Inequalities
Supplement for Math 121, Pierce College
I.1. Solving Linear Inequalities Algebraically
Here are some examples of linear inequalities:
x 3 5
2x  1  2
4x  x  2
Solutions to an inequality consist of any value(s) of the variable that make the inequality
true. For example, x = 2 and x = 0 are solutions to the inequality x – 3 < 5, while x = 10
is not a solution to this inequality. The set of all the solutions for an inequality is called
the solution set. In order to solve inequalities we need to follow a step-by-step process
similar to the process we use to solve linear equations with one important difference.
Rules to follow when solving inequalities
The first two rules are the same rules we use to solve equations:
1) We may add or subtract any quantity from both sides of the inequality and keep
the same set of solutions to the inequality.
2) We may multiply or divide both sides of the inequality by any positive number
and keep the same set of solutions.
The third rule is unique to inequalities
3) If we multiply or divide both sides of the inequality by a negative number, we
need to reverse the direction of the inequality to keep the same solutions.
Let’s look at an example to see that the third rule is true. Consider the basic
inequality 2  5 . If we were to simply multiply both sides by -1, we would be left with
 2  5 , which is not true. By reversing the inequality sign, we get the true statement
 2  5 .
Writing Solutions to Inequalities
The solution to an inequality is a set of numbers. For example, x  5 describes the set of
all numbers that are strictly greater than 5. This set would include numbers like 6, 237,
11
5.01, and . When the solution to an inequality is written, there are two standard ways
2
of writing the solution.
Set-Builder Notation
To write x  5 formally in set notation, we would write {x | x  5} . This is read as “the
set of x-values such that x is greater than 5.” The curly brackets denote we’re writing a
set of numbers. The vertical line (|) is read as “such that.”
Interval Notation
To write x  5 in interval notation, we would write (5, ) . The problem with this
notation is that you have to be careful not to mistake it for a coordinate point. With this
notation, writing (a,b) means {x | a  x  b }. Notice that with x  5 , there was no upper
value, so infinity (  ) was used to denote that the value of x could be as big as we want.
If the inequality is not a strict inequality, then we use square brackets. For example
2  x  4 would be written [2,4). Notice the square bracket is used for the inequality that
includes the “or equal to,” and the round bracket is used for the strict inequality.
Solving linear inequalities:
Example 1. Solve x  3  5
x 3 5
x 8
Add three to both sides of the equation
We now write the answer formally in set notation or interval notation.
Set notation: {x | x  8}
Interval notation: (,8)
8
Example 2. Solve 4  2x  10
4  2x  10
 2x  6
x  3
Subtract 4 from both sides
Divide by -2. Don’t forget to change the direction
of the inequality.
Set notation: {x | x  3}
Interval notation: [3, )
-3
Compound Inequalities
A compound inequality like 2  x  4 describes a set of x-values between two numbers.
You can think of this as saying 2  x and x  4 . The only values for x that satisfy this
compound inequality are numbers that satisfy both parts of the inequality.
When you solve compound inequalities, you can either carefully work with all three
parts at once, or you can break the compound inequality into two pieces, work with each
separately, then rewrite your answer as a compound inequality in the end.
Example 3. Solve 2  3x  1  6
Method 1:
2  3x  1  6
3  3x  7
7
1 x 
3
Add 1 to all parts of the inequality
Divide all parts by 3
Method 2: Start by breaking the inequality into two pieces:
2  3x  1
3  3x
1 x
3x  1  6
3x  7
7
x
3
Add 1 to both sides of each inequality
Divide both sides of each by 3
Either way, we put the answer together in the end as:
7
Set notation: {x | 1  x  }
3
7
Interval notation: (1, )
3
1
7/3
Important note: Suppose3 the shaded area on the number line below represents the set of
values you’re trying to describe with an inequality
2
4
Writing 4  x  2 is not correct notation. First of all, it would suggest that 4 < 2, which
is clearly not true. More importantly, 4  x  2 would mean “the set of x values for
which 4 is less than x and x is less than 2.” It is not possibly for x to be both. Here, the
correct way to write this inequality is: {x | x  2 or x  4} .
