1.7 Linear Inequalities
Inequalities: Definitions and Properties
Imagine a number line with two points on it labeled a and b. Then there are three possibilities,
1. a is to the right of b, denoted a > b
2. a is to the left of b, denoted a < b
3. a is on top of b, denoted a = b.
We are going to look at the properties of inequalities so that we can solve them.
Properties: Let a, b, c and d be real numbers.
1. If a < b, then b > a.
2. If a < b and c < d, then a + c < b + d.
3. If a < b, then a + c < b + c and a – c < b – c.
𝑎
𝑏
𝑎
𝑏
4. If a < b and c > 0, then ac < bc and 𝑐 < 𝑐 .
5. If a < b and c < 0, then ac > bc and 𝑐 > 𝑐 .
6. If a < b and b < c, then a < c.
1
1
7. If 0 < a < b, then 𝑎 > 𝑏.
8. If 0 < a < b and n > 0, then 𝑎𝑛 < 𝑏 𝑛 .
We will use these properties to help us solve linear inequalities. A linear inequality is equivalent to
ax+b<0 for some a and b. The < could also be a >, ≤, 𝑜𝑟 ≥. A solution to the inequality is all the numbers
that make the statement true. Usually solutions to inequalities contain infinitely many numbers. Solving
linear inequalities is similar to linear equations with a few differences. Let’s work a few examples.
Examples:
1. 8𝑥 − 9 ≥ 12𝑥 + 3
2. −5(𝑥 + 2) + 3 < 11
4
3
3. 5 (𝑥 + 3) > 4 𝑥 + 2
4. 3(𝑥 + 6) − 9 ≤ 13 − (−7 − 3𝑥)
5. 2(3 − 4𝑥) + 2 ≥ 5(𝑥 + 7) − 10
7
5
6. − (𝑥 + 1) − 15 ≥ 𝑥 + 7
3
3
Solutions
1. As with linear equations, we want to get the x’s on one side. Mainly we use property 3 to do
that. We can subtract 12x from both sides and add 9 to both sides to get
8𝑥 − 9 ≥ 12𝑥 + 3
−4𝑥 ≥ 12
We can divide both sides by -4, but according to rule 5, we need to switch the inequality symbol.
This is one of the differences between solving inequalities and equations.
𝑥 ≤ −3
So 𝑥 ≤ −3 is our solution, we will learn another form to write the solutions after these
examples.
Notice if we subtracted 8x from both sides instead of 12x, the x-coefficient would have been
positive and we could have avoided needing rule 5.
8𝑥 − 9 ≥ 12𝑥 + 3
−12 ≥ 4𝑥
−3 ≥ 𝑥
We got the same solution before just turned around, but we avoided using rule 5.
2. We need to distribute to collect like terms before we can get the x by itself.
−5(𝑥 + 2) + 3 < 11
−5𝑥 − 10 + 3 < 11
−5𝑥 − 7 < 11
−5𝑥 < 18
18
𝑥>− .
5
3. We have fractions in this inequality. Usually it is wise to get rid of them. We can use the LCD of
20 to do so.
4
3
(𝑥 + 3) > 𝑥 + 2
5
4
4
3
20 ⋅ ( (𝑥 + 3)) > ( 𝑥 + 2) ⋅ 20
5
4
3
16(𝑥 + 3) > ( 𝑥) ⋅ 20 + 2 ⋅ 20
4
16𝑥 + 48 > 15𝑥 + 40
𝑥 > −8
4. We need to distribute first. Also moving the x’s to the right side avoids needed to use rule 5.
2(3 − 4𝑥) + 2 ≥ 5(𝑥 + 7) − 10
6 − 8𝑥 + 2 ≥ 5𝑥 + 35 − 10
−8𝑥 + 8 ≥ 5𝑥 + 25
−17 ≥ 13𝑥
17
−
≥𝑥
13
5. We need to distribute a to get the x’s on one side.
3(𝑥 + 6) − 9 ≤ 13 − (−7 − 3𝑥)
3𝑥 + 18 − 9 ≤ 13 + 7 + 3𝑥
3𝑥 + 9 ≤ 20 + 3𝑥
Notice we have the same term on both sides, 3x. If we subtract 3x from both sides, we get
9 ≤ 20
This statement is always true no matter what x is, so the solution to this inequality is all real
numbers.
