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Math Review Objectives
At the end of this chapter you should:


Be able to solve basic algebraic problems
Rearrange equations algebraically
Math Review
Orders of Operation:
It is important when solving mathematic equations that a uniform set of rules be followed. If you
were given the following equation:
2+5x3=?
you might realize that two different answers were possible depending upon the order in which you
performed the mathematical operations.
2 + 5 x 3 = 21
(if the addition was done first and then the multiplication)
2 + 5 x 3 = 17
(if the multiplication was done first and then the addition)
Clearly, there cannot be more than one correct answer to the equation. Which answer is correct? To
prevent ambiguities like this from happening, a set of rules for the order of operations is followed. They
are:
a)
b)
c)
d)
First do all operations inside parenthesis first,
Then do all exponential functions,
Then do all multiplication and division operations,
Lastly, do all addition ad subtraction operations.
Based on the above rules, the answer to 2 + 5 x 3 is 17 (do the multiplication 5 x 3 = 15, and then add 2).
Example 2.1
Solve the following mathematical equations:
a) (3 + 1)  2
b) 4 x 3 – 7
c) (6 – 2)  1
d) (3 + 2) x 2
e) (10 – 3) x 3
f) (8 – 7) + 2
g) 9 x 2  3
h) (3 x 4) – 10
ANSWER:
a) 2
b) 5
c) 4
d) 10
Exponents:
e) 21
f) 3
g) 6
h) 2
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Exponential expressions can be broken down into two parts: the base and the exponent. An
exponent simply tells you how many times to multiply the base by itself.
102 (the 10 is the base, the 2 is the exponent). We say, “Ten raised to the second power.”
This example tells us to multiply 10 twice (in other words 10 x 10 = 100). Thus 10 2 = 100
35 (the 3 is the base, the 5 is the exponent). We say, “Three raised to the third power.”
This example tells us to multiply 3 five times (in other words 3 x 3 x 3 x 3 x 3 = 243) Thus 3 5 = 243
Example 2.2
Solve the following exponential expressions:
a) 34
b) 43
c) 103
d) 52
e) 64
f) 82
g) 53
h) 92
ANSWER:
a) 81
b) 64
c) 1000
d) 25
e) 1296
f) 64
g) 125
h) 81
Sometimes an exponent may be negative, as in 10-2. This has a very different meaning that of the
example above.
10-2 = 1  1  0.01
102 100
Example 2.3
Solve the following exponential expressions:
a) 2-4
b) 4-3
c)10-3
d) 5-2
ANSWER:
a) 0.0625
b) 0.015625
c)0.001
d) 0.04
Thus you can see, that when exponents are positive, the resulting value is larger than 1. If an
exponent is negative, the resulting value is less than 1.
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There is one special case you should be aware of. Any number with a zero as an exponent is
defined as 1.
100 = 1
30 = 1
50 = 1
Fractions:
Fractions are nothing more than the ratios. A ratio compares a part of something to the total parts.
For example, if you had 8 pieces of fruit: 1 apple and 7 oranges, the apple would represent:
1part apple 1
  0.125 of the fruit
8 parts fruit 8
Every fraction can be represented as a decimal number by carrying out the required division. In a
fraction, the number on top is called the numerator, while the number on the bottom is called the
denominator.
Mathematics Involving Fractions:
Cross Multiplication:
a c

b d
If
then
axd=bxc
Example 2.4
1 3

2 6
then
1x6=2x3
Multiplication:
a c

b d
a c
b d

Example 2.5
1 5

2 8
15
28


5
16
Division:
a
c
b

d
a c

b d

a d

b c
Example 2.6
3
2
4

5
Addition:
3 2

4 5

3 5

4 2

35
42

15
8
10
a c

b d

(a  d) + (c  b)
b d
Example 2.7
4 4

6 5

(4  5) + (4  6)
6 5
20 + 24
30

44
30

The last thing to check when working with fractions is to see if they can be simplified. In other
words, to see if there is a common factor in both the numerator and the denominator which can be factored
out. Since both 44 and 30 are divisible by two, the fraction can be reduced to:
44
30

2 22

2 15
since
2
22 .
is the same as “1”, it can be removed, and the simplified answer is
2
15
Subtraction:
a c

b d

(a  d)  (c  b)
b d
Example 2.8
4 4

5 6

(4  6)  (5  4)
5 6

24 - 20
30

4
30
Remember to check if the fraction can be simplified. Since both 4 and 30 are divisible by two, the
fraction can be reduced to:
4
30