I.2. Solving Absolute Value Inequalities
The absolute value of a quantity measures it’s magnitude, or size. Another way to think
of it is the absolute value gives a distance from zero on the number line. Because of this,
the absolute value is always positive. The absolute value is denoted using vertical bars
surrounding the quantity.
Example 1. Find 3 and  3
In this case, both quantities have magnitude 3, and both are located 3 units away from
zero on the number line, so
3  3 and  3  3
Now lets look at how absolute values work in inequalities. Consider the inequality
x  3 . This is asking what values of x have magnitude less than three. Numbers like 1,
2, 1.5, and 2.9 would work, but numbers like –1, -2, -1.5, and –2.9 would also work,
since their magnitude is also less than three. We can also think of this problem as asking
what numbers are within 3 units of zero on the number line. This is going to extend both
left and right from zero.
From this, we can conclude that x  3 can be rewritten as {x | 3  x  3} , since x has be
less than three, but if it were smaller than –3, the magnitude would no longer be less than
three.
Example 2. Rewrite x  4 as an inequality without using absolute values.
In this case, the magnitude of x need to be greater than or equal to 4, so x needs to be 4
units or further away from zero on a number line. So x could be greater than 4, or x could
be less than –4, since the magnitude would then be larger than 4.
So x  4 can be rewritten as {x | x  4 or
x  4}
General Rule for Working with Absolute value inequalities
In these rules, A and B can be variables, numbers, or expressions
1) A  B can be rewritten as  B  A  B
2) A  B can be rewritten as A  B or
A  B
Example 3. Rewrite x  2  5 without absolute values.
Using the first rule, we can rewrite this as  5  x  2  5
Notice that we’re saying that the quantity we’re taking the absolute value of is less than
5, so the quantity is less than 5 or greater than –5.
If a problem involves absolute values, but the absolute value is not by itself on one side
of the inequality, we first want to use our basic inequality rules to isolate the absolute
value. In other words, we want to get the absolute value by itself on one side of the
inequality before rewriting it.
Example 4. Solve  x  2  8  3
 x 2 8  3
 x  2  5
x2 5
x  2  5 or
x  7 or
Subtract 8 from both sides
Divide by -1. Don’t forget to change the direction
of the inequality.
Now we can rewrite the inequality with the
absolute value
x  2  5 Add 2 to both sides of each inequality
x  3
In set notation, the solution is: {x | x  7 or
x  3}
I.3. Solving Inequalities Graphically
1
Consider the inequality x  1  0 . We can fairly easily solve this algebraically to see
2
the solution is {x | x  2} . Let’s look at different, graphical method for solving
1
inequalities. We will start by graphing the equation y  x  1 .
2
3
2
1
0
-1
-1
0
1
2
3
4
-2
-3
1
1
x  1 , the inequality x  1  0 is equivalent to writing y  0 . In
2
2
other words, we’re looking for the set of x-values for which the y-value of the graph is
greater than 0 (that is, the x-values for which the graph of the line is above the x-axis).
Looking at the graph, we can confirm our earlier solution; when x>2, the y value of the
graph is greater than 0.
Notice that since y 
1
x  1  1 .
2
The inequality given is the same as writing y  1 . Looking at the graph, we can see
that y  1 for all x-values greater than 0. So the solution is {x | x  0}
Example 1. Using the graph drawn above, solve
Example 2. Using a graph, solve the inequality 2x  1   x  5
Start by graphing y1  2 x  1 and y 2   x  5
5
y1
4
3
2
y2
1
0
-1
-1
0
1
2
3
4
We’re looking for the set of x-values so that y1  y 2 . In other words, we’re looking for
the x-values where the first line is below (or equal to) the second line. From the graph we
see that this occurs for x  2 .
