Sat math Essentials

Subtraction 
Multiplication
Scalar Multiplication
k
a
1
a
2
a
3
a
4
=
k
a
1
k
a
2
k
a
3
k
a
4
a
1
a
2
a
3
a
4
×
b
1
b
2
b
3
b
4
=
a
1
b
1
+
a
2
b
3
a
1
b
2
+
a
2
b
4
a
3
b
1
+
a
4
b
3
a
3
b
2
+
a
4
b
4
a
1
a
2
a
3
a
4
−
b
1
b
2
b
3
b
4
=
a
1
−
b
1
a
2
−
b
2
a
3
−
b
3
a
4
−
b
4
a
1
a
2
a
3
a
4
+
b
1
b
2
b
3
b
4
=
a
1
+
b
1
a
2
+
b
2
a
3
+
b
3
a
4
+
b
4
a
1
a
2
a
3
a
4
–
N U M B E R S A N D O P E R AT I O N S R E V I E W
–
6 6

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Practice Question
Which of the following shows the correct solution to the problem above? 
a.
b.
c.
d.
e.
Answer
e.
4
3
7
1
+
6
2
5
2
=
4
+
6
3
+
2
7
+
5
1
+
2
=
10
5
12
3
10
5
12
3
24
6
35
2
−
2
1
2
−
1
11
11
4
4
7
8
8
7
4
3
7
1
+
6
2
5
2
=
–
N U M B E R S A N D O P E R AT I O N S R E V I E W
–
6 7

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E q u a t i o n s
To solve an algebraic 
equation
with one variable, find the value of the unknown variable.
Rules for Working with Equations
1.
The equal sign separates an equation into two sides.
2.
Whenever an operation is performed on one side, the same operation must be performed on the
other side.
3.
To solve an equation, first move all of the variables to one side and all of the numbers to the other. Then
simplify until only one variable (with a coefficient of 1) remains on one side and one number remains on
the other side.
C H A P T E R
Algebra Review
This chapter reviews key skills and concepts of algebra that you need
to know for the SAT. Throughout the chapter are sample questions in
the style of SAT questions. Each sample SAT question is followed by
an explanation of the correct answer.
6
6 9

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Example
7
x
11 
29 
3
x
Move the variables to one side.
7
x
11 
3
x
29 
3
x
3
x
Perform the same operation on both sides.
10
x
11 
29
Now move the numbers to the other side.
10
x
11 
11 
29 
11
Perform the same operation on both sides.
10
x
40
Divide both sides by the coefficient.
1
1
0
0
x
4
1
0
0
Simplify.
x
4
Practice Question
If 13
x
28 
22 
12
x
, what is the value of
x
?
a.
6
b.
2
6
5
c.
2
d.
6
e.
50
Answer
c.
To solve for 
x
:
13
x
28 
22 
12
x
13
x
28 
12
x
22 
12
x
12
x
25
x
28 
22
25
x
28 
28 
22 
28
25
x
50
x
2
Cross Products
You can solve an equation that sets one fraction equal to another by finding 
cross products
of the fractions. Find-
ing cross products allows you to remove the denominators from each side of the equation by multiplying each side
by a fraction equal to 1 that has the denominator from the opposite side.
Example
a
b
d
c
First multiply one side by 
d
d
and the other by 
b
b
. The fractions 
d
d
and 
b
b
both
equal 1, so they don’t change the equation.
a
b
d
d
d
c
b
b
a
bd
d
b
b
d
c
The denominators are now the same. Now multiply both sides by the
denominator and simplify.
bd
a
bd
d
bd
b
b
d
c
ad
bc
The example above demonstrates how finding cross products works. In the
future, you can skip all the middle steps and just assume that 
a
b
d
c
is the
same as 
ad
bc.
–
A L G E B R A R E V I E W
–
7 0

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Example
6
x
1
3
2
6
Find cross products.
36
x
6 
12
36
x
72
x
2
Example
4
x
x
16
12
Find cross products.
16
x
4(
x
12)
16
x
4
x
48
12
x
48
x
4
Practice Question
If
9
y
y
1
2
7
, what is the value of
y
?
a.
28
b.
21
c.
6
1
3
1
d.
7
3
e.
28
Answer
b.
To solve for 
y
:
9
y
y
1
2
7
Find cross products.
12
y
9(
y
7)
12
y
9
y
63
12
y
9
y
9
y
63 
9
y
3
y
63
y
21
Checking Equations
After you solve an equation, you can check your answer by substituting your value for the variable into the orig-
inal equation.
Example
We found that the solution for 7
x
11 
29 
3
x
is 
x
4. To check that the solution is correct, substitute 4
for 
x
in the equation:
7
x
11 
29 
3
x
7(4) 
11 
29 
3(4)
28 
11 
29 
12
17 
17
This equation checks, so 
x
4 is the correct solution!
–
A L G E B R A R E V I E W
–
7 1

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Equations with More Than One Variable
Some equations have more than one variable. To find the solution of these equations, solve for one variable in terms
of the other(s). Follow the same method as when solving single-variable equations, but isolate only one variable.
Example
3
x
6
y
24
To isolate the 
x
variable, move 6
y
to the other side.
3
x
6
y
6
y
24 
6
y
3
x
24 
6
y
3
3
x
24
3
6
y
Then divide both sides by 3, the coefficient of
x
.
x
8 
2
y
Then simplify. The solution is for 
x
in terms of
y
.
Practice Question
If 8
a
16
b
32, what does 
a
equal in terms of
b
?
a.
4 
2
b
b.
2 
1
2
b
c.
32 
16
b
d.
4 
16
b
e.
24 
16
b
Answer
a.
To solve for a in terms of
b
:
8
a
16
b
32
8
a
16
b
16
b
32 
16
b
8
a
32 
16
b
8
8
a
32
8
16
b
a
4 
2
b
7 2
Special Tips for Checking
Equations on the SAT
1.
If time permits, check all equations.
2.
For questions that ask you to find the solution to an equation, you can simply substitute each answer
choice into the equation and determine which value makes the equation correct. Begin with choice 
c
.
If choice 
c
is not correct, pick an answer choice that is either larger or smaller.
3.
Be careful to answer the question that is being asked. 
S
ometimes, questions require that you solve
for a variable and then perform an operation. For example, a question may ask the value of 
x
2. You
might find that 
x
= 2 and look for an answer choice of 2. But the question asks for the value of 
x
2
and the answer is not 2, but 2 
2. Thus, the answer is 0.

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M o n o m i a l s
A 
monomial
is an expression that is a number, a variable, or a product of a number and one or more variables.
6
y
5
xy
2
19
a
6
b
4
P o l y n o m i a l s
A 
polynomial
is a monomial or the sum or difference of two or more monomials.
7
y
5
6
ab
4
8
x
y
3
8
x
9
y
z
Operations with Polynomials
To add polynomials, simply combine like terms.
Example
(5
y
3
2
y
1) 
(
y
3
7
y
4)
First remove the parentheses:
5
y
3
2
y
1 
y
3
7
y
4
Then arrange the terms so that like terms are grouped together:
5
y
3
y
3
2
y
7
y
1 
4
Now combine like terms:
Answer: 6
y
3
5
y
3 
Example
(2
x
5
y
8
z
) 
(16
x
4
y
10
z
) 
First remove the parentheses. Be sure to distribute the subtraction correctly to all terms in the second set of
parentheses:
2
x
5
y
8
z
16
x
4
y
10
z
Then arrange the terms so that like terms are grouped together:
2
x
16
x
5
y
4
y
8
z
10
z