes. In the next chapter, you will discover a third way to approach this problem that you may find even easier than the methods above.
Multiplying Binomials
Multiplying binomials is easy. Just be sure to use FOIL (First, Outer, Inner, Last).
Combine Similar Terms First
When manipulating long, complicated algebraic expressions, combine all similar terms before doing anything else. In other words, if one of the terms is 5x and another is −3x, simply combine them into 2x. Then you won’t have as many terms to work with. Here’s an example:
(3x^{2} + 3x + 4) + (2 − x) − (6 + 2x) =
3x^{2} + 3x + 4 + 2 − x − 6 − 2x =
3x^{2} + (3x − x −2x) + (4 + 2 − 6) =
3x^{2}
TERMinology
Remember: A term is a
number, variable, or a
number and variable that
are combined by multiplication
or division. So in
the expression 6x + 10 = y,
6x, 10, and y are all terms.
6x + 10, however, is not
a term because a term
includes only variables and
their coefficients.
Evaluating Expressions
Sometimes you will be given the value of one of the variables in an algebraic expression and asked 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 2x = −1, then (2x − 3)^{2} = ?
Solution: Don’t solve for x; simply plug in −1 for 2x, like this:
(2x − 3)^{2} = (−1 − 3)^{2}
= (−4)^{2}
= 16
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}, what is the value of x ?
Here’s How to Crack It
Expand both sides of the equation using FOIL:
(x + 3)(x + 3) = x^{2} + 6x + 9
(x − 2)(x − 2) = x^{2} − 4x + 4
x^{2} + 6x + 9 = x^{2} − 4x + 4
Now you can simplify. Eliminate the x^{2} terms, because they are on both sides of the equal sign. Now you have 6x + 9 = − 4x + 4, which simplifies to
10x = −5
x = −
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} + 7x + 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} + 7x + 12 = (x )(x )
Factoring
When factoring an equation
like x^{2} + bx + c, think
“A.M.” Find two numbers
that Add up to the middle
term (b) and Multiply to
give the last term (c).
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} + 7x + 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} + 7x + 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
Here’s How to Crack It
Since the question told you that k is a negative integer, you can immediately eliminate (D) because it is a positive integer. 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} − 13x + 12
(x − 2)(x − 6) = x^{2} − 8x + 12
(x − 3)(x − 4) = x^{2} − 7x + 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.
SAT Favorites
The test writers play 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} + 2xy + y^{2}
(x − y)^{2} = x^{2} − 2xy + y^{2}
Here’s an example of how these equations will likely be tested on the SAT. Try it:
11. If 2x − 3y = 5, what is the value of 4x^{2} − 12xy + 9y^{2} ?
A)
B)12
C)25
D)100
Here’s How to Crack It
We are given a quadratic equation that contains two variables. ETS is testing one of its favorites.
In this case, work with 2x − 3y = 5. If you square the left side of the equations, you get
(2x − 3y)^{2} = 4x^{2} − 12xy + 9y^{2}
That’s precisely the expression for which you need to find the 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, (C).
Did you notice that this question was just another version of being asked to solve for the value of an expression rather than for a variable? Quadratics are one of ETS’s 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
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.
x = x (x + 7)
3x − 3 = x^{2} + 7x
Now, just rearrange the terms to set the quadratic equal to 0. You’ll get x^{2} + 4x + 3 = 0. Now, it’s time to factor:
x^{2} + 4x + 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 (B), that’s the one you want.
In addition to your knowing how to solve easily factorable quadratics, ETS would also like to see you demonstrate your understanding of the quadratic formula. I know what you are thinking: “Not that thing AGAIN! Can’t I just solve it with that nifty program I have on my graphing calculator?” Why yes, yes you can, but only if the problem appears in the calculator-permitted section of the test. Trust us on this one—the test writers are not always going to put these types of problems in the calculator section. Knowing the quadratic formula is an easy way to gain points on a question the test writers intend to be “hard.”
For a quadratic equation in the form y = ax^{2} + bx + c, the quadratic formula is:
To find the roots of a quadratic, or the points where y = 0, simply plug your values for a, b, and c into the quadratic formula.
