MULTIPLICATION AND DIVISION
Terms under different radicals can be combined under a common radical if one term is
multiplied or divided by theother. But if the radicals are different (for example, square
root and cube root), combining terms is a bit tricky—as in the last example below.
=
x
=
x 5~
=
x!
2
or x
=
x
=
y 5
=
xy
=
x
=
y
5
Î
x
y
=
3
x
=
x 5 x
1/3
x
1/2
5 x
1/3 1 1/2
5 x
5/6
14. ~2
=
2a!
2
5
(A) 4a
(B) 4a
2
(C) 8a
(D) 8a
2
(E) 6a
The correct answer is (C). Square each of the two terms, 2 and
=
2a, sepa-
rately. Then combine their squares by multiplication: ~2
=
2a!
2
5 2
2
3~
=
2a!
2
5
4 3 2a 5 8a.
Roots You Should Know
For the GRE, memorize the roots in the following table. If you encounter one of these
radical terms on the exam, chances are you’ll need to know its equivalent integer to
answer the question.
In the table, notice that the cube root of a negative number is negative and the cube
root of a positive number is positive.
Chapter 10: Math Review: Number Theory and Algebra 233
www.petersons.com
Square Roots of “Perfect Square”
Integers
Cube Roots of “Perfect Cube”
Integers (Positive and Negative)
=
121 5 11
=
144 5 12
=
169 5 13
=
196 5 14
=
225 5 15
=
625 5 25
=
3
~2!8 5~2!2
=
3
~2!27 5~2!3
=
3
~2!64 5~2!4
=
3
~2!125 5~2!5
=
3
~2!216 5~2!6
=
3
~2!343 5~2!7
=
3
~2!512 5~2!8
=
3
~2!729 5~2!9
=
3
~2!1,000 5~2!10
Roots and the Real Number Line
As with exponents, the root of a number can bear a surprising relationship to the
magnitude and/or sign (negative vs. positive) of the number. This is another of the test
makers’ favorite areas. Here are three rules you should remember:
If n . 1, then 1 ,
=
3
n ,
=
n , n (the greater the root, the lesser the value).
However, if n lies between 0 and 1, then n ,
=
n ,
=
3
n , 1 (the greater the
root, the greater the value).
n 5 64
1 ,
=
3
64 ,
=
64 ,64
1 , 4 , 8 , 64
n 5
1
64
1
64
,
Î
1
64
,
Î
3
1
64
, 1
1
64
,
1
8
,
1
4
, 1
Every negative number has exactly one cube root, and that root is a negative
number. The same holds true for all other odd-numbered roots of negative
numbers.
=
3
2 27 523
(23)(23)(23) 5227
=
5
232 522
(22)(22)(22)(22)(22) 5232
PART IV: Quantitative Reasoning234
NOTE
The square root (or other
even-number root) of any
negative number is an
imaginary number, not a real
number. That’s why the
preceding rules don’t cover
these roots.
www.petersons.com
Every positive number has only one cube root, and that root is always a positive
number. The same holds true for all other odd-numbered roots of positive
numbers.
15. Which of the following inequalities, if true, is sufficient alone to show that
=
3
x ,
=
5
x ?
(A) 21 , x , 0
(B) x . 1
(C) |x| ,21
(D) |x| . 1
(E) x ,21
The correct answer is (E). If x ,21, then applying a greater root yields a
lesser negative value—further to the left on the real number line.
LINEAR EQUATIONS WITH ONE VARIABLE
Algebraic expressions are usually used to form equations, which set two expressions
equal to each other. Most equations you’ll see on theGRE are linear equations, in
which the variables don’t have exponents. To solve any linear equation containing one
variable, your goal is always the same: Isolate the unknown (variable) on one side of
the equation. To accomplish this, you may need to perform one or more of the
following four operations on both sides, depending on the equation:
Add or subtract the same term from both sides.
Multiply or divide by the same term on both sides.
Clear fractions by cross-multiplication.
Clear radicals by raising both sides to the same power (exponent).
Performing any of these operations on both sides does not change the equality; it
merely restates the equation in a different form.
Let’s take a look at each of these four operations to see when and how to use each one.
Add or subtract the same term from both sides of the equation.
To solve for x, you may need to either add or subtract a term from both sides
of an equation—or do both.
16. If 2x 2 6 5 x 2 9, then x 5
(A) 29
(B) 26
(C) 23
(D) 2
(E) 6
The correct answer is (C). First, put both x-terms on the left side of the
equation by subtracting x from both sides. Then combine x-terms:
Chapter 10: Math Review: Number Theory and Algebra 235
ALERT!
The operation you perform on
one side of an equation must
also be performed on the
other side; otherwise, the two
sides won’t be equal.
www.petersons.com
2x 2 6 2 x 5 x 2 9 2 x
x 2 6 529
Next, isolate x by adding 6 to both sides:
x 2 6 1 6 529 1 6
x 523
Multiply or divide both sides of the equation by the same non-zero term.
