Download Unit 2 - WUSD-ALgebra-I-and

Unit 2
Solving Linear Equations and Inequalities
Introduction
Formulas are one of the most useful tools we find in mathematics. We have all used
formulas from geometry to find perimeters, areas and volumes. There are two ways to use
formulas. The first way in which we use formulas involves substitution and simplification.
To illustrate this we will apply the formula for the volume of a sphere as given below.
V=
4
 r3
3
In this formula the ‘V’ stands for volume and the ‘r’ stands for radius. If we are given the
size of the radius we can use the formula to find the volume. An example is shown below.
If r = 10 in. then
So
V=
4
 (10) 3
3
( we substitute 10 for ‘r’ )
4
 (1000)
3
V = 4188.8 in3
V=
And using our calculator
Follow the above example and find the volume for spheres with a radius as given below:
Problem A: Find the volume of a
sphere with r = 15”
Problem B: Find the volume of a
sphere with r = 8 cm
V = ___________
V = ____________
The other way a formula is used requires solving an equation. In this situation we know how
big the sphere should be in terms of its volume and we need to find the radius. We can use
the formula with the method of guess and check to find the radius required to produce a
volume of 30 m3. We start by substituting the 30 for the ‘V’ in the volume formula.
30 =
4
 r3
3
Find the value of the radius to the nearest hundredth using the method of guess and check.
r = _____________
If we compare the two ways to use a formula which do you think is easiest?
Do you think we need a better method than “guess and check” to solve equations?
Unit 2
Vocabulary and Concepts
Equation
An equation is two equal expressions.
Ex: 3x + 2 = 5x – 24
Inequality
An inequality is a mathematical sentence that says two expressions
are not equal. Inequality symbols are >, <, >, <, ≠
Ex: 3x + 1 > 7
Isolate
To isolate a variable means to get the variable by itself on one side
of the equation.
Solution
A solution is a value that makes an equation true.
Solve
To solve an equation means to find the number that makes the equation
true.
Verify
To verify a solution is to check your answer by substituting the answer
into the original equation.
Key Concepts for Solving Equations
Goal:
The goal of solving an equation is to isolate the variable.
Method:
The method for solving an equation is to reverse the operations we see
in the original equation and the order of operations. (The keyword here
is reverse. If the equation has addition in it then we will need to subtract.)
Process:
The process for solving an equation is doing the same operation with the
same values on both sides of the equation. This means using a Property
of Equality.
Properties of Equality
Addition Property of Equality
This is used when we add the same value to both sides of an equation.
Subtraction Property of Equality
This is used when we subtract the same value to both sides of an equation.
Multiplication Property of Equality
This is used when we multiply the same value to both sides of an equation.
Division Property of Equality
This is used when we divide by the same value on both sides of an equation.
Unit 2 Section 1
Objectives

The student will explain the process of solving an equation.

The student will use the properties of equality to solve one step equations
The solving of an equation is the reverse of evaluation. To illustrate this, let’s examine the
problems below.
Example A
We start knowing the value of a variable and an expression, then we must find the overall
value of the expression by substituting and simplifying.
Evaluation:
given x = 4
with 3(x + 6)
then we substitute and simplify
3(4 + 6)
3(10)
30
Example B
When we solve an equation we start knowing the expression and the overall value of the
expression, then we find the value of the variable. An example is shown below.
Solving:
3(x + 6) = 30
…
x=4
(Notice the expression and overall value.)
(We will follow the solving process here.)
(We finish with a value for x.)
If we look at the two examples we can see that what we start and what we end with are in
reverse order!
Evaluation:
Solving:
Starts with “x =” and expression
Starts with overall value and expression
Finishes with the overall value
Finishes with “x =”
So solving means we will REVERSE the operations and the order of operations used in the
process of evaluation.
Definition:
An Equation is a mathematical sentence that says two expressions are equal.
Equations can be true or false. To illustrate these possibilities we can examine the
equations below.
Example A
42 – 1 = 3(6 – 1)
42 – 1 = 3(5)
16 – 1 = 3(5)
16 – 1 = 15
15 = 15
Example B
True!
2(4 + 1)2 = 72
2(5)2 = 72
2(5)2 = 49
2(25) = 49
50 = 49
False!
Definition:
A Solution to an equation is a value that makes the equation true.
The examples below show us how we can determine when a value is a solution to an
equation and when it is not.
Example: Determine which if either of the values below is a solution to the equation.
x2 + 1 = 3x + 5 with 3, 4
First we can try 3
Second we try 4
32 + 1 = 3(3) + 5
9 + 1 = 3(3) + 5
9+1=9+5
10 = 14
42 + 1 = 3(4) + 5
16 + 1 = 3(4) + 5
16 + 1 = 12 + 5
17 = 17
This is false, ‘3’ is not a solution!
This is true, ‘4’ is a solution!
There will be times when an equation can have two or more solutions but there will also
be times when an equation will have no solutions. (For our purposes at this time we will
see problems for which “none” of the listed values is a solution and problems where there
can be more than one solution in the list of possible values.)
Definition:
Solving an equation is the process of finding a value that makes the equation true.
Solving an equation involves a goal, a method and a process. These are:
Goal:
The goal of solving an equation is to isolate the variable.
Method:
The method for solving an equation is to reverse the operations we see in the
original equation and the order of operations.
Process:
The process for solving an equation is doing the same operation with the same
value on both sides of the equation.
First we need to understand the process using the Properties of Equality. In the
process, if we start with a true equation, then the equation must remain true at each
step from the start to the end, otherwise our result will be invalid. The examples below
illustrate the Properties of Equality.
In each example we will start with a true equation and then do the same operation on both
sides of the equation.
Example A
Example B
Example C
Example D
12 = 12
+3
+3
21 = 21
-4
-4
8 = 8
5(8)
5(8)
30 = 30
15 = 15
17 = 17
40 = 40
Addition Property
Of Equality
6 = 6
Subtraction Property
Of Equality
Multiplication Property
Of Equality
5
5
Division Property
Of Equality
If we were to add only to one side of the equation, or if we added different values to the left
and right sides of the equation, then we would end up with a false equation. So we would
not find the solution to the equation.
Solving one-step equations means we need to recognize the operation that is in the
original equation and then perform the reverse operation with the same value. The
operations of addition and subtraction are reverses and multiplication and division are
reverses. Below are examples with explanations of this process.
Original equation
Original equation
x + 12 = 21
-12 -12
(This equation has addition of 12 in it.)
(So we subtract 12 from both sides.)
x = 9
(12 and -12 add up to 0 and are gone)
3.5y = 23.45
3.5
3.5
(This equation has multiplying by 3.5 in it.)
(So we divide by 3.5 on both sides.)
y = 6.7
Original equation
a
= 18
3
(This equation has division by -3 in it.)
a .
-3 = 18 . -3
3
(So we multiply by -3 on both sides.)
a = -54
VIDEO LINK: Khan Academy Algebra: Linear Equations 4
Exercises Unit 2 Section 1
Define or complete the following
1. Define Equation
__________________________________________________
2. If we want to solve an equation we want to _______________ the operations
we see in the original equation.
3. The goal of solving an equation is to _______________________________.
4. If an equation is true then to keep it true we should _____________________.
5. Decide whether each equation is true or false.
a. _____ -2 + 7 = 32 - 4
b. _____ 6 - 16 = -5(3) + 5
c. _____ 1 + 42 = 1 + (3)(5)
d. _____
-6 + (3)(3) = 22
6. What does it mean to SOLVE an equation? ___________________________.
7. Decide which of the listed numbers is a solution to the equation. If none of
the listed values is a solution write 'none'.
a. _______
2x + 4 = 14
c. _______
3a - 1 = 3a + 5
5 or 6
b. _______
1 or 3
x2 + 5 = 9
d. _______ x + 2 = 2x – 1
1, 2, or -2
2, 3, or 4
Solve the following equations and list the property that is used. Show your WORK.
8. y + 6 = 23
9.
z - 11 = 33
10. 34 = z + 15
11. -17 = a - 9
12. w - 15 = 22
13. a + 6 = -20
14. a + 24 = 24
15. 5 = z – 5
16. 28 = 4w
17.
-4x = 36
18.
20.
-18 =
23.
1
2
y
2 .5
= a -
1
4
26. y - 1.09 = -1.09
a
2
= 18
19. 11 =
21.
3t = -20
24.
-10 = -2x
27.
-24 = 5a
z
9
22. y - 4.56 = -6.2
25.
2
v
=
5
3
Solve the following equations and CHECK your solutions. Show your WORK.
28.
12a = 108
29. -23 = x – 4
30.
55 = -10y
31.
a
4
= -11
Translate and solve:
32.
6 less than a number is 18. Find the number.
33.
When a student finds of
34.
The quotient of a number and 5 is -6. What is the number?
35.
Four times a number is 50. Find the number.
1
a number, the result is 7. Find the number.
3
Unit 2 Section 2
Objectives