The Signs They
Are a Changin’
The quadratic formula
works for quadratics in
the form y = ax^{2} + bx + c.
There is only addition in
that form, so be careful
when your quadratic has
negative signs in it.
Here’s an example:
7x^{2} − 5x − 17 = 0
So a = 7, b = −5, and c = −17. Plugging the constants into the quadratic equation you get
Let’s put your quadratic skills to work with a problem you may see on the SAT.
12. What is the product of all the solutions to the equation 3z^{2} − 12z + 6 = 0 ?
A)
B)2
C)4
D)4
Here’s How to Crack It
Using the quadratic formula x = you would do the following:
So. x = 2 + or 2 − . “Product” means to multiply, so use FOIL to multiply (2 + ) × (2 − ) to get 4 − 2 + 2 − ()^{2} = 4 − 2 = 2, which is (B).
Wow, that was a lot of work! Wouldn’t it be great if there were a shortcut? Actually, there is! When a quadratic is in the form y = ax^{2} + bx + c, the product of the roots is equal to the value of c divided by the value of a. In this case, that’s 6 ÷ 3 = 2! It’s the same answer for a lot less work. (See the inset “The Root of the Problems” for this and another handy trick—they’re worth memorizing.)
The Root of the Problems
ETS will sometimes ask you to solve for the sum or the product of the roots of a quadratic equation. You can use the quadratic formula and then add or multiply the results, but it’s quicker to just memorize these two expressions.
sum of the roots: −
product of the roots:
IMAGINARY AND COMPLEX NUMBERS
So far you have been working with real numbers, which are any numbers that you can place on a number line. The SAT will also ask you to do mathematical operations with imaginary or complex numbers.
Imaginary Numbers
An imaginary number, very simply, is the square root of a negative number. Since there is no way to have a real number that is the square root of a negative number, mathematicians needed to come up with a way to represent this concept when writing equations. They use an italicized lowercase “I” to do that: i = , and the SAT will likely tell you that in any problem involving imaginary numbers.
Another common piece of information you will need to know about i is how it behaves when it is raised to a power. Here is i raised to the powers 1 through 8. Can you complete the next four values of i in the series?
Did you notice anything about the answer? If you said that there is a repeating pattern, then you are correct. This pattern will be helpful in answering questions containing imaginary and complex numbers.
Complex Numbers
Complex numbers are another way in which the SAT may test the concept of imaginary numbers. A complex number is one that has a real component and an imaginary component connected by addition or subtraction. 8 + 7i and 3 − 4i are two examples of complex numbers.
You may be tested on complex numbers in a variety of ways. ETS may ask you to add or subtract the complex numbers. When you are completing these operations, you can treat i as a variable. Just combine the like terms in these expressions and then simplify (don’t forget to distribute the subtraction sign).
Here’s an example:
2.For i = , what is the result of subtracting (2 + 4i) from (−5 + 6i) ?
A)−7 + 2i
B)−3 − 10i
C) 3 + 2i
D) 7 − 10i
Here’s How to Crack It
Set up the subtraction necessary.
(−5 + 6i) − (2 + 4i)
Distribute the negative sign to both terms in the second set of parentheses to get
−5 + 6i − 2 − 4i
Combine like terms to get −7 + 2i, which is (A).
Since you never ended up with an i^{2} term, you never even needed to worry about the fact that. You just treat i = as a regular variable.
The SAT may also ask you to multiply complex numbers. Again you can treat i as a variable as you work through the multiplication as if you were multiplying binominals. In other words, use FOIL to work through the problem. The only difference is that you substitute −1 for i^{2}.
Finally, ETS may ask you about fractions with complex numbers in the denominator. Don’t worry—you won’t need polynomial or synthetic division for this. You just need to rationalize the denominator, which is much easier than it may sound.
To rationalize the denominator of a fraction containing complex numbers, you need to multiply the numerator and denominator by the conjugate. To create the conjugate of a complex number, you simply need to switch the addition or subtraction sign connecting the real and imaginary parts of the number for its opposite.