Note: if the inequality had been the other way around, 9 ≥ 20, this would always be a false
statement. Meaning the inequality would have no solution.
6. We should clear the fractions first by multiplying by 3.
7
5
− (𝑥 + 1) − 15 ≥ 𝑥 + 7
3
3
7
5
(− (𝑥 + 1) − 15) ⋅ 3 ≥ ( 𝑥 + 7) ⋅ 3
3
3
−7(𝑥 + 1) − 45 ≥ 5𝑥 + 21
−7𝑥 − 7 − 45 ≥ 5𝑥 + 21
−7𝑥 − 52 ≥ 5𝑥 + 21
−73 ≥ 12𝑥
73
−
≥𝑥
12
Graphical Representation of Inequalities
We can represent the solutions to inequalities as intervals on a number line. Here are several intervals,
their number line representation and their inequality equivalent. With interval notation, we usually
want all the numbers in between two other numbers and potentially including those two numbers. For
example, [-3,2), means all numbers in between -3 and 2 including the 3 but not the 2. The “[“ or “]”
symbols means to include the number next to the symbol. The “(“ or “)” means to not include the
number. Sometimes instead of two numbers we want one number and infinity. We always use
parenthesis with infinity, because in a sense including it would mean we could reach infinity. Here are
some visual representations.
Here are the solutions to the previous examples in interval notation.
1. We want x to the left of -3 including -3, (−∞, −3].
2. We want x larger than −
18
5
but not including −
18
,
5
(−
18
, ∞).
5
3. We want x larger than -8 but not -8, (−8, ∞).
17
17
17
4. We want x smaller than − 13 including − 13, (−∞, − 13).
5. We want all real numbers, (−∞, ∞).
73
73
73
6. We want x smaller than − 12 including − 12, (−∞, − 12).
Compound Inequalities
Sometimes one inequality is not enough for a given situation. If there are more than one inequality, we
call them a compound inequality. There are two main ways to combine inequalities, with “and” or with
“or.” Combining with “and” means we want both inequalities to be satisfied at the same time. “Or”
means we only need one of the inequalities to be satisfied. To solve compound inequalities, we solve
them individually and try to combine the results. To see how this works, let’s do a few examples.
Examples: Solve the compound inequalities.
𝑥
1. 2𝑥 − 5 ≥ −3 𝑎𝑛𝑑 3 − 7 < 3
2.
3.
𝑦
+ 3 < −1 𝑜𝑟 3 + 3𝑦
4
3
𝑥 + 2 ≥ −7 𝑎𝑛𝑑 − 𝑥
2
≥9
+2≥3
4. 4𝑥 + 2 ≥ 18 𝑎𝑛𝑑 2𝑥 + 3 ≤ 9
4
5. 3 𝑥 + 2 ≥ 3 𝑜𝑟 6 − 2𝑥 ≥ 4
6.
2
𝑥
3
− 4 ≥ 2 𝑜𝑟 3𝑥 − 10 ≤ 20
Solutions
1. We solve each inequality separately and we will worry about the “and” later.
First inequality
2𝑥 − 5 ≥ −3
2𝑥 ≥ 2
𝑥≥1
Second inequality
𝑥
−7<3
3
𝑥
< 10
3
𝑥 < 30
The compound inequality had an “and,” which means we want x to be bigger than or equal to 1
AND x to be less than 30. Interval notation can nicely describe this as [1,30).
2. Solving each inequality separately.
𝑦
+ 3 < −1
4
𝑦
< −4
4
𝑦 < −16
Second inequality.
3 + 3𝑦 ≥ 9
3𝑦 ≥ 6
𝑦≥2
The compound inequality has an “or,” so we want y to be less than -16 OR y to be bigger than or
equal to 2. In interval notation this is (−∞, −16) ∪ [2, ∞).
