This chapter explores linear inequalities. Before attempting this chapter you must have prior knowledge of solving linear equations. The chapter covers solving linear inequalities, displaying the solutions on a number line, solving double-ended inequalities, and look at the intersection of two solutions.
Equations and inequalities connection
Example:
Answer:
24 – x = 3(2x + 1)
24 – x = 6x + 3
24 + 3 = 6x + x
21 = 7x
x = 3
Explanation:
You must already know how to solve linear equations for example below is an equation to solve naturally.
24 – x = 3(2x + 1)
First we expand the equation as shown below.
24 – x = 6x + 3
Arrange x terms on one side as shown below.
24 + 3 = 6x + x
Now we can simplify as shown below
21 = 7x
We divide both sides by 7 and the result becomes.
x = 3
Example:
Answer:
24 – x ≤ 3(2x + 1)
24 – x ≤ 6x + 3
24 – 3 ≤ 6x + x
21 ≤ 7x
Explanation:
We can use the same principle used above to solve the following inequality.
24 – x ≤ 3(2x + 1)
First we expand as shown below.
24 – x ≤ 6x + 3
Rearrange
24 – 3 ≤ 6x + x
…and simplify…
21 ≤ 7x
…then divide both sides by 7 to simplify again. We get that;
We can also indicate this on a number line to get a better visual as shown below.
Negative coefficient of x
Example: You may have a negative number of xs. For example below is an inequality that we could solve.
16 – 4x ≤ 22
Answer:
16 < 22 + 4x
16 – 22 ≤ 4x
-6 ≤ 4x
-1.5 ≤ x
x ≥ -1.5
x ≥ -1.
Explanation: There are two ways we could solve the equality; we could move the x term to the right to make it positive as shown below.
16 < 22 + 4x
And then move the integer to the other side as shown below.
16 – 22 ≤ 4x
The result becomes;
-6 ≤ 4x
We divide both sides by 4 and the result becomes;
-1.5 ≤ x
or
x ≥ -1.5
The other alternative is not moving the term with x and move 16 instead.
We divide both sides by -4 and the result becomes;
x ≥ -1.
The solution can also be shown on a number line as shown below.

You must have noticed that when you divide or multiply by a negative, the inequality changes direction.
Double-ended inequalities
Example:Below is what a double-ended inequality looks like.
6 – x ≤ 2x + 3 < 15
Answer:
6 – x ≤ 2x + 3 2x + 3 < 15
6 – x ≤ 2x + 3 2x < 15 – 3
3 ≤ 3x 2x < 12
1 ≤ x x < 6
1 ≤ x < 6
Explanation: There are two parts to the inequality so we split it into two parts as shown below.
6 – x ≤ 2x + 3 2x + 3 < 15
Now we can solve each part separately as have been done below.
6 – x ≤ 2x + 3 2x < 15 – 3
3 ≤ 3x 2x < 12
1 ≤ x x < 6
Now we can put the solution back together to result.
1 ≤ x < 6
The solution has been shown on the number line below.
Solving inequalities in a pair
Example:
You could also work out where two solutions overlap. For example suppose we had to find the values which satisfy both inequalities shown below.
3x + 2 > x – 6 and 2x > 5x – 9
Answer:
3x + 2 > x – 6
3x – x > -6 – 2
2x > -8
x > -4
2x > 5x – 9
2x – 5x > – 9
-3x > -9
x > 3
x > -4 and x < 3
Explanation:
First we solve the first inequalities.
3x + 2 > x – 6
3x – x > -6 – 2
2x > -8
x > -4
Then we solve the second inequality
2x > 5x – 9
2x – 5x > – 9
-3x > -9
x > 3
We have;
x > -4 and x < 3
We can look at both solutions on a number line.

The solutions as we can see above overlap between -4 and 3.
Exam question
Example:ABC is a triangle with perimeter greater than 28cm and less than 36cm. The length of AB is an integer. AC is 2cm longer than AB. BC is 5cm shorter than AB.
Answer:
Perimeter = x + (x + 2) + (x – 5)
= 3x – 3
28 < 3x – 3 < 36
28 < 3x – 3 3x – 3 < 36
31 < 3x 3x < 39
101/3 < x < 13
101/3 < x < 13
x = 11, 12
Explanation:Find all possible values for the length of AB.
First we find an expression for the perimeter.
Perimeter = x + (x + 2) + (x – 5)
= 3x – 3
The perimeter is greater than 28cm and less than 36cm.
28 < 3x – 3 < 36
The inequality contains two parts so we solve each part separately.
28 < 3x – 3 3x – 3 < 36
31 < 3x 3x < 39
101/3 < x < 13
The range of values x can take are;
101/3 < x < 13
We know that AB is an integer so we look for whole numbers in the range that is;
x = 11, 12
The length of AB is either 11cm or 12cm.
13 has not been included in the final answer.