For example, the conjugate of 8 + 7i is 8 − 7i, and the conjugate of 3 − 4i is 3 + 4i.
Just like when you expand the expression (x + y)(x − y) to get x^{2} − y^{2}, you can do the same with a complex number and its conjugate. The Outer and Inner terms will cancel out, giving you (x + yi)(x − yi) = (x^{2} − i^{2}y^{2}) = (x^{2} + y^{2}).
(8 + 7i) × (8 − 7i) = 8^{2} − 7^{2}i^{2}, and substituting i^{2} = −1 gives you 8^{2} + 7^{2} = 113.
Here’s an example of how the SAT will use complex numbers in a fraction.
3. If i = , which of the following is equivalent to ?
A)2 +
B)2 − i
C)
D)28 − 14
Multiplying by One
(in Disguise)
In order to keep a fraction
the same, you must multiply
by 1. If you multiply the
numerator of a fraction by
the conjugate, you must do
that same operation to the
denominator or you have
changed the fraction:
is the same thing as
multiplying × 1.
Here’s How to Crack It
None of the answer choices have i in the denominator, if they have a denominator at all, so you need to get rid of that i. Roots in denominators are generally not accepted, either. To get rid of both things, you need to multiply the whole fraction by the conjugate of the denominator. This means keep the same terms, but switch the sign between them. Be very careful to not make a sign error as you work through the problem. The expression becomes
Multiply the two fractions, using FOIL on the denominators to get
Combine like terms in the denominator, and the expression becomes
There is still an i in the denominator, but you can get rid of it. Since i = , i^{2} = −1, so substitute that into the fraction to get
Reducing the fraction gives you 2 + , which is (A).
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.
SAT Smoke and Mirrors
When you’re asked to
solve an equation involving
an absolute value, it is
very likely that the correct
answer will be the negative
result. Why? Because
the test writers know that
you are less likely to think
about the negative result!
Another way to avoid mistakes
is to do all the math
inside the absolute value
symbols first, and then
make the result positive.
Here’s an example:
9.
|x + 3| = 6
|y − 2| = 7
For the equations shown above, which of the following is a possible value of x − y ?
A)−14
B)−4
C)−2
D)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 (B).
Algebra Drill 1: No Calculator Section
Work these questions without your calculator using the skills you’ve learned so far. Answers and explanations can be found on this page.
5.
y = 3x − 1
y + x = 1
In the system of equations above, if (x, y) is the solution to the system, what is the value of ?
A)
B)
C)
D)
8.For the equation = x + 3, the value of m is −3. What is the solution set for the equation?
A){−3, 3}
B){−2}
C){−2, −7}
D){3, 6}
11.If i = , what is the product of (4 + 7i) and ?
A)16 − i
B)14 + i
C)2 − 8i − 14i^{2}
D)
14.
rx^{2} = x + 3
A quadratic equation is provided above, where r and s are constants. What are the solutions for x ?
A)
B)
C)
D)
Algebra Drill 2: Calculator-Permitted Section
Work these questions using your calculator as needed and applying the skills you’ve learned so far. Answers and explanations can be found on this page.
4.If x + 6 > 0 and 1 − 2x > −1, then x could equal each of the following EXCEPT
A)−6
B)−4
C)0
D)
7.If , what is the value of x ?
A)−
B)
C)0
D)2
10.If the product of x and y is 76, and x is twice the square of y, which of the following pairs of equations could be used to determine the values of x and y ?
A)xy = 76
x = 2y^{2}
B)xy = 76
x = (2y)^{2}
C)x + y = 76
x = 4y^{2}
D)xy = 76
x = 2y
12.If − 6 < − 4r + 10 ≤ 2, what is the least possible value of 4r + 3 ?
A)2
B)5
C)8
D)11
16.How many solutions exist to the equation |x| = |2x − 1| ?
A)0
B)1
C)2
D)3
25.The sum of three numbers, a, b, and c, is 400. One of the numbers, a, is 40 percent less than the sum of b and c. What is the value of b + c ?