The student will explain the process of solving a two-step equation.

The student will use the Properties of Equality to solve two-step
equations.
The solving of an equation is the reverse of evaluation. In this section we will examine
how the order of operations must also be reversed to solve two step equations.
In the order of operations used to evaluate expressions we always do any multiplication
or division before we perform the addition or subtraction. Below is a list of steps we will
follow to solve multi-step equations. This list will be expanded as we continue through
Unit 2. This is our first list of steps there will be others that are more complete.
The Algorithm for Solving Multi-Step Equations
1. Reverse any addition or subtraction we see in the original equation.
2. Reverse any multiplication or division we have left in the equation.
When we solve equations we will often be asked to list the properties that we use
to transform the equation step by step. The examples below show the method and the
listed properties in the solution process.
Example A
2x + 11 = 37
- 11
-11
Subtraction Property of Equality
2x = 26
2
2
Division Property of Equality
x = 13
Checking the answer
2(13) + 11 = 37
26 + 11 = 37
37 = 37
In the above example the addition of 11 in the original equation was reversed by
subtracting 11 from both sides. This means we used the Subtraction Property of Equality.
And the step is labeled appropriately. The second step in solving the equation is to
reverse the multiplication by 2. We did this by dividing both sides of the equation by 2.
This step is also labeled appropriately as the Division Property of Equality. Note also that
the equal signs are lined up to help keep our work straight.
Example B
w
- 3.6 = -4.5
 2 .2
+ 3.6
+ 3.6
w
= -0.9
 2 .2
w
-2.2 .
= -0.9
 2 .2
w = 1.98
Addition Property of Equality
.
-2.2
Multiplication Property of Equality
Checking the answer
1.98
- 3.6 = -4.5
 2 .2
 0.9 - 3.6 = -4.5
-4.5 = -4.5
Example C
7
x +2 = 6
3
-2
-2
3.
7
x
3
= 4
7
x
3
=4.3
7x
= 12
7
7
Subtraction Property of Equality
Multiplication Property of Equality
Division Property of Equality
12
7
Checking the answer
x =
7 . 12
+2 = 6
3
7
4 + 2 = 6
6 = 6
Note that there are other ways to solve Example C but clearing the denominator
by using the Multiplication Property of Equality is one of the most straight forward.
VIDEO LINK:
Khan Academy Solving ax + b = c
Exercises Unit 2 Section 2
Solve the following equations. Show all the WORK. List the Property of Equality for
each step and check your answers for the equations in this section.
1. 3x + 2 = 23
4. 5x - 7 = 3
7. 8 = -2y - 16
2.
14 = 2x - 12
3.
4=
z
-2
7
z
+ 4 = -9
2
5. -33 = 6x + 3
6.
8. 4x - 14 = 0
9. 3z + 31 = 39
Solve the following equations. Show all the WORK.
10. 13 =
13.
a
+3
9
w
+ 3.2 = 2.6
 1.3
16. a - 71 = 5
11. 34 = -x - 2
14. -
4
2
16
x=
5
3
3
17. -3.8 =
12.
48 = 11y + 48
15.
-9x = 18
a
 6 .5
18. Define EQUATION.
19. What is the expression on the right side of the equation?
5x2 – x = -11x + 3
20. Decide whether each equation is true or false.
a. _____ (2)(4) + 1 = 32
b. _____ 40 - 6 = 5(7)
21. What does it mean to SOLVE an equation?
22. What is the SOLUTION of an equation?
Write mathematical expressions for each problem.
23.
seven less than a number 'x'
24.
five more than twice a number 'y'
25. half the sum of eight and a number 'z'
Translate and solve the following problems:
26. The product of 3 and a number decreased by 7 is 11. Find the number.
27. When a student divided a number by 2 and increased that by 5, the result is -8.
What is the number?
28. Peter subtracted 3 from the product of 4 and a number, the result was 17.
What was the number?
29. Sarah's work in solving the equation 3x - 12 = 36 is shown below.
3x - 12 = 36
3
3
x - 12 = 12
+ 12 +12
x = 24
a. Check Sarah's solution. Show your work.
b. Solve the equation and check your solution.
c. Explain what Sarah did incorrectly.
29. James' work in solving the equation 24 = -2x + 6 is shown below.
24 = -2x + 6
+6
+6
30 = -2x
-2
-2
-15 = x
a. Check James' solution. Show your work.
b. Solve the equation and check your solution.
c. Explain what James did incorrectly.
Unit 2 Section 3
Objectives

The student will list the algorithm for solving a multi-step equation
and explain the steps.

