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Math 71
Chapter 1
Use the order of operations to
simplify each expression.
8  322  5  48  6
What is your first step?
8-3
2-5
and/or
8-6
This is not the correct first step. The order of operations gives
the first step as clearing the grouping symbols -- the
parenthesis. Review the order of operations.
8  322  5  4 8  6
8  323  4 2
Now what?
83
23
and /or
4 2
You must clear the grouping symbols first. Review the
order of operations.
8  322  5  4 8  6
8  323  4 2
8  36  8
8  32 Remember, addition and/or subtraction
86
14
is generally the last step in simplifying
an expression -- not usually the first.
Now, you try some problems.
12  3 5 2  3
2
a)
2
7  3  62
b) 24  3 5  2  1 3
2
Answers
Solve the linear equation.
9y  36  5y   y  23y  9
Step 1: Simplify each side of the equation using the
distributive property and combining like terms.
9y  36  5y   y  23y  9
9y 18  15y  y  6y 18
Be careful here -- watch your signs . Check your work with
mine. Are your signs the same as mine?
9y  36  5y   y  23y  9
9y 18  15y  y  6y 18
24 y 18  5y 18
Now, isolate the variable.
9y  36  5y   y  23y  9
9y 18  15y  y  6y 18
24 y 18  5y 18
29y  0
Next, divide both sides of the equation by 29.
9y  36  5y   y  23y  9
9y 18  15y  y  6y 18
24 y 18  5y 18
29y  0
y 0
Check the answer.
9y  36  5y   y  23y  9
90  36  50  0  230  9
0  36  29
18  18
The solution checks. Therefore, the answer is {0}.
Solve:
3x  32  x   6x 1
You try this problem and then check your solution.
Go ahead.
Solution
Answers
a) 10
b) 2


Solution
3x  32  x   6x 1
3x  6  3x  6x  6
6x  6  6x  6
add 6 to both sides
6x  6x
subtract 6x from both sides
00
always true
x can be any real number
Problem Solving
• 1. Read the problem carefully. Let x
represent one of the quantities in the problem
• 2. If necessary, write expressions for any
other unknown quantities in the problem in
terms of x.
• 3. Write an equation in x that describes the
verbal conditions of the problem
• 4. Solve the equation and answer the
problem’s question in a complete sentence.
• Check the solution.
Example
• When two times a number is decreased
by 3, the result is 11. What is the
number?
Solution
Let x represent the number.
Two times a number = 2x
Two times a number is decreased by 3 = 2x-3
The result is 11.
2x  3  11
Example
• The length of a rectangular pool is 6
meters less than twice the width. If the
pool’s perimeter is 126 meters, what are
its dimensions?
You try to set up and solve this problem.
Go Ahead.
Solution
l  2w  6
P  2l  2w
Solution
126  22w  6  2w
126  4w 12  2w
126  6w 12
138  6w
23  w
l  2w  6
l  223  6
l  40
The dimensions are 23 meters by
40 meters.
Properties of Exponents
bn  b  b  b 
b n times
b n  b m  b n m
bn
nm

b
,b0
m
b
b 0  1, b  0
1
n
b  n ,b0
b
b

n m
 b m n
ab  a n b n
n
a n a n
   n , b  0
b  b
Example
7x
2
y

5 2
There are several ways you
can start this problem.
7x y 
2 5 2
2
4 10
7 x y
Distributed the power.
7x
2
2
y
4
7 x y
10
y
2 4
7 x
10
y
4
49x

5 2
10
Example
3 4 5 2
20a b c 

5 2 
2a b c 
There is more than one way to start this problem.
I am going to simplify the inside first.
5 2
20a b c 

5 2 
2a b c 
3 4
2
10a b b c 


3
a


5 4
2 4
2
10a b c 


1


2 6 4
2
10a b c 


1


2 6 4


1

2 6 4 
10a b c 
2
2
10a b c 


1


2 6 4


1

2 6 4 
10a b c 
1
4 12 8
100a b c
2
Now you try some problems.
a)
2x
4

5 3 4
y z
4 3 1 4
2a b c 
b)  2 5 2 
 3a b c 
Solution
Solution
2x
4
2
4

5 3 4
y z
x16 y 20z 12
x16
20 12
2
y
  z
4
16
x
20 12
16y z

4
3 1 4
4
16 4
2a b c 
 2 5 2 
 3a b c 
24 a16b12c 4 

 34 a 8b 20c 8 



3 a c
2
4
8 12 20 8
ab b c
81a 8
16b 32c 4