A)40
B)60
C)150
D)250
CHAPTER DRILL ANSWERS AND EXPLANATIONS
Algebra Drill 1: No Calculator Section
5. CStart by multiplying the second equation by 2 to clear the fractions. The equation becomes y + 2x = 2. To get it into the same form as the other equations, subtract 2x from both sides to get y = −2x + 2. Set the two x expressions equal to get 3x − 1 = −2x + 2. Add 2x and 1 to both sides, so the equation becomes 5x = 3, then divide by 5 to find that x = . Plug this value into the y = 3x − 1 to get . Finally, find the value of: , which is (C).
8. BSince the question gives the value of m, the first step is to plug that value into the original equation to get = x + 3. Now square both sides of the equation to remove the square root= (x + 3)^{2}: or −3x − 5 = x^{2} + 6x + 9. Now combine like terms. If you combine the terms on the right side of the equation, you can avoid having a negative x^{2} term. The equation becomes 0 = x^{2} + 9x + 14. Factor the equation to find the roots: 0 = (x + 2)(x + 7). The possible solutions to the quadratic are −2 and −7. Don’t forget to plug these numbers back into the original equation to check for extraneous solutions. Begin by checking x = −2. When you do this, you get = (−2) + 3, or = 1, or = 1, which is true. Now, check x = −7. Set it up as = (−7) + 3, and start simplifying to get = −4. You can technically stop simplifying here, as there is a negative number on the right-hand side of the equal sign. Remember, when taking a square root with a radical provided, it will yield the positive root only. So −7 cannot be part of the solution set. Be very careful of trap answer (C), and select (B).
11. AUse FOIL to multiply the two binomials together. The expression becomes 4 − 8i + 7i − 14i^{2}. Simplify the result by multiplying through where you can to get 2 − 8i + i − 14i^{2}. To combine the i terms, multiply 8 by to get . Now the expression is 2 − i + i − 14i^{2}, which can be further simplified to 2 − i − 14i^{2}. Substitute −1 for i^{2} and combine like terms: 2 − i + 14 = 16 − i, which is (A).
14. AThe first step to answering this question is to get the equation into the standard form of a quadratic equation by moving all the terms to the left or right side of the equation and setting it equal to zero, like this: rx^{3} − x − 3 = 0. Now that you have the equation in standard form, you can begin to solve for the roots. Since you are given variables instead of numbers, factoring this quadratic would require higher-level math, if it were even possible. You may have noticed the familiar form of the answer choices. They are in a form similar to the quadratic equation. Remember that a quadratic in standard form is represented by the equation ax^{2} + bx + c = 0, and the quadratic formula is x = . In this equation, a = r, b = −, and c = −3. Therefore, . This exact format is not present in the answer choices, but the root part only matches the one in (A), so that is likely the answer. You will have to do a little more manipulation before you can get the equations to match exactly. The fractions need to be split up, so rewrite the equation as or .
Algebra Drill 2: Calculator-Permitted Section
4. ASolve the first inequality by subtracting 6 from each side so that x > −6. You are looking for values that won’t work for x, and x cannot equal −6. Therefore, the answer must be (A). Just to be sure, solve the next inequality by subtracting 1 from each side to get −2x > −2. Divide by −2, remembering to switch the sign because you are dividing by a negative number, to get x < 1. The values in (B), (C), and (D) fit this requirement as well, so they are values for x and not the correct answer.
7. BTo solve this equation, use cross multiplication to get (2x)(x + 2) = (x^{2} + 1)(2). Expand the equation to get 2x^{2} + 4x = 2x^{2} + 2. Once you combine like terms, the result is 2x^{2} − 2x^{2} + 4x = 2 or 4x = 2. Solve for x by dividing both sides by 4 to get x = , which is (B).
10. ATranslate each statement, piece by piece. The first part tells us that “the product of x and y is 76.” Since product means multiplication, the first equation must be xy = 76, so you can eliminate (C). The second part says that “x is twice the square of y,” which translates to x = 2y^{2}, so eliminate (B) and (D), and (A) is the only choice left. Notice that only the y needs to be squared, which is why (B) is wrong. The second equation for (B) would be written as “the square of twice y,” which is not what the problem stated.