To solve for x, you may need either to multiply or divide a term from both
sides of an equation. Or you may need to multiply and divide.
17. If 12 5
11
x
2
3
x
, then x 5
(A)
3
11
(B)
1
2
(C)
2
3
(D)
11
12
(E)
11
3
The correct answer is (C). First combine the x-terms:
12 5
11 2 3
x
Next, clear the fraction by multiplying both sides by x:
12x 5 11 2 3
12x 5 8
Finally, isolate x by dividing both sides by 12:
x 5
8
12
or
2
3
If each side of the equation is a fraction, your best bet is to cross-multiply.
Where the original equation equates two fractions, use cross-multiplication
to eliminate the fractions. Multiply the numerator from one side of the
equation by the denominator from the other side. Set the product equal to
the product of the other numerator and denominator. (In effect, cross-
multiplication is a shortcut method of multiplying both sides of the equation
by both denominators.)
PART IV: Quantitative Reasoning236
www.petersons.com
18. If
7a
8
5
a 1 1
3
, then a 5
(A)
8
13
(B)
7
8
(C) 2
(D)
7
3
(E) 15
The correct answer is (A). First, cross-multiply as described above:
(3)(7a) 5 (8)(a 1 1)
Next, combine terms (distribute 8 to both a and 1):
21a 5 8a 1 8
Then isolate a-terms on one side by subtracting 8a from both sides and
combine the a-terms:
21a 2 8a 5 8a 1 8 2 8a
13a 5 8
Finally, isolate a by dividing both sides by its coefficient 13:
13a
13
5
8
13
a 5
8
13
Square both sides of the equation to eliminate radical signs. Where the variable is
under a square-root radical sign, remove the radical sign by squaring both sides of
the equation. (Use a similar technique for cube roots and other roots.)
19. If 3
=
2x 5 2, then x 5
(A)
1
18
(B)
2
9
(C)
1
3
(D)
5
4
(E) 3
The correct answer is (B). First, clear the radical sign by squaring all
terms:
~3
2
!~
=
2x!
2
5 2
2
~9!~2x!54
18x 5 4
Next, isolate x by dividing both sides by 18:
x 5
4
18
or
2
9
Chapter 10: Math Review: Number Theory and Algebra 237
www.petersons.com
LINEAR EQUATIONS WITH TWO VARIABLES
What we’ve covered up to this point is pretty basic. If you haven’t quite caught on, you
should probably stop here and consult a basic algebra workbook for more practice. On
the other hand, if you’re with us so far, let’s forge ahead and add another variable.
Here’s a simple example:
x 1 3 5 y 1 1
Quick . . . what’s the value of x? It depends on the value of y, doesn’t it? Similarly, the
value of y depends on the value of x. Without more information about either x or y,
you’re stuck—but not completely. You can express x in terms of y, and you can express
y in terms of x:
x 5 y 2 2
y 5 x 1 2
Let’s look at one more:
4x 2 9 5
3
2
y
Solving for x in terms of y:
4x 5
3
2
y 1 9
x 5
3
8
y 1
9
4
Solving for y in terms of x:
4x 2 9
3
2
5 y
2
3
~4x 2 9!5y
8
3
x 2 6 5 y
To determine numerical values of x and y, you need a system of two linear equations
with the same two variables. Given this system, you can find the values of x and y by
using either the substitution method or the addition-subtraction method.
Next, we’ll apply each method to determine the values of two variables in a two-
equation system.
PART IV: Quantitative Reasoning238
ALERT!
You can’t solve one equation
if it contains two unknowns
(variables). You either need to
know the value of one of the
variables or you need a
second equation.
www.petersons.com
The Substitution Method
To solve a system of two equations using the substitution method, follow these four
steps (we’ll use x and y here):
In either equation, isolate one variable (x) on one side.
Substitute the expression that equals x in place of x in the other equation.
Solve that equation for y.
Now that you know the value of y,plugitintoeither equation to find the value
of x.
20. If
2
5
p 1 q 5 3q 2 10, and if q 5 10 2 p, then
p
q
5
(A)
5
7
(B)
3
2
(C)
5
3
(D)
25
6
(E)
36
6
The correct answer is (D). First, combine the q-terms in the first
equation:
2
5
p 5 2q 2 10
Next, substitute (10 2 p)forq (from the second equation) in the first
equation:
2
5
p 5 2~10 2 p!210
2
5
p 5 20 2 2p 2 10
2
5
p 5 10 2 2p
Move the p-terms to the same side, then isolate p:
2
5
p 1 2p 5 10
12
5
p 5 10
p 5
S
5
12
D
~10!
p 5
25
6
Chapter 10: Math Review: Number Theory and Algebra 239
www.petersons.com
The Addition-Subtraction Method
Another way to solve for two unknowns in a system of two equations is with the
addition-subtraction method. Here are the five steps:
Make the coefficient of either variable the same in both equations (you can
disregard the sign) by multiplying every term in one of the equations.
