Sat math Essentials

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8 3

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Practice Question
What is the equation represented in the graph above?
a.
y
x
2
3
b.
y
x
2
3
c.
y
(
x
3)
2
d.
y
(
x
3)
2
e.
y
(
x
1)
3
Answer
b.
This graph is identical to a graph of
y
x
2
except it is moved down 3 so that the parabola intersects the
y
-axis at 
3 instead of 0. Each 
y
value is 3 less than the corresponding 
y
value in 
y
x
2
, so its equation
is therefore 
y
x
2
3.
x
y
1
2
3
4
5
6
1
2
3
4
5
–1
–2
–6
–5
–4
–3
–1
–2
–3
–4
–5
–6
x
y
1
2
3
4
5
6
7
1
2
3
4
5
–1
–2
–3
–1
–2
–3
–4
–5
–6
–7
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A L G E B R A R E V I E W
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8 4

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R a t i o n a l E q u a t i o n s a n d I n e q u a l i t i e s
Rational numbers are numbers that can be written as fractions (and decimals and repeating decimals). Similarly,
rational equations
are equations in fraction form.
Rational inequalities
are also in fraction form and use the sym-
bols <, >,
≤
, and 
≥
instead of
.
Example 
Given 
30, find the value of
x
.
Factor the top and bottom:
30
You can cancel out the (
x
5) and the (
x
2) terms from the top and bottom to yield:
x
7 
30
Now solve for 
x
:
x
7 
30
x
7 
7 
30 
7
x
23
Practice Question
If
17, what is the value of
x
?
a.
16
b.
13
c.
8
d.
2
e.
4
Answer
e.
To solve for 
x,
first factor the top and bottom of the fractions:
17
17
You can cancel out the (
x
8) and the (
x
2) terms from the top and bottom:
x
13 
17
Solve for
x
:
x
13 
13 
17 
13
x
4
(
x
+ 8)(
x
+ 13)(
x
2)
(
x
+ 8)(
x
2)
(
x
+ 8)(
x
2
+ 11
x
26)
(
x
2
+ 6
x
16)
(
x
+ 8)(
x
2
+ 11
x
26)
(
x
2
+ 6
x
16)
(
x
+ 5)(
x
+ 7)(
x
2)
(
x
+ 5)(
x
2)
(
x
+ 5)(
x
2
+ 5
x
14)
(
x
2
+ 3
x
10)
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8 5

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R a d i c a l E q u a t i o n s
Some algebraic equations on the SAT include the square root of the unknown. To solve these equations, first iso-
late the radical. Then square both sides of the equation to remove the radical sign.
Example
5
c
15 
35
To isolate the variable, subtract 15 from both sides:
5
c
15 
15 
35 
15
5
c
20
Next, divide both sides by 5:
5
5
c
2
5
0
c
4
Last, square both sides:
(
c
)
2
4
2
c
16
Practice Question
If 6
d
10 
32, what is the value of
d
?
a.
7
b.
14
c.
36
d.
49
e.
64
Answer
d.
To solve for 
d,
isolate the variable:
6
d
10 
32 
6
d
10 
10 
32 
10
6
d
42
d
7
(
d
)
2
7
2
d
49
42
6
6
d
6
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A L G E B R A R E V I E W
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8 6

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S e q u e n c e s I n v o l v i n g E x p o n e n t i a l G r o w t h
When analyzing a sequence, try to find the mathematical operation that you can perform to get the next number
in the sequence. Let’s try an example. Look carefully at the following sequence:
2, 4, 8, 16, 32, . . .
Notice that each successive term is found by multiplying the prior term by 2. (2 
2 
4, 4 
2 
8, and so
on.) Since each term is multiplied by a constant number (2), there is a constant ratio between the terms. Sequences
that have a constant ratio between terms are called 
geometric sequences
.
On the SAT, you may be asked to determine a specific term in a sequence. For example, you may be asked
to find the thirtieth term of a geometric sequence like the previous one. You could answer such a question by writ-
ing out 30 terms of a sequence, but this is an inefficient method. It takes too much time. Instead, there is a for-
mula to use. Let’s determine the formula:
First, let’s evaluate the terms.
2, 4, 8, 16, 32, . . .
Term 1 
2
Term 2 
4, which is 2 
2
Term 3 
8, which is 2 
2 
2
Term 4 
16, which is 2 
2 
2 
2
You can also write out each term using exponents:
Term 1 
2
Term 2 
2 
2
1
Term 3 
2 
2
2
Term 4 
2 
2
3
We can now write a formula:
Term 
n
2 
2
n
1
So, if the SAT asks you for the thirtieth term, you know that:
Term 30 
2 
2
30 
1
2 
2
29
The generic formula for a geometric sequence is Term 
n
a
1
r
n
1
, where 
n
is the term you are looking
for,
a
1
is the first term in the series, and 
r
is the ratio that the sequence increases by. In the above example,
n
30
(the thirtieth term),
a
1
2 (because 2 is the first term in the sequence), and 
r
2 (because the sequence increases
by a ratio of 2; each term is two times the previous term).
You can use the formula Term 
n
a
1
r
n
1
when determining a term in any geometric sequence.
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8 7

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Practice Question
1, 3, 9, 27, 81, . . .
What is the thirty-eighth term of the sequence above?
a.
3
38
b.
3 
1
37
c.
3 
1
38
d.
1 
3
37
e.
1 
3
38
Answer
d.
1, 3, 9, 27, 81, . . . is a geometric sequence. There is a constant ratio between terms. Each term is three
times the previous term. You can use the formula Term 
n
a
1
r
n
1
to determine the 
n
th term of
this geometric sequence.
First determine the values of
n
,
a
1
, and 
r
:
n
38 (because you are looking for the thirty-eighth term)
a
1
1 (because the first number in the sequence is 1)
r
3 (because the sequence increases by a ratio of 3; each term is three times the previous term.)
Now solve:
Term 
n
a
1
r
n
1
Term 38 
1 
3
38 
1
Term 38 
1 
3
37
S y s t e m s o f E q u a t i o n s
A system of equations is a set of two or more equations with the same solution. If 2
c
d
11 and 
c
2
d
13
are presented as a system of equations, we know that we are looking for values of
c
and 
d
, which will be the same
in both equations and will make both equations true.
Two methods for solving a system of equations are