The student will combine like terms on each side of an equation before
solving a two-step equation.
Definition:
Terms are considered to be "like" when they have the same variables with
the same exponents.
The examples below contain terms that are considered to be like and some that are not
like. We will apply the definition above in finding the sets of like terms in each
expression.
Example A
3x + 2y - 5x + y
In the above expression the "3x" and the "-5x" are like since they have the
same variable. Also the "2y" and the "-y" are like for the same reason.
Example B
9 - 2.5a + 6 - a - 4
In the above expression the "-2.5a" and the "-a" are like since they have
the same variable. Also the "9", "6" and the "-4" are like because they are
constants or plain numbers. ( They do not have a variable. )
Example C
4z2 + 12 - 3z + 7z2 + 8z
In the above expression the "4z2" and the "7z2" are like since they have
the same variable with the same exponent. The "-3z" and the "-5z" are
like because they also have the same variable with the same exponent (the
understood exponent of one ). The constant 12 does not have a matching
like term. The terms with "z2" and "z" are not like since they have
different exponents.
Combining terms is done by adding the coefficients of like terms. In Example C the
simplifying would be:
4z2 + 12 - 3z + 7z2 + 8z
11z2 + 5z + 12
The 4z2 plus the 7z2 is 11z2 and the -3z + 8x is 5z with the 12 remaining unchanged.
We will frequently need to simplify equations as we solve them by combining like
terms. This will add a step to our algorithm for solving a multi-step equation.
The Algorithm for Solving Multi-Step Equations
1. Simplify the equation by combining like terms as needed.
2. Reverse any addition or subtraction we see in the original equation.
3. Reverse any multiplication or division we have left in the equation.
The examples below show equations solved including the combining of like terms.
Example A
3x + 11 - x = -17
2x + 11 = -17
- 11 - 11
( We could insert the understood coefficient: 3x + 11 - 1x = -17 )
( 3 - 1 is 2 so we have 2x )
Subtraction Property of Equality
2x = -28
2 = 2
Division Property of Equality
x = -14
Example B
x + x + 15 +2x +4 = 3x + 14 - x - 2x
4x + 19 = 14
- 19 - 19
( We combined all the like terms on each side. )
Subtraction Property of Equality
4x = -5
4 = 4
x =-
Division Property of Equality
5
4
In example B it is critical to note that we cannot combine terms across the equal sign.
Each side of the equation MUST be simplified separately!
VIDEO LINK: Khan Academy Solving Multi-Step Equations
TenMarks Youtube Properties of Equality
Exercises Unit 2 Section 3
Solve the following equations. Show all your WORK CORRECTLY
1. 4x - 7 = 21
2. 24 = -3y - 9
3. 4 =
z
- 7
5
4. –x + 4 = 15
5. -8z + 4 + z = -24
6. 10 =
1
x+4
3
7. 8 = y + 9 + y
8. 1.7x - 15 + .8x = 10
9. 16 = a - 2 - 8a
10. 21 = -x + 5 + 3x
11. z + z + z + z - 3 = -1
12.
13. 20 = 4y + 12 + 8 + y
14. 6x + 15 = x + 27 - x
15. 2a + 7 - 3a = 11
16. 2x + 2 - x - x = 15
17. 2x2 + 3x - x2 - x2 + 8 = 5
18. 18.7 = 3.4a + 7.5 - 2.1a + 6
1
1
1
y+ y-4 + y=5
2
2
2
20. 4a + 17 + 2a + 1 - 11a = 28
19.
2
1
= 6z 5
5
List the properties of equality you would use to make the top equation into the bottom
equation in each problem. Write the name of the property in FULL.
a
21. 12x + 5 = 41
22.
= -5
4
12x = 36
a = -20
________________________________
________________________________
23. 5x - 13 = 70
24.
5x = 83
________________________________
-24 = -3x
8=x
________________________________
25. Describe how we can check our answer after we solve an equation.
26. What does it mean to SOLVE an equation?
27. What is the goal of solving an equation?
28. Solve the following equation and explain what operation or property you are
using at each step in the process.
2x + 7 + x - 1 = 36
29. Solve the following equation and explain what operation or property you are
using at each step in the process.
2y + 18 - y - y = 5y + 10 + y - 1
30.
Sam's work in solving the equation 4x - 2 - 6x = x + 14 - x is shown below.
4x - 2 - 6x = x + 14 - x
2x - 2 = 14
+2
+2
2x = 16
2
2
x = 8
a. Check Sam's solution. Show your work.
b. Solve the equation and check your solution.
c. Explain what Sam did incorrectly.
Unit 2 Section 4
Objectives

The student will state and apply the Distributive Property

The student will use Distributive Property in solving multi-step equations.
The Distributive Property is one of the most useful concepts we find in Algebra. When
we evaluate expressions we must do operations that are in parentheses first. However when
one of the values in the parentheses is a variable it is not possible to perform the operation
in the parentheses first. If we are multiplying the parentheses by a value we can still simplify
the expression using this Distributive Property. The property is shown below.
The Distributive Property
a(b +c) = ab + ac
The Distributive Property allows us to multiply through parentheses.
Below are examples of how the Distributive Property can be used.
Example A
3(x + 5)
The arcs show our multiplication through the parentheses.
3x + 15
We first multiply 3 times 'x' and get '3x', then 3 times 5 for 15.
Example B
8(x - 11)
8x - 88
8 time ‘x’ is ‘8x’
8 times -11 produced the -88.
Example C
-2(3a - 7)
-6a + 14
-2 times ‘3a’ is ‘6a’
The process is the same but we must be careful of the signs.
Example D
z(z - 2y)
z2 - 2yz
z time ‘z’ is z2
We can multiply by variables as well as constants.
Example E
-(4w - 9)
-1(4w - 9)
-4w + 9
In this example we used the understood coefficient of one
in order to make our multiplying easier. The overall effect
of the negative is to change the signs of all the terms in the
parenthetical expression.
There will be times when our equations will contain parentheses. This will mean once again
that we need to add to our list of steps in the algorithm for solving equations.
The Algorithm for Solving Multi-Step Equations
1.
2.
3.
4.
Use the Distributive Property to remove parenthetical expressions.
Simplify the equation by combining like terms as needed.
Reverse any addition or subtraction we see in the original equation.
Reverse any multiplication or division we have left in the equation.
The examples below show how we use the Distributive Property in solving equations.
Example A
4(x + 12) = 36
4x + 48 = 36
- 48 -48
4x = -12
4
4
x = -3
This is where we used the Distributive Property
Subtraction Property of Equality
Division Property of Equality
Example B
-2(x - 5) + 3x = 17
-2x +10 + 3x = 17
x + 10 = 17
- 10 -10
x= 7
This step is where we used the Distributive Property.
Here we simplified the left hand side by combining terms.
Subtraction Property of Equality
Example C
3(2x + 7) + x - 1 = -x -(4 - x)
6x + 21 + x - 1 = -x - 4 + x
We used the Distributive Property on both sides!
7x + 20 = -4
We simplified both sides.
- 20
-20
we used the Subtraction Property of Equality.
7x = -24
7
7
We used the Division Property of Equality.
24
x=7
VIDEO LINK: Kahn Academy: Solving Equations with the Distributive Property
Exercises Unit 2 Section 4
Solve the following equations. Show all your work correctly. Show also the checks
for your answers in this section.
1. -4(m + 12) = 36
2. 3(y + 8) - 7 = 33
4.
-35 = 5(3k + 2)
5.
7.
6x -2(3x + 1) = x + 4
1
(4x - 8) = 17
2
3. -(a + 4) + 6 = 10
6. -24 = 2a - 6(a - 12)
8. 12 = 2(z - 5)
9. x + 9 - x = 3(2x -1)
Solve the following equations. Show all your work correctly.
10. -2(x - 3) + 2x - 6 = x - 1
11. 5 = 3(4a + 2)
12. -8 = 2(y - 5) + 10
13. -(2x - 21) + 2x = 4x - 5
14. 2a + 5(a + 3) = 32
15. 2.5(x + 7) = 28.75
16.
15 =
3
1
(z - 1) + z
2
2
Translate and solve:
17. The sum of a number and 2 multiplied by 3 is 18. Find the number.
18. Tina solves the equation 8(x - 2) + 2x = 74. What is the value of x?
a) 10
b) 5.8
c) 9
d) 7.6
19. Solve the following equation and explain what operation or property you are
using at each step in the process.
2(a + 3) + 5a - 31 = 17
20.
Check your solution to problem 19. Show your work.
21. Juan's work in solving the equation 4(x - 2) - 3x + 13 = -15 is shown below.
4(x - 2) - 3x + 13 = 7
4x - 2 - 3x + 13 = 7
x + 11 = 7
- 11
-11
x = -4
a. Check Juan's solution. Show your work.
b. Solve the equation and check your solution.
c. Explain what Juan did incorrectly.
22. Mike's work in solving the equation x - (3x + 4) = -16 is shown below.
x - (3x + 4) = -16
x - 3x + 4 = -16
-2x + 4 = -16
-4
-4
-2x = -20
-2
-2
x = 10
a. Check Mike's solution. Show your work.
b. Solve the equation and check your solution.
c. Explain what Mike did incorrectly.
Unit 2 Section 5
Objective