Make sure the equations list the same variables in the same order.
Place one equation above the other.
Add the two equations (work down to a sum for each term), or subtract one
equation from the other, to eliminate one variable.
You can repeat steps 1 to 3 to solve for the other variable.
21. If 3x 1 4y 528, and if x 2 2y 5
1
2
, then x 5
(A) 212
(B) 2
7
5
(C)
1
3
(D)
14
5
(E) 9
The correct answer is (B). To solve for x, you want to eliminate y.You
can multiply each term in the second equation by 2, then add the
equations:
3x 1 4y 528
2x 2 4y 5 1
5x 1 0y 527
x 5
2
7
5
Since the question asked only for the value of x, stop here. If the question had asked
for both x and y (or for y only), you could have multiplied both sides of the second
equation by 3, then subtracted the second equation from the first:
3x 1 4y 528
3x 2 6y 5
3
2
0x 1 10y 5
29
1
2
10y 5
2
19
2
y 5
2
19
20
PART IV: Quantitative Reasoning240
NOTE
If a question requires you to
find values of both unknowns,
combine the two methods. For
example, after using addition-
subtraction to solve for x,
substitute the value of x into
either equation to find y.
www.petersons.com
Which Method Should You Use?
Which method—substitution or addition-subtraction—you should use depends on
what the equations look like to begin with. To see what we mean, look again at this
system:
2
5
p 1 q 5 3q 2 10
q 5 10 2 p
Notice that the second equation is already set up nicely for the substitution method.
You could use addition-subtraction instead; however, you’d just have to rearrange the
terms in both the equations first:
2
5
p 2 2q 5210
p 1 q 5 10
Now, look again at the following system:
3x 1 4y 528
x 2 2y 5
1
2
Notice that the x-term and y-term already line up nicely here. Also, notice that it’s easy
to match the coefficients of either x or y: multiply both sides of the second equation by
either 3 or 2. This system is an ideal candidate for addition-subtraction. To appreciate
this point, try using substitution instead. You’ll discover that it takes far more number
crunching.
SIMPLE ALGEBRAIC FUNCTIONS
In a function (or functional relationship), the value of one variable depends upon the
value of—or is “a function of”—another variable. In mathematics, the relationship is
expressed in the form y = f(x), where y is a function of x. To find the value of the
function for any value x, simply substitute the x-value for x wherever it appears in the
function. For instance, suppose that f(x)=x
2
+3x + 4. Here’s how you’d find f(2 + a):
f~2 1 a!5~2+a!
2
+3~2+a!24
5 4+4a + a
2
+6+3a 2 4
5 a
2
+7a +6
A relatively simple GRE function problem amounts to a single plug-in exercise like
the above example. But a more complex function problem might require that you
apply the same function twice.
Chapter 10: Math Review: Number Theory and Algebra 241
TIP
To solve a system of two linear
equations with two variables,
use addition-subtraction if you
can quickly and easily
eliminate one of the variables.
Otherwise, use substitution.
www.petersons.com
22. If f(x)=x + 1, then
1
f~x!
3 f
S
1
x
D
5
(A) x(x +1)
(B)
x+1
x
(C) x
(D)
x+1
2
(E)
1
x
The correct answer is (E). To determine
1
f~x!
, substitute x +1forf(x):
1
f~x!
=
1
x +1
To determine f
S
1
x
D
, substitute
1
x
for x in the function:
f
S
1
x
D
=
1
x
+1=
1+x
x
To solve the problem, multiply as follows:
1
x +1
3
1+x
x
=
1
x
LINEAR EQUATIONS THAT CANNOT BE SOLVED
Never assume that one linear equation with one variable is solvable. If you can reduce
the equation to 0 5 0, then you can’t solve it. In other words, the value of the variable
could be any real number.
23. 3x 2 3 2 4x = x 2 7 2 2x +4
Column A
Column B
x 0
(A) The quantity in Column A is greater.
(B) The quantity in Column B is greater.
(C) The quantities are equal.
(D) The relationship cannot be determined from the information given.
The correct answer is (D). All terms on both sides cancel out:
3x 2 3 2 4x = x 2 7 2 2x +4
2x 2 3=2x 2 3
0=0
∴x = any real number
In some cases, what appears to be a system of two equations with two variables is
actually one equation expressed in two different ways.
PART IV: Quantitative Reasoning242
www.petersons.com
. (the greater the root, the lesser the value).
However, if n lies between 0 and 1, then n ,
=
n ,
=
3
n , 1 (the greater the
root, the greater the value).
n. eliminate the fractions. Multiply the numerator from one side of the
equation by the denominator from the other side. Set the product equal to
the product of the