Set notation: {x | x  2}
Interval notation: (,2]
Example 3. Solve 4 x  x 2  5  0
The graphical method works also for higher-order inequalities. We graph
y  4x  x 2  5
10
8
6
4
2
0
-2
-1
-2
0
1
2
3
4
5
6
From the graph, we can see that y is greater than zero (above the x-axis) for values of x
between -1 and 5. Our solution then is:
Set notation: {x | 1  x  5}
Interval notation: (1,5)
Example 4. Solve 4 x  x 2  5  0
Using the graph from example 3, we see that y is less than zero for values of x that are
less than -1 or greater than 5.
In set notation, we’d write: {x | x  1 or x  5}
Note: Again, it would be incorrect to write {x | 5  x  1}
Interval notation is a little trickier here. We write (,1)  (5, ) . The funny U-shape
is the union symbol, which combines the sets to indicate that x can lie in either set.
I.4. Solving Higher Order Inequalities Algebraically
You may be wondering if inequalities like 4 x  x 2  5  0 can be solved algebraically.
While the technique is not quite as straightforward as the method for linear inequalities,
there is a way.
When you consider the function y  4 x  x 2  5 , there are only three possibilities for y:
y can be less than zero, greater than zero, or equal to zero. Think about how the function
could change from being less than zero to being greater than zero. On a graph, this
would mean the graph changing from being below the x-axis to being above the x-axis.
As you might guess, the only ways this can happen are:
1) The graph crosses the x-axis (y is equal to zero)
2) The graph has a break (discontinuity) in it
The first way is probably the most common, and the only thing you’ll see with
polynomial functions. The second way can occur with rational functions, piecewise
functions, and some other functions.
Our algebraic technique uses this idea to help us solve inequalities. We’ll start by
determining where the function is equal to zero or is undefined.
Example 1. Determine where x 2  x  6 is equal to zero or undefined
We set the function equal to zero and factor
x2  x  6  0
( x  3)( x  2)  0
From the factored form, we see that the expression is equal to zero when x is 3 or -2.
Since it’s a quadratic, we know that it is defined for all x.
Now we will place these x values on a number line.
-2
3
Notice that these values divide the number line into three segments. On any one of these
segments, the expression cannot change sign. As we saw before, the expression can only
change sign when x is zero or undefined.
Now we need to determine whether x 2  x  6 is positive or negative on each of these
intervals. We can determine this by evaluating the expression at a value inside the
interval. Note: any x-value in the interval may be used; in the table below, one x-value
from each interval was arbitrarily chosen.
Interval
x  2
2 x 3
x3
x-value inside
interval
-5
1
4
Value of x 2  x  6
at that x-value
24
-6
6
Since the expression cannot change value inside each interval, and since the expression is
positive on one value in the interval x  2 , we can conclude that on the interval x  2 ,
the expression must be positive on the whole interval x  2 . Likewise the expression is
negative on the interval  2  x  3 , and positive on the interval x  3 .
Sometimes, we might show this on the number line we drew
3
-2
2
Example 2. Solve x  x  6  0
From the work we just did, we saw that x 2  x  6 was negative on the interval
 2  x  3 . So our solution is:
Set notation: {x | 2  x  3}
Interval notation: (2,3)
Example 3. Solve ( x  1)( x  2)( x  3)  0
Start by finding where ( x  1)( x  2)( x  3)  0 . Since the polynomial is in factored form,
we can see that it will be zero when x  1,2,3 . Putting these values onto a number line
reveals our intervals
-2
1
3
Now we pick a value inside each interval and evaluate the expression to determine if it is
positive or negative in that interval.
Interval
x-value inside
interval
x  2
 2  x 1
1 x  3
x3
-3
0
2
4
Value of
( x  1)( x  2)( x  3) at
that x-value
-24
6
-4
18
For easy reference, we’ll record the results on the number line
-2
1
3
Now we can easily answer the question ( x  1)( x  2)( x  3)  0
Set notation: {x | 2  x  1 or x  3}
Interval notation: [2,1]  [3, ) .