The student will solve equations with variables terms on both sides.
There will be equations in which there are variable terms on both sides of the equal sign.
When variable terms occur on the same side of the equal sign we can combine them
because the terms will be either positive or negative and their coefficients can be added.
However, when the terms are on opposite sides of the equal sign we have a different
situation, as illustrated below.
Example A
Example B
x+x=8
2x = 8
x=x
In example A the plus sign between the variables tells us exactly what to do, so we can
add the x's and solve the equation. In example B the equal sign creates an equation that
can be true or false but there is NO arithmetic to do! So we can say the equation is true
but we can't add or subtract the x's.
The goal of solving an equation is to isolate the variable on one side of the equation.
When an equation has variable terms on both sides we will need to find a way to eliminate
one of these terms. We can use the properties of equality to make one of the two terms
go away. The example below shows us how to use the Subtraction Property of Equality
to turn a more complicated equation into a simple two-step equation.
5x + 7 = 3x + 25
- 3x
- 3x
2x + 7 = 25
-7
-7
2x = 18
2
2
x=9
Subtraction Property of Equality
Subtraction Property of Equality
Division Property of Equality
In the example above the positive ‘3x’ and the ‘-3x’ add up to zero and are gone. The ‘5x’
minus the ‘3x’ is ‘2x’ and this creates a two-step equation which we can routinely solve.
To solve equations with variable terms on both sides of the equal sign we should always
use the Addition or Subtraction Properties of Equality to eliminate the smaller of the
two variable terms.
One of the keys to solving equations with variables on both sides is recognizing which term
is smaller. To illustrate this idea we will examine the pairs of terms below, some are simple
but others we need to examine closely.
Example A
3x
5x
Example B
Example C
Example D
-8x 2x
-4a -11a
-6y 6y
In example A, both of the coefficients are positive and the smaller term is '3x'. In example B,
we must keep in mind that any negative number is less than a positive value so '-8x' is smaller.
With example C we have two negative coefficients but -11 is smaller than -4 so the '-11a' is
the smaller of the two terms. The final example D shows numbers that initially look similar
but again the negative is always less than the positive so '-6y' is the smaller term.
The example below shows another equation with variable terms on both sides being solved.
2y - 31 = -9y + 24
+ 9y
+9y
11y - 31 = 24
+31 +31
11y = 55
11
11
y=5
Addition Property of Equality
Addition Property of Equality
Division Property of Equality
In the above example the '-9y' was smaller than the '2y' so to cancel the -9y out we had to
use the Addition Property of Equality and add a positive 9y to both sides. By doing this we
eliminate the ‘y’ term on the right hand side of the equation.
The situation with variable terms on both sides of the equal sign means we need to add one
more step to our algorithm for solving multi-step equations. The new list is given below.
The Algorithm for Solving Multi-Step Equations
1.
2.
3.
4.
5.
Use the Distributive Property to remove parenthetical expressions.
Simplify the equation by combining like terms as needed.
Use Properties of Equality to eliminate extra variable terms.
Reverse any addition or subtraction we see in the original equation.
Reverse any multiplication or division we have left in the equation.
The example below shows us how to apply the algorithm for solving multi-step equations.
4(2a + 1) - 6a + 3 = -(a + 19)
8a + 4 - 6a + 3 = -a - 19
2a + 7 = -a - 19
+a
+a
3a + 7 = -19
- 7
- 7
3a = -26
3
3
26
a=3
VIDEO LINK:
First we apply the Distributive Property
Second we combine like terms
Third use the Addition Property of Equality
Fourth use the Subtraction Property of Equality
Fifth use the Division Property of Equality
Khan Academy Variables on Both Sides of an Equation
Exercises Unit 2 Section 5
Solve the following. Show ALL STEPS. If there is no solution, write NO SOLUTION.
1. 4x + 5 = 7x - 28
2.
5a + 7 = a - 41
3. 2 = -5(a + 2)
4.
5(a + 3) = 27 - a
5. 2(x + 7) = 2x + 14
6.
5y - 2 = 2(3y + 9) - 1
7. z + 7 = z - 12
8.
-2w + 33 = 4(3w - 3) + w
9. -y + 16 = 5(y - 4)
10. 4x + 25 = 7(x - 2) + x - 3
11. 3(a + 15) = 2(a + 1)
12. x - 21 = x + x + x + 49
13. 3a + 85 + a = 85
14.
-11 = y + 20
a
16. 11 =
+7
3
18. -(x + 4) + x = x + 15
15. -3a + 51 = 0
17.
14 - 3x = 23 - 7x
19. Which of the values below is the solution to the equation 3x + 4 = 25
a) 6
b)
7
c) -7
d) no solution
20. Which of the values below is a solution to the equation
a) 5
b) -5
c) 15
d)
-2x + 8 = x - 7
15
2
21. Which of the values below is a solution to the equation 4x - (x + 2) = 3(5 - x) - 12
a)
29
6
b) -
5
6
c)
5
6
d)
7
6
22. Solve the following equation and explain what operation or property you are
using at each step in the process.
5x + 7 = -x - 53
23. Solve the following equation and explain what operation or property you are
using at each step in the process.
-2(x - 3) - 4x = -(x + 5) + 11
24. Check your solution to problem #23. Show your work.
25. Solve the equation 2(x + 4) = x + 7 + x and explain the meaning of your result.
26. Solve the equation -3(x - 1) = -3x + 3 and explain the meaning of your result.
27. April's work and steps in solving the equation 9x + 7 = -3x + 13 are shown below.
9x + 7 = -3x + 13
6x + 7 = 13
-7 -7
Combining like terms
Subtraction Property of Equality
6x = 6
6
6
x=1
a. Check April's solution. Show your work.
b. Solve the equation and check your solution.
c. Explain what April did incorrectly.
Unit 2 Section 6
Objectives