12. DNotice that this question is asking for an expression instead of a variable, so manipulate the inequality to so that you get 4r + 3 in the inequality. Treat each side of the inequality separately to avoid confusion. Starting with the −6 < −4r + 10 part, multiply both sides of the inequality by −1, remembering to switch the sign, to get 6 > 4r − 10. Add 13 to each side to get 19 > 4r + 3. Then solve the right side of the inequality. Again, multiply both sides of the inequality by −1, switching the sign to get 4r − 10 ≥ −2. Now add 13 to each side of the equation: 4r + 3 ≥ 11. Finally, combine the equations to get the range for 4r + 3. Since the question asks for the least possible value of the expression, 11, (D), is the correct answer to the question. If you see the answer before the last step above, you don’t need to combine the equations.
16. CIf |x| = |2x − 1|, either x = 2x − 1 or −x = 2x − 1. The solutions to these equations are 1 and , respectively. However, the only thing you need to recognize is that the equation has two different solutions to establish that the answer is (C).
25. DThis is a system of equations question in disguise. First, locate a piece of information in this question that you can work with. “The sum of three numbers, a, b, and c, is 400,” seems very straightforward. Write the equation a + b + c = 400. Now the question tells you that “one of the numbers, a, is 40 percent less than the sum of b and c.” Translate this piece by piece to get a = (1 − 0.4) (b + c), or a = 0.6(b + c). Distribute the 0.6 to get a = 0.6b + 0.6c. Arrange these variables so they line up with those in the first equation as a − 0.6b − 0.6c = 0. To solve for b + c, stack the equations and multiply the second equation by −1:
a + b + c = 400
−1(a − 0.6b − 0.6c) = 0(−1)
Now solve:
Simplify by dividing both sides by 1.6 to get b + c = 250. The correct answer is (D).
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 rarely 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. If you don’t have the answer, look for multiples of your solutions. When the simultaneous equation question asks for a single variable and addition and subtraction don’t work, try to make something disappear. Multiply the equations to make the coefficient(s) of the variable(s) you don’t want go to zero when the equations are added or subtracted.
○Some SAT problems require algebraic manipulation. Use tricks when you can, but if you have to manipulate the equation, take your time and work carefully to avoid unnecessary mistakes. You don’t get partial credit on the SAT for getting the problem mostly correct.
○When working with inequalities don’t forget to flip the sign when you multiply and divide by negative numbers.
○When working with inequalities over a range of values, treat each side of the inequality as a separate problem. Then combine the problems in a logical order, making sure the “arrows” are pointing to the correct numbers.
○When writing a system of equations, start with the most straightforward piece of information. You can also use the equations in the answer choices to help you narrow down the possibilities for your equations. Eliminate any answers in which an equation doesn’t match your equation.
○When a question asks for an extraneous solution, first solve your equation, and then plug the answers back into the equation. If the equation is not true when solved with the solution, then that solution is extraneous.
○When solving quadratic equations, you may need to FOIL or factor to get the equation into the easiest form for the question task. Don’t forget about the common equations that ETS uses when writing questions about quadratics.
○To solve for the roots of a quadratic equation, set it equal to zero by moving all the terms to the left side of the equation, or use the quadratic formula:
x =
When solving for the sum or product of the roots, you can also use these formulas:
sum of the roots: −
product of the roots:
○The imaginary number i = , and there is a repeating pattern when you raise i to a power: i, −1, −i, 1. When doing algebra with i, treat it as a variable, unless you are able to substitute −1 for i^{2} when appropriate.
○A complex number is a number with a real and an imaginary component joined by addition or subtraction. In order to rationalize a complex number, you need to multiply it by its conjugate, or the same complex number with the addition or subtraction sign switched to the opposite sign.
○The absolute value of a number is its distance from zero; distances are always positive. When working inside the | |, remember to consider both the positive and the negative values of the expression. Also remember that | | work like ( ); you need to complete all the operations inside the | | before you can make the value positive.