Exercises
I.1
Solve each of the following inequalities algebraically. Write your answer in your
instructor’s preferred notation.
1. x  1  2
11. 2  x  1  7
2. 2x  3  4
12.  3  x  5  2
3. 2  0.2x  1
13. 3  2x  1  9
1
4. 4  x  6
14. 1  5  3x  2
4
5. 4  2x  8
15. 2  6  2x  4
6. 3  6  3x
16. 3  4  x  8
7. x  1  3x  4
For the following two problems, break
apart the inequality, solve each piece,
then put your answers together at the
end.
8. 2x  3  4x  2
9. x 
1
x2
2
10. 0.4x  x  6
17. x  3  2x  1  4x  1
18. 2x  3  x  5  3x  1
I.2.
Rewrite each of the following inequalities without using absolute values.
1. x  3  5
3. 4 x  1  2
2. 2 x  1  6
4. x  3  4
Solve each of the following inequalities algebraically. Write your answer in your
instructor’s preferred notation.
5. x  3  5
9. x  3  2  5
6. 2 x  1  6
10. x  1  3  4
7. 4 x  1  2
11. 4  2 x  1  2
8. x  3  4
12. 3 x  5  2  7
I.3.
Solve each of the following inequalities graphically. Write your answer in your
instructor’s preferred notation.
1. 2x  4  2
7. x 2  2 x  1  1
2. 4  x  1
8.  x 2  4 x  1  4
3. x  1 
1
x 1
2
4. 4  x  2x  2
5. x 2  2 x  8  0
6. x  7 x  12  0
2
9. x 2  2 x  8  0
10. x 2  7 x  12  0
11. ( x  2)( x  1)( x  3)  0
12. x 3  1  0
I.4.
Solve each of the following inequalities algebraically. Write your answer in your
instructor’s preferred notation.
1. x 2  2 x  3  0
7. ( x  1) 2 ( x  2)  0
2. x 2  3x  2  0
8. ( x  2) 2 ( x  3)  0
3. x 2  2 x  3  0
9. x 2  2 x  1  0
4. x 2  3x  2  0
10. x 2  x  4  0
5. ( x  1)( x  2)( x  4)  0
11. x 2  2 x  1  0
6. ( x  3)( x  1)( x  1)  0
12. x 2  x  4  0
Solutions to Odd Exercises
I.1.
{x | x  3}
1.
[3, )
7.
{x | x  5 2}
(5 2 , )
13.
{x | 1  x  4}
(1,4)
{x | x  4}
(,4]
15.
{x | 1  x  2}
[1,2)
17.
{x | x  1}
(1, )
3.
{x | x  15}
(,15)
9.
5.
{x | x  2}
(,2]
11.
I.2.
1.  5  x  3  5
{x | 1  x  6}
[1,6)
7.
{x | x  3 4 or x  1 4}
(,1 4]  [3 4 , )
9.
{x | 0  x  6}
(0,6)
3. 4x  1  2 or 4x  1  2
{x | 2  x  8}
5.
(2,8)
11.
I.3.
{x | x  3}
1.
(3,  )
7.
{x | 2  x  0}
(2,0)
{x | x  2 or x  4}
(,2)  (4, )
3.
{x | x  4}
(,4]
9.
5.
{x | 2  x  4}
[2,4]
11.
I.4.
{x | 3  x  1}
1.
(3,1)
3.
5.
7.
{x | x  3 or x  1}
(,3)  (1, )
{x | 2  x  1 or
(,3)  (1, )
x  4}
{x | x  2 or x  0}
(,0]  [2, )
{x |  2  x  1 or
(2,1)  (3, )
{x | x  2}
(,2)
9. no solutions
11.
{x | x  1}
(,1)  (1, )
x  3}