Algebra 2: Chapter 2 Guideline on Equations, Inequalities, and Absolute Values
Example 1: Clearing Fractions and Decimals
1a)
14

1
 z

3
14

5
14



To clear fractions, find the common denominator among them and multiply
14 z

7
42
14
2 z

3

5

60
14 it by the common term.
Notice that the equation was
2 z

2 z

1
After subtracting 3 to both sides of the equation and then dividing by two, you get the final answer.
1b) 0.5
a

0.75
a

1.2

1.45
multiplied by 14. The equation has now been transformed through distribution and then simplified.
1.25
a

1.2

1.45
1.25
a

0.25
There are two ways this problem can go: clearing the decimal or combining like terms first. The best method is to combine like terms first since not all the numbers have two decimal places.
Combining like terms and simplifying makes it easier to clear the decimals as we can multiply by 100.
100 1.25
a

0.25

125 a

25 a

25
125

1
5
After multiplying by 100, the decimals are cleared and you can solve for ‘a’ with no problem. You can leave the answer in simplified form or as a decimal which is 0.20.
Example 2: Zero Product Property
A simple and non complex property that people get confused with that they decide to either distribute or even use the F.O.I.L. The key to this property is that if the equation is equal to zero, you can just separate the two products, as you will see.
2a) x

5
 
5
 
0 x

5

0 and x

5

0 x
 
5 and x

5
2b)
3 x x

12
 
0
3 x

0 and x

12

0 x

0 and x
 
12
Notice how the binomials are separated that you end up getting two solutions. There’s nothing much to this problem. No need to use F.O.I.L. at all.
Just like example 2a, you just separate the binomials and solve for ‘x’ separately. Don’t distribute the ‘3x’ as it will make the problem more complex.

Example 3: Using Equations to Solve Word Problems
The perimeter of a rectangular field is 606 meters. The length is 42 meters more than twice the width. Find the dimensions (length and width).
P

2 L

2 W Method: The general equation to finding the perimeter of a rectangle is
Given(s):

606
L

2 W

42 Will be substituted for L in the general equation.
Solving for ‘W’
2(2 W

42)

2 W

606 Rewrite the general equation for perimeter to fit the givens.
4 W

84

2 W

606 Distribute and combine like terms.
6 W

522 Subtracted by 84 to both sides.
W

87
Solving for ‘L’
Divided both sides by 6.
As you get your final answers, rewrite in this sentence: “The length of the field is 216 meters and the width of the field is 87 meters.”
Example 4: Solving Formulas
4a) Solve A

LW for W
A

LW
Divided both sides by L
L L
W

A
L
Final solution
4b) Solve A

P

Pr t for r
A

P

Pr t
A

P
Pt

Pr t
Pt
Subtract P to both sides. r

A

P
Pt
Divide both sides by Pt
Example 5: Graphing and Solving Inequalities
When an equation has < or >, then your graph will be an open dot going left or right
When an equation has ≤ or ≥, then your graph will be a closed dot going left or right
When an equation has =, then your graph will just be closed dot of the solution

5a) x

7
 
5 x

2 –4 –3 –2 –1 1 2 3 4
Notice the inequality is x < 2, so there’s an open dot on 2, showing that it’s not part of the solution.
5b)
5c)

3 x

3 x

2

6
3
–4
–4
–3
–3
–2
–2
–1
–1
1
1
2
2
3
3
4
4
The inequality is x > -3 because when you divide by negative numbers, the sign changes.
Same concept as example 5b, but this time using a closed dot to express that 2 is part of the inequality.
Theorem
Example 6: Theorems Involving Absolute Value (Theorem 2-5 in book)
Example (with steps) Graph
If x b , then x
  b and x
 b
–4 –3 –2 –1 1 2 3
If x b , then
 b
 x
 b
If x b , then x
 b or x
  b
 x

3
 
5

5
 x

3

5 Use second theorem

8
 x
 
2 Subtracted 3 to all sides.
–10 –8 –6 –4 –2 2 4 6
4
8 10
–10 –8 –6 –4 –2 2 4 6 8 10
*Note that the second theorem involves connected lines, called a compound inequality.
That means that the solutions in between -8 and -2 are all considered the solutions of the inequality.
* The third theorem has two separate lines, called a disjunction. That means that any number greater than 8 or less than -6 satisfies the equation.
*Absolute value inequalities will ALWAYS have TWO solutions.