The student will define and find the absolute value of real numbers.

The student will solve equations with absolute values.
There are problems that we will need to solve that must have only positive values as
answers.
In Algebra we have an operation called the absolute value function that produces
positive answers.
Definition:
The absolute value of a number is the positive distance between the number
and zero.
We use two vertical bars (
|| )to indicate when we want to find the absolute value of a
number. Below are examples of the notation used for the absolute value.
Example A
Example B
Example C
|10|
|-2.4|
|x |
Example D
|z - 3 |
The examples above show the notation used to tell us to find the absolute value of '10',
'-2.4', 'x' and the expression 'z - 3'. One way to find the absolute value of a number is
to count the number of units between the number and zero. Below are examples of
using the real number line to find the absolute value.
Example A Finding |4|
-7
-6
-5
-4
-3
-2
-1
as we count
So the |4| = 4
0
1
1
2
2
3
3
4
4
5
6
7
Example B
Finding the |-5|
-7
-6
-5
as we count
Example C
-4
1
-3
2
-2
3
-1
4
0
1
2
3
4
5
6
7
6
7
So the |-5| = 5
5
Finding the |2.5|
-7
-6
-5
-4
-3
-2
-1
0
as we count
1
1
2
2
3
4
5
So the |2.5| = 2.5
.5
From these examples we can conclude two things. First the absolute value of any number
is always POSITIVE. Second it does not matter if the value is an integer, fraction, decimal,
or irrational number, the answer maintains all the digits. This means finding
the absolute value of number is relatively straight forward.
To find the absolute value of a number we must make the value in the vertical
bars positive. Whether the number starts as positive or negative, the answer is
positive.
The examples below illustrate this principle.
|11| = 11
|-43| = 43
|-8.25| = 8.25
|3 1 | = 3 1
3
3
|-| = 
We will find at times that there are absolute values in the equations we need to solve.
An absolute value in an equation can cause an equation to have more than one solution
or it can also mean there will be no solution. The examples below illustrates these
possibilities.
Example A
Given the equation |x| = 9, both x = 9 and x = -9 are solutions
since both values make the equation true. We can demonstrate
this by substitution as shown below.
|-9| = 9
Example B
and
|9 | = 9
Given the equation |x| = -3, we CAN NOT find a solution. Since
the answer to an absolute value problem is always positive there
is no way to get a negative answer.
The fact that we can have two solutions means we must actually treat an equation that
contains an absolute value function as two separate equations. We create these two
separate equations because the number in the absolute value could be positive or
negative and make the equation true. The examples below show how to create the two
equations.
|2x + 3|
Example A
= 13
2x + 3 = 13
-3
-3
2x + 3 = -13
-3
-3
2x = 10
2x = -16
2
2
2
x = 5
2
x = -8
The solutions are x = 5 and x = -8. Frequently we will show the answer in
what is called a solution set {-8, 5}.
These answers should also be checked as with all solutions to equations.
|2x + 3|
|2(5) + 3|
|10 + 3|
|13|
|2x + 3| = 13
|2(-8) + 3| = 13
|-16 + 3| = 13
|-13| = 13
= 13
= 13
= 13
= 13
13 = 13
13 = 13
When an equation has terms or factors outside the absolute value sign, these must be
reversed before we create our two separate equations. (These outside operations would
be done last when evaluating and so must be reversed first when solving.)
Example B
3|z| + 5 = 26
-5
-5
Subtraction Property of Equality
3|z| = 21
3
3
|z| =
7
Division Property of Equality
The absolute value is isolated so
we create the two separate
equations.
z=7
x = -7
So the solution set is { -7, 7 }
( We should check the solutions. )
Example C
In this example we must first isolate the absolute value before we can
create our two equations!
|x - 5| + 3x =
-3x
0
-3x
|x - 5| = -3x
x - 5 = 3x
-x
-x
x - 5 = -3x
-x
-x
- 5 = 2x
- 5 = -4x
2
2
x = -
-4
5
2
-4
5
= x
4
so the solution set could be
{- 5 , 5 }
2
4
BUT we still need to check our answers.
|- 5
2
|x - 5| - 3x = 0
. 5
- 5| - 3 - = 0
|x - 5| - 3x
| 5 - 5| - 3 . 5
2
|- 15 | + 15 = 0
2
2
15
15
+
=0
2
2
30 = 0
2
4
=0
=0
4
|  15 | - 15 = 0
4
4
15 15
=0
4
4
0 = 0
15 = 0
Only one of the solutions is valid! The - 52- did not give us a true
equation when we checked the answer! It is called an "extraneous"
solution and cannot be part of our solution set.
So the solution set is
{5}
4
VIDEO LINK: Khan Academy Solving Absolute Value Equations
Exercises Unit 2 Section 6
Solve the following equations. Show all the work correctly.
List your answers as solution sets. If there is no solution, answer NO SOLUTION.
1. |x| = 3
2.
|x| = -2
3.
4. |y| - 2 = 8
5.
|z| + 10 = 28
6. -7 + |s| = 0
7.
8.
|a| - (-2) = 4
9. | 2y + 1 | = 7
10. | -3x + 2 | = 5
11.
2 = 6 + |t|
13. –(|x| + 2) = -6
14.
16. -6|4x - 1| + 7 = -179
17. -3 – (15 - |a|) = 12
18. 2|x| + 5 = 19
19.
20. |w| + 3 = 17
2
21. 10 – 4|x| = -2
3
23.
24. |3/4 – x| = 6
6 - |t| = 14
5 = -3|a| + 17
22. 7 – 3|x| = 1
2
12. -|x| = -14
4 – (2 - |n|) = 2
15. 3|x + 1| = 36
4|t| = 16
7
25. (9|x| + 3) – 5|x| - 3 = 12
|t| = 13
26. |x + 2| = 3x
27. |3x - 1| + x = 2x + 13
28.
|4x - 1| = |5x + 2|
29. The solution set for the equation 2|x + 5| = 16 is
a) {3,9}
b) {3, -13}
c) {13}
d) {9, -13}
30. The solution set for the equation 5 - |2x + 7| = x is
a)
{2,

