2 Solving linear inequalities algebraically

When we are asked to solve an inequality , the inequality will contain an unknown variable, say x . Solving means obtaining all values of x for which the inequality is true. In a linear inequality the unknown appears only to the first power, that is as x , and not as x 2 , x 3 , x 1 ∕ 2 and so on.

Consider the following examples.

Example 35

Solve the inequality 4 x + 3 > 0 .

Solution

4 x + 3 > 0 4 x > − 3 ,  by subtracting 3 from both sides x > − 3 4  by dividing both sides by 4.

Hence all values of x greater than − 3 4 satisfy 4 x + 3 > 0 .

Example 36

Solve the inequality − 3 x − 7 ≤ 0 .

Solution

− 3 x − 7 ≤ 0 − 3 x ≤ 7  by adding 7 to both sides x ≥ − 7 3  dividing both sides by  − 3  and reversing the inequality

Hence all values of x greater than or equal to − 7 3 satisfy − 3 x − 7 ≤ 0 .

Task!

Solve the inequality 17 x + 2 < 4 x + 1 .

This is done by making x the subject and obtain it on its own on the left-hand side.

Start by subtracting 4 x from both sides to remove quantities involving x from the right:

13 x + 2 < 1

Now subtract 2 from both sides to remove the 2 on the left:

13 x < − 1 . Finally, the range of values of x are x < − 1 ∕ 13

Example 37

Solve the inequality 5 x − 2 < 4 and depict the solution graphically.

Solution

5 x − 2 < 4 is equivalent to − 4 < 5 x − 2 < 4

We treat each part of the inequality separately:

− 4 < 5 x − 2 − 2 < 5 x  by adding 2 to both sides − 2 5 < x  by dividing both sides by 5

So x > − 2 5 . Now consider the second part: 5 x − 2 < 4 .

5 x − 2 < 4 5 x < 6  by adding 2 to both sides x < 6 5  by dividing both sides by 5

So x < 6 5 .Putting both parts of the solution together we see that the inequality is satisfied when

− 2 5 < x < 6 5 . This range of values is shown in Figure 11.

Figure 11 :

{ $|5x-2|$ less than $4$ which is equivalent to $2/5$ less than $x$ less than $6/5$}

Task!

Solve the inequality 1 − 2 x < 5 .

First of all rewrite the inequality without using the modulus sign:

− 5 < 1 − 2 x < 5

Then treat each part separately. First of all consider − 5 < 1 − 2 x . Solve this:

x < 3

The second part is 1 − 2 x < 5 . Solve this.

x > − 2

Finally, give the solution as one statement:

− 2 < x < 3 .

Exercises

In the following questions solve the given inequality algebraically.

1.   4 x > 8 2.   5 x > 8 3.   8 x > 5 4.   8 x ≤ 5
5.   2 x > 1 6.   3 x < − 1 7.   5 x > 2 8.   2 x > 0
9.   8 x < 0 10.   3 x ≥ 0 11.   3 x > 4 12.   3 4 x > 1
13.   4 x ≤ − 3 14.   3 x ≤ − 4 15.   5 x ≥ 0 16. 4 x ≤ 0
17.   5 x + 1 < 8 18.   5 x + 1 ≤ 8 19.   7 x + 3 ≥ 0
20.   18 x + 2 > 9 21.   14 x + 11 > 22 22. 1 − 5 x ≤ 0
23.   2 + 5 x ≥ 1 24.   11 − 7 x < 2 25. 5 + 4 x > 2 x + 1
26.   7 x − 3 > 1 27.   2 x + 1 ≥ 3 28.   5 x < 1
29.   5 x ≤ 0 30.   1 − 5 x > 2 31.   2 − 5 x ≥ 3
1.   x > 2 2.   x > 8 ∕ 5 3.   x > 5 ∕ 8 4.   x ≤ 5 ∕ 8
5.   x > 1 ∕ 2 6.   x < − 1 ∕ 3 7.   x > 2 ∕ 5 8.   x > 0
9.   x < 0 10.   x ≥ 0 11.   x > 4 ∕ 3 12.   x > 4 ∕ 3
13.   x ≤ − 3 ∕ 4 14.   x ≤ − 4 ∕ 3 15.   x ≥ 0 16.   x ≤ 0
17.   x < 7 ∕ 5 18.   x ≤ 7 ∕ 5 19.   x ≥ − 3 ∕ 7 20. x > 7 ∕ 18
21.   x > 11 ∕ 14 22.   x ≥ 1 ∕ 5 23.   x ≥ − 1 ∕ 5 24.   x > 9 ∕ 7
25. x > − 2 26.   x > 4 ∕ 7 or x < 2 ∕ 7 27.   x ≥ 1 or x ≤ − 2 28.   − 1 ∕ 5 < x < 1 ∕ 5
29.   x = 0 30.   x < − 1 ∕ 5 , x > 3 ∕ 5 31.   x ≤ − 1 ∕ 5 , x ≥ 1