3. Solving each inequality.
3
𝑥 + 2 ≥ −7 𝑎𝑛𝑑 − 𝑥 + 2 ≥ 3
2
3
𝑥 ≥ −9 𝑎𝑛𝑑 − 𝑥 ≥ 1
2
𝑥 ≥ −6 𝑎𝑛𝑑 𝑥 ≤ −1
In interval notation, we get [−6, −1].
4. Solving each inequality.
4𝑥 + 2 ≥ 18 𝑎𝑛𝑑 2𝑥 + 3 ≤ 9
4𝑥 ≥ 16 𝑎𝑛𝑑 2𝑥 ≤ 6
𝑥 ≥ 4 𝑎𝑛𝑑 𝑥 ≤ 3
Notice it says we want x to bigger than 4 and less than 3 at the same time. This is not possible,
so the answer is no solution.
5. Solving each inequality.
4
𝑥 + 2 ≥ 3 𝑜𝑟 6 − 2𝑥 ≥ 4
3
4
𝑥 ≥ 1 𝑜𝑟 − 2𝑥 ≥ −2
3
3
𝑥 ≥ 𝑜𝑟 𝑥 ≤ 1
4
3
In interval notation, this is (−∞, 1] ∪ [ , ∞). But notice that these intervals overlap, so the
4
simplified solution is all real numbers, or (−∞, ∞).
6. Solving each inequality.
2
𝑥 − 4 ≥ 2 𝑜𝑟 3𝑥 − 10 ≤ 20
3
2
𝑥 ≥ 6 𝑜𝑟 3𝑥 ≤ 30
3
𝑥 ≥ 9 𝑜𝑟 𝑥 ≤ 10
Notice we want x bigger than 9 OR x less than 10, this would be everything or (−∞, ∞ ).
Some compound inequalities can be condensed down into a single three-part inequality. For example,
the compound inequality of 6 < 2𝑥 + 5 𝑎𝑛𝑑 2𝑥 + 5 < 20 can be condensed to 6 < 2𝑥 + 5 < 20. Only
certain compound inequalities can be condensed like this as the inequalities need to share an expression
and the inequalities need to be in the right direction (you can’t have something like 4 > 3x < 5). The
advantage to the three-part inequality is that it automatically combines both answers at the end and
both answers can be found simultaneously.
Let’s solve the compound inequality. We can perform the same operation to all three parts at once. Our
goal is to get the middle expression to be an x by itself. We start by subtracting 5 from each part.
6 < 2𝑥 + 5 < 20
1 < 2𝑥 < 15
To get the middle part to be just x, we need to divide by 2.
1
15
<𝑥<
2
2
. 5 < 𝑥 < 7.5
Notice we can immediately write the solution in interval notation, (. 5, 7.5).
Examples
1. 5 < 4𝑥 − 11 < 13
2. 1 ≤
3. −3
2𝑥−5
<5
3
−4𝑥+1
< −5 ≤
2
Solutions
1. Solving
5 < 4𝑥 − 11 < 13
16 < 4𝑥 < 24
4<𝑥<6
(4, 6)
2. Solving by clearing fraction first.
2𝑥 − 5
<5
3
3 ≤ 2𝑥 − 5 < 15
8 ≤ 2𝑥 < 20
4 ≤ 𝑥 < 10
[4, 10)
3. Clearing fractions and remembering to flip symbols.
−4𝑥 + 1
−3 <
≤2
−5
15 > −4𝑥 + 1 ≥ −10
14 > −4𝑥 ≥ −11
−7.5 < 𝑥 ≤ 2.75
(−7.5, 2.75]
1≤
Absolute Value Inequalities
What would the solution to |x|<5 look like? Recall, we can think of the absolute value of a number as
the distance between a number and zero. We can interpret the inequality |𝑥| < 5 to mean the distance
between x and zero is less than 5. So from zero, x can go less than 5 units to the right and less than 5
units to the left.