2
}
3
b)
{-12}
c)
{-12,

3
}
2
d)
{-12,

2
}
3
{-4,
22
}
7
31. The solution set for the equation |6x + 1| = |x + 21|
a)
{4,

22
}
7
b)
{4 }
c)
{  22 }
7
d)
32. Explain why |x + 1| = -2 has no solution.
33. Solve the following equation and explain what operation or property you are
using at each step in the process.
-2|3x + 5| - 7 = -9
34. Check your solutions to problem #33. Show your work. Explain how you know
your solutions are correct.
35. Given the equation |x + 2| = a – 5, what values of ‘a’ will produce at least
one real solution.
36. Explain your answer to problem #35.
Unit 2 Section 7
Objectives

The student will explain why solving equations for a given variable is helpful.

The student will isolate a variable in formulas using the properties of equality.
The formula for the perimeter of a rectangle is given as P  2l  2w . There is a
formula from basic geometry which we have used in the past. If we know that the
length is 4 ft and the width is 3 ft then we can substitute these values into the formula
easily and find the perimeter as shown below.
P  2l  2w
P  2(4)  2(3)
P  86
P  14 ft
P
l

 1.
2w
w
In this form finding the perimeter is much more difficult.
However this formula could also be written as
P
l

1
2w
w
P
4
 1
2(3)
3
P
4

1
6
3
P
7

6
3
P  14 ft
When we need to use a formula it is very helpful if one of the important variables
is isolated. In this section we will be using the Properties of Equality to isolate a
variable in a formula or equation. Once we have isolated the desired variable we
can then use the new equation to find quantities by substitution.
The process of isolating a given variable in a formula or equation is called "solving a
literal equation". The examples below show us how to use the Properties of Equality.
Example A
Given the equation
Isolate h
( Volume = length . width . height )
V  lwh
V  lwh
V
lwh

lw
lw
V
 h
lw
Using the Division Property of Equality
Example B
C  p  tp
Given the equation
Isolate t
C  p  tp
p
( Cost = price + tax rate . price )
Using the Subtraction Property of Equality
p
C  p  tp
p
Using the Division Property of Equality
p
Cp
t
p
Example C
V 
Given the equation
 r 2h
3
Isolate h
V 
3 V 
( Volume = (  . radius squared . height) / 3 )
( This is the Volume of a cone. )
r h
2
3
 r 2h
3
3
Using the Multiplication Property of Equality
3V   r 2 h
3V  r 2 h

 r2  r2
3V
h
 r2
Using the Division Property of Equality
The equations we will be given do not have to be from geometry or finance, however.
The example below is of a more general nature.
Example C
Given the equation
y=
2
x+c
3
Isolate x
2
x+c
3
-c
-c
2
y-c= x
3
2
3 . (y - c) = x . 3
3
3y - 3c = 2x
y=
Using the Subtraction Property of Equality
Using the Multiplication Property of Equality
Using the Distributive Property
3y - 3c = 2x
2
2
Using the Division Property of Equality
3 y  3c
=x
2
If after we isolate x were we told that y = 4 and c = 8 we could use substitution
and find the value of x as shown below.
3 y  3c
=x
2
3( 4)  3(8)
=x
2
12  24
=x
2
 12
=x
2
-6 = x
Since we had x isolated the arithmetic to find its value was very straight forward.
VIDEO LINK:
Khan Academy Manipulating Formulas
Exercises Unit 2 Section 7
Solve for 'x' in each equation. Show your work correctly.
1.
x–k=g
2. x + 3c = b
4.
x
z
a
5.
7.
10.
r = sx + t
x
3