2 2

2 15
since 2 is the same as “1”, it can be removed, and the simplified answer is 2 .
2
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Isolating Variables (Algebraic Rearrangement):
It is sometimes necessary to rearrange an equation so that the variable of concern is alone on one
side (isolated). The rearrangement must be performed following the correct mathematic rules. The most
important rule is that whatever is done to one side of an equality MUST be done to the other side of the
equality, or the resulting equation is no longer valid.
Suppose you were given the following equation to start with:
x=y
(1)
Obviously, whatever the value of “x” is, it must also be the same value as “y”. Suppose I add 5 to
the lefthand side:
x+5=y
(2)
I have clearly altered the meaning of equation 2 from the intended meaning of equation 1. The
quantity on the lefthand side is no longer the same as the quantity on the righthand side as stated in
equation 1. In order to maintain the integrity of the original equation, I must clearly add 5 to the righthand
side:
x+5=y+5
(3)
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While equation 3 looks different than equation 1, the overall integrity (meaning) the equation has
been preserved. When rearranging equations, you must always be certain that the manipulations you
perform do not alter the meaning of the equation. The manipulations will alter the appearance of the new
equation you arrive at, but the new equation must always be a valid representation of the original equation.
Let’s look at another example. Suppose you were given the following equation:
y = mx + b
Suppose you were asked to solve for “x” (isolate the “x” variable). In this example there are two
things being done to the variable “x”. “X” is being multiplied by “m”, and “b” is being added to the
product of “mx”. Isolating variables depends on two things: determining the order of operations being
done on the variable of interest, and then undoing those operations. To get rid of the operations, you start
with the last mathematical operation and go backwards through the order until you have undone them all.
The easiest way to undo a mathematical operation is to perform its opposite operation. The table below
lists them.
Table 2.1
Mathematical Operation
Opposite Operation
Addition
Division
Multiplication
Subtraction
Exponents
Subtraction
Multiplication
Division
Addition
Square root, cube root, etc
Going back to our example, based on the order of operations the equation tells us to
1) multiply “m” and “x” together
2) then add “b” to product of “mx”
To isolate “x” we must:
1) subtract b
2) divide by m
Remember what ever we do to one side of the equality (the equal sign), we must also do to the
other side.
First, we get rid of the “b”. Since it is being added to the quantity “mx” we should subtract it
(remember to do the opposite operation).
Subtract “b” from both sides:
y – b = mx + b – b
Since b – b = 0, this leaves:
y – b = mx
Since “x’ is being multiplied by “m”, we should divide both sides by “m” (the opposite operation):
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y - b mx

m
m
You might realize that mx  m  x and that m  1
m
m
1
m
Thus, the rearranged equation in terms of “x” is”
y-b
 x
m
Let’s work through one more example. Solve the following equation for “y”.
5y
 2  z
x
List the correct order of operations done on “y”
1) “y” is being multiplied by 5.
2) the product “5y” is being divided by x.
3) 2 is being subtracted from “ 5y ”.
x
To solve for “y” we must:
1) add 2.
2) multiply by x.
3) divide by 5.
Remember we go in reverse order. Also remember that what ever we do to one side of the
equality (the equal sign), we must also do to the other side.
First, we get rid of the “2”. Since it is being subtracted from the quantity “ 5y ” we should add it to both
x
sides.
5y
 2  2  z + 2
x
Simplifying the equation gives us:
5y
 z + 2
x
Next we get rid of the “x”. Since it is being divided into “5y” we should multiply both sides by it. Be
careful here, the entire right side is being multiplied by “x”.
x 
5y
 (z + 2)  x
x
Simplifying the equation gives us:
5y  (z + 2)  x
The last step is to get rid of the “5”. Since it is being multiplied by “y” we should divide both sides by it.
Be careful here, the entire right side is being divided by “5”.
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5y
(z + 2)  x

5
5
Simplifying the equation gives:
y 
(z + 2)  x
5
End of chapter questions
1.
Solve the following fractions:
2.
Solve the following equations:
a) (1 + 2)  3
e) (8 – 3) x 2
b) 3 x 4 – 5
f) (9 – 1) + 2
c) (6 – 2)  2
g) 6 x 3  2
d) (7 + 3) x 3
h) (6 x 5) – 10
3.
Solve the following exponents:
5
e) 6-4
f) 72
g) 53
h) 9-2
a) 2
b) 3-3
c) 104
d) 6-2
4.
Isolate the variable indicated:
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