We have that x could be anything in the shaded region from (-5, 5). What about |𝑥| > 5? This would
mean that x was more than 5 away from zero on the right and on the left. It would look like this:
We have that x could be anything in the shaded region. The solution to |𝑥| < 5 looks like the solutions
we found for compound AND inequalities and the solution to |𝑥| > 5 looks like solutions for compound
OR inequalities.
|𝑠𝑜𝑚𝑒𝑡ℎ𝑖𝑛𝑔| < 𝑘 means − 𝑘 < 𝑠𝑜𝑚𝑒𝑡ℎ𝑖𝑛𝑔 < 𝑘
|𝑠𝑜𝑚𝑒𝑡ℎ𝑖𝑛𝑔| > 𝑘 means 𝑠𝑜𝑚𝑒𝑡ℎ𝑖𝑛𝑔 > k OR 𝑠𝑜𝑚𝑒𝑡ℎ𝑖𝑛𝑔 < −k
To solve absolute value inequalities, we convert them to compound inequalities and the solve. But
before we convert, we may need to get the absolute value by itself first.
Examples
1.
2.
3.
4.
5.
6.
7.
2|5 − 2𝑥| − 3 < 7
−4 ≥ −2|5𝑥 + 7|
|2𝑥 + 12| < −3
3|4 − 6𝑧| + 2 < 14
−18 ≥ −3|2𝑥 + 12|
3|2𝑥 − 7| − 1 > 26
|7𝑥 − 17| > −5
Solutions
1. Before we can use the box, we need to get the absolution value by itself. We can add 3 to both
sides and divide by 2.
2|5 − 2𝑥| − 3 < 7
2|5 − 2𝑥| < 10
|5 − 2𝑥| < 5
Now that we have the absolute value by itself we can use the conclusions from the box. We
have the first case.
−5 < 5 − 2𝑥 < 5
We have a three-part inequality to solve.
−10 < −2𝑥 < 0
5>𝑥>0
The solution is (0,5).
2. Again, isolate the absolute value.
−4 ≥ −2|5𝑥 + 7|
2 ≤ |5𝑥 + 7|
|5𝑥 + 7| ≥ 2
The third step isn’t technically needed but it does put it in the same form as the box. Using the
conversion from the box we get:
5𝑥 + 7 ≥ 2 𝑂𝑅 5𝑥 + 7 ≤ −2
5𝑥 ≥ −5 𝑂𝑅 5𝑥 ≤ −9
9
𝑥 ≥ 1 𝑂𝑅 𝑥 ≤ −
5
9
(1,
The solution is (−∞, − 5) ∪
∞).
3. |2𝑥 + 12| < −3. Notice we are trying to set an absolute value less than a negative number.
Absolute values have to be positive or zero. This is impossible, so we have no solution.
4. Isolate absolute value.
3|4 − 6𝑧| + 2 < 14
3|4 − 6𝑧| < 12
|4 − 6𝑧| < 4
This is the first case from the box.
4
3
−4 < 4 − 6𝑧 < 4
−8 < −6𝑧 < 0
4
>𝑧>0
3
The solution is (0, ).
5. Isolate.
−18 ≥ −3|2𝑥 + 12|
6 ≤ |2𝑥 + 12|
|2𝑥 + 12| ≥ 6
This is the second case from the box.
2𝑥 + 12 ≥ 6 𝑂𝑅 2𝑥 + 12 ≤ −6
2𝑥 ≥ −6 𝑂𝑅 2𝑥 ≤ −18
𝑥 ≥ −3 𝑂𝑅 𝑥 ≤ −9
The solution is (−∞, −9) ∪ (−3, ∞).
6. Isolate.
3|2𝑥 − 7| − 1 > 26
3|2𝑥 − 7| > 27
|2𝑥 − 7| > 9
Second case.
2𝑥 − 7 > 9 𝑂𝑅 2𝑥 − 7 < −9
2𝑥 > 16 𝑂𝑅 2𝑥 < −2
𝑥 > 8 𝑂𝑅 𝑥 < −1
(−∞,
(8,
The solution is
−1) ∪
∞).
7. |7𝑥 − 17| > −5. Notice we want an absolute value to be bigger than a negative number. By
definition, this has to be true no matter what the x is. The solution is all real numbers.