a
4

b
6
–3 =
1
xy
3. –bx = d
6.
ax + b = c
8.
nx – m = -p
9. 5x – 8a = 2x + 7a
11.
b b x
  0
4 5 9
12. 4x + b = 0
Solve the following problems, show all work and answer all questions in full sentences
where appropriate.
13. Given the formula for the area of a triangle is A = 1/2 bh with 'A' the area, 'b' the
base and 'h' the height, answer the questions that follow.
a. Solve the formula for 'b' and explain how you found the answer.
b. If the area (A) of a triangle is 50 cm2 and the height is 4 cm use the equation
from part 'a' above and find the base (b). Show your work algebraically and
explain how you found your answer.
c. Check your work for part 'b' algebraically by using the area and the height in
the original equation and find the base. Show your work.
14. Given the formula for the converting from Fahrenheit to Celsius is C = 5/9 (F - 32)
with C the temperature measured in Celsius and F the temperature measured in
Fahrenheit, answer the questions that follow.
a. Solve the formula for 'F' and explain how you found the answer.
b. If the temperature is 40o Celsius us the equation from part 'a' to convert it into
a temperature in Fahrenheit. Show your work algebraically and explain how
you found the answer.
c. Check your work algebraically by using the 40o and the original equation and
solving for 'F'. Show your work.
15. Given the formula for the Circumference of a circle is C = 2r with 'C' the
Circumference and 'r' the radius, answer the questions that follow.
a. Solve the formula for 'r' and explain how you found the answer.
b. If the Circumference of a circle is 320 cm use the equation from part 'a'
above and find the radius. Show your work algebraically and explain how
you found your answer.
c. Check your work algebraically by using the circumference in the original
equation and find the radius. Show your work.
16. The surface area (A), of a right circular cylinder with radius r and height h is
given by
A = 2r 2 + 2rh.
a. Solve the formula for h. Show your work algebraically, and explain how
you found your answer.
b. Use your equation from part 'a' to find the height of a right circular cylinder
with surface area of 30 cm2 and radius 3 cm. Show your work algebraically,
and explain how you found your answer.
c. Check your answer to part 'b' by using the original equation to find the height
of the cylinder. Show your work algebraically.
Solve the following equations. Show all the work correctly.
17. y + 12 = -3(y - 4)
18. 2x + 15 = 9(x - 2) - x - 3
19. 2(a - 18) = 4(a + 3)
20. x + x - 1 = x + 2x + x + 8
21. 3a + 5 + 2a = 4a + 5
22.
23. -2a + 41 = 39
2
1
= y 3
5
a
24. 17 =
+ 21
3
25. |z| + 5 = 19
26. |2x + 4| = 14
27.
28.
-3|x + 2| + 1 = -29
|x + 3| = |3x - 7|
Unit 2 Section 8
Objective
 The student will solve and graph linear inequalities in one variable.
There are times when the problems we solve do not require equations but inequalities
instead. These kinds of problems would be similar to:
If Sam is paid $200 dollars a week and $15 for every car stereo he installs, find the
number of stereos he must install to make more than $650 a week.
The phrase “more than” means that there will be a range of values beginning with 31 and
going up from there for which Sam can make more than $650. We could say that Sam
must install more than 30 stereos. This can be expressed symbolically with x > 30.
There are four inequality symbols that we will use in Algebra I, these are:
Symbol
>
<
>
<
Meaning
greater than
less than
greater than or equal to
less than or equal to
Inequalities can be true or false just like equations can be true or false. The examples
below show these possibilities.
Example A
9>5
We translate this into words as “nine is greater than five” .
This inequality is true.
Example B
-6 > 5
Example C
5>5
We translate this into words as “five is greater than five” .
This inequality is false.
Example D
2<5
We translate this into words as “two is less than or equal to five” .
This inequality is true.
Example E
6<6
We translate this into words as “six is less than or equal to six” .
This inequality is true.
We translate this into words as “negative six is greater than five” .
This inequality is false.
We found out in Unit 1 that graphs help us to interpret many ideas and relationships in
mathematics. When we work with inequalities graphs can be an important way to
visualize our answers. Inequalities are graphed on the real number line. We will use
a method called “split point and test point” to graph our inequalities. Below are examples
of this method.
Example A
Given the inequality x > 3 the ‘3’ is the split point. The ‘3’ will
divide the real number line into two regions, a region to the left
of ‘3’ and a region to the right of ‘3’. One of these regions
contains numbers that will make the inequality true and the other
region has values that will make the inequality false.
-7
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
7
Our first step in the process is to find ‘3’ on the real number line and place
an open circle on it. This circle is open in order to indicate that the ‘3’ is
the split point but it will not make the inequality true. ( 3 > 3 is false. )
Next we pick a test point. This can be any point other than ‘3’, but we
should choose a number that is easy to work with like ‘1’. Then we
substitute this value into the inequality.
1>3
This inequality is false, so the left region does not make the inequality
true. We can now pick a point from the right side region like ‘5’ and
substitute the ‘5’ into the inequality.
5>3
The inequality is now true, so the right region contains all the values that
make the inequality true. So we shade the right side of the real number
line as shown below.
-7
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
7
The shading is a dark line with an arrow indicating that all the numbers to
right from 3 to infinity will make the inequality true.
Example B
Given the inequality y < -2 the ‘-2’ is the split point. The ‘-2’ will
divide the real number line into two regions, a region to the left
of ‘-2’ and a region to the right of ‘-2’. One of these regions
contains numbers that will make the inequality true and the other
region has values that will make the inequality false.
-7
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
7
Our first step in the process is to find ‘-2’ on the real number line and place
a closed circle on it. This circle is closed in order to indicate that the ‘-2’ is
the split point and it will make the inequality true. ( -2 < -2 is true. )
Next we pick a test point. This can be any point other than ‘-2’, but we
should choose a number that is easy to work with like ‘1’. Then we
substitute this value into the inequality.
1 < -2
This inequality is false, so the right region does not make the inequality
true. We can now pick a point from the left side region like ‘-4’ and
substitute the ‘-4’ into the inequality.
-4 < -2
The inequality is now true, so the left region contains all the values that
make the inequality true. So we shade the left side of the real number
line as shown below.
-7
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
7
The shading is a dark line with an arrow indicating that all the numbers to the
left from -2 to negative infinity will make the inequality true.
Example C
Given the inequality a < 1.5 the graph for this inequality is shown below.
-7
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
7
We estimate the location of the circle for the split point ‘1.5’ as accurately
as possible.
Example D Given the inequality w > -4
-7
-6
-5
-4
-3
-2
2
the graph for this inequality is shown below.
3
-1
0
1
2
3
We estimate the location for the split point ‘-4
4
5
6
7
2
’ as accurately as possible.
3
We have looked at graphing the simplest of inequalities. Inequalities however can have
expressions as well as isolated variables. Below is an example of an inequality that has
more complicated expressions.
3(2y -1) + 5 > 7y + 6
In order to graph this inequality we must first isolate the variable. When the variable is
by itself the inequality will look exactly like the problems in the previous examples A
through D. In the previous examples the split point was easy to locate. To change the
complicated expression into a simple inequality we use the Properties of Inequality.
The Properties of Inequality operate in a similar way to the Properties of Equality. This
means that whatever we do to one side of the inequality we must do the other side as
well. There is one difference in the properties of Inequality that we must investigate.
The inequality -2 < 3 is true. We can demonstrate this by plotting the points on the
real number line as shown below.
-7
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
7
If we take the original inequality and we multiply -1 times both values, let’s see where
they end up on the real number line.
(-1)(3)=-3
(-1)(-2)=2
-7
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
7
-7
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
7
Our two points ended up switching places on the real number line which means when we
multiply or divide by any negative number the values switch places and so we must
also switch the inequality symbol as shown below.
-2 < 3
(-1)(-2) < (3)(-1)
2 > -3
The Properties of Inequality tell us that we must always perform the same operation
on both sides of the inequality and IF we multiply or divide by a negative value then
we must switch the inequality symbol.
Below are some examples of solving inequalities using the Properties of Inequality. Note
how similar this process is to solving an equation.
Example A
3x + 11 > -4
- 11
-11
3x > -15
3
3
x > -5
Subtraction Property of Inequality
Division Property of Inequality
When we are graphing a more complicated inequality it is always a good
practice to substitute our test points into the original inequality. We will
put an open circle on '-5' and then test the value '1' as below.
3(1) + 11 > -4
14 > -4
is true so we shade the right side of the real number line. and our graph is:
-7
-6
Example B
-5
-4
-3
-2
-1
0
-4(x + 1) -2 > -14
-4x - 4 - 2 > -14
-4x - 6 > -14
+6
+6
-4x > -8
-4
-4
x<2
-7
-6
-5
-4
-3
-2
-1
1
2
3
4
5
6
7
Distributive Property
Combining like terms
Addition Property of Inequality
Division Property of Inequality
We divided by a negative value so
we had to switch the sign.
0
1
2
3
4
5
6
7
Example C
5x - 6 > 7x - 4
-5x
-5x
-6 > 2x - 4
+4
+4
-2 > 2x
-2
-2
-1 > x
-7
-6
-5
-4
-3
-2
-1
Subtraction Property of Inequality
Addition Property of Inequality
Division Property of Inequality
0
1
2
3
4
5
6
7
VIDEO LINK: Khan Academy Inequalities on a Number Line
Exercises Unit 2 Section 8
Decide which of the listed values from the given set makes each inequality true and list
that as your answer.
1.
{-9, 4, 8}
3. {-2, 0, 1, 5}
x>6
2.
{-7, -1, 0, 3}
a<0
4. {-2, -1, 0, 1}
-1 > y
-1.2 > z
Graph the inequalities below.
5. x > 4
6. y < -2
7. -1.25 > a
9. z > -1
10. z < -5
11. 0.9 < a
1
< x
2
1
12. - > w
4
8. 3
Solve and graph the following inequalities. Show your work correctly. If there is a false
inequality, show your work to prove it is false.
13. 4x > 12
14. 16 > y + 10
15.
2x - 7 < 4
16. -2x < 8
17.
2y - 5 + y > 10
18.
-8 < 4(w + 4)
19. x + 1 > x
20.
y-1>y+1
21.
5x + 2 < -2(x + 6)
22. 3x + 1 < x
23.
1 < 5y + 1
24.
3x + 4 > -(x + 1) + 2x
25. -4x < 0
26.
41 < 3y + 2
27.
7(x - 4) > 3(x + 2) + 2x
28. Which of the following is the graph of the solution set for
-6x + 14 > 2
a.
0
1
2
3
4
b.
0
1
2
3
4
c.
0
1
2
3
4
d.
-4
-3
-2
-1
0
Unit 2 Section 9
Objective
 The student will solve and graph inequalities with conjunctions and
disjunctions.
There are two key words in logic and mathematics. These are words we frequently use in
every day conversation. Their meanings and uses in mathematics are very special however.
The two words are "or" and "and". These two words can be used to link two inequalities
together. The mathematical sentence created by using the word "or" is called a disjunction.
The mathematical sentence created by using the word "and" is called a conjunction. We
will begin by investigating the disjunction.
A disjunction is formed by linking two inequalities with the word "or". A disjunction
will be true when a value makes either of the two inequalities true.
To illustrate this principle we need to examine the statement below.
The assembly will be on Thursday or the assembly will be on Friday.
Here we have two statements that can be true or false joined by the word "or". This is a
disjunction. As long as one of the statements is true, the overall sentence is true. If both
of the statements are false, then the overall sentence is false. A similar sentence in
Algebra would be:
x > 7 or x < 2
This sentence will be true or false based on the value of 'x'. Let’s look at how different
values for 'x' effect the outcome of the sentence.
if x = 0 then we have the sentence
less than or equal to two.
0 > 7 or 0 < 2 this is true because zero is
if x = 5 then we have the sentence
inequalities are false.
5 > 7 or 5 < 2 this is false because both
if x = 9 then we have the sentence
greater than seven.
9 > 7 or 9 < 2 this is true because nine is
In mathematics we use symbols to represent words. For instance, ">" means "greater than".
We have symbols for all the inequalities. In Algebra we use the symbol "  " for the word "or".
The disjunction from our example above can be written in two ways as shown below.
x > 7 or x < 2
x>7  x< 2
Like a simple inequality a disjunction can be graphed on the real number line. Below is an
example of solving and graphing a disjunction.
Example
Given 2x + 1 < 3  22 < 5x + 2 we can solve each inequality separately.
- 1 -1
-2
2x < 2
2
-2
20 < 5x
2
5
x<1 
5
4< x
These two inequalities give us two split points which we must plot on the real
number line.
-7
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
7
These two split points create three regions on the real number line. We will need
to pick a test point from each of the regions and substitute the point into the
disjunction to see which region or regions to shade.
Left Region
(test point is 0)
0<1 
0< x
This is true since zero
is less than one.
Middle Region
(test point is 2)
Right Region
(test point is 5)
2<1 
5<1 
4<2
This is false since
both are false.
4<5
This is true since four is
less than or equal to five
Because the left region and the right region are true we must shade these two
sections of the real number line as shown below.
-7
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
7
The word "and" forms a conjunction of inequalities and must be evaluated differently.
A conjunction is formed by linking two inequalities with the word "and". A
conjunction will be true only when a value makes both of the two inequalities true.
To illustrate this principle we need to examine the statement below.
The assembly will be on Thursday and we will need to change the bell schedule.
Here we have two statements that can be true or false joined by the word "and". This is a
conjunction. If both of the statements are true then the overall sentence will be true. If
either of the statements are false then the overall sentence is false. A similar sentence in
Algebra would be:
-1 < x and x < 6
This sentence will be true or false based on the value of 'x'. Let’s look at how different
values for 'x' effect the outcome of the sentence.
if x = -3 then we have the sentence -1 < -3 and -3 < 6 this is false because negative
three is NOT greater than negative one.
if x = 4 then we have the sentence -1 < 4 and 4 < 6 this is true because both
inequalities are true.
if x = 8 then we have the sentence -1 < 8 and 8 < 6
NOT less than six.
this is false because eight is
The symbol we use to represent the word "and" is "  ". The above conjunction can
be represented in the two ways shown below.
-1 < x and x < 6
-1 < x  x < 6
There is also a third way a conjunction can be shown. A conjunction is usually a set of
values that is between two split points. So it could be shortened to:
-1 < x < 6
If we were to translate the sentence into words it would be "x is greater than negative one
and x is less than six". Which would mean x is between the two numbers.
Graphing this conjunction would result in:
-7
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
7
We would apply the same method as in the example for disjunction by picking points from
all three regions and testing them to see which region to shade.
VIDEO LINK: Youtube conjunctions and Disjunctions
Exercises Unit 2 Section 9
Decide which of the listed values from the given set makes each conjunction or
disjunction true and list that as your answer.
1.
{-3, 5, 10}
x > 6 or x < -2
3. {-4, -2, -1, 5}
5. {-4, -1,0, 3
2.
a < -2  a > 4
1
}
8
-1 < w < 3
{1, -1, 0, 4}
5 > y and y > 1
4. {-4, -3, -2, 2.4} z > -6  -2.4 > z
1
4
6. {-5, -4, -2, 3} z > 0 or -4
1
>z
2
Graph the inequalities below.
7. x > 3 or x < -1
8.
4 > y and y > -2
10. a < 2.5  a > -1.7
11. -5 < z < -1
9. a < -2  a > 2
Solve and graph the following inequalities. Show your work correctly.
12. 3x > 12 or x + 2 < 3
13. 3 > 2y + 5  -4y < -16
14. 5x - 7 < 23 and x + 9 < 2x + 10
15. -2(x + 1) < 6 
16.
1
x<3
2
9 > 4x - 1 > -9
17. Which of the following is the graph of the inequality 3x - 8 > 4 and
a.
c.
2
2
3
3
4
4
5
5
6
6
b.
2
d.
2
3
3
4
4
5
5
6
6
18. Which of the following is the graph of the inequality 2x + 6 < 14 or
a.
c.
2
2
3
3
4
4
5
5
6
6
b.
2
d.
2
3
3
4
4
5
5
x
+ 1.5 < 4
2
6
6
x
+ 1.5 > 4
2