Solving linear inequalities · Inequalities

An inequality says one side is bigger or smaller than the other, instead of equal to it. Its answer isn’t one number. It is a whole range of numbers.

Sign	Means	In words
$x > 2$	greater than 2	“more than 2”
$x \ge 2$	greater than or equal to 2	“at least 2”, “not less than 2”
$x < 2$	less than 2	“less than 2”
$x \le 2$	less than or equal to 2	“at most 2”, “not more than 2”

On a number line, an open circle means the end number is left out, and a filled circle means it’s included.

x > 2Open circle: 2 itself is left out

x ≤ 3Filled circle: 3 is included

−1 ≤ x < 4Between −1 and 4: −1 in, 4 out

Solving: like an equation, with one rule

Solve an inequality the same way as an equation: expand brackets, then collect the $x$ terms on one side and the numbers on the other. You can add or subtract anything on both sides. You can multiply or divide both sides by a positive number. There is one extra rule.

Try it

Doing the same to both sidesAdd or multiply, then change the number

2 < 5before−4 > −10the sign has turned round

multiply by = −2

Multiplying by a negative number reflects both numbers across 0. Now 5 × (−2) = −10 is to the left of 2 × (−2) = −4, so the sign must turn round: −4 > −10. The same happens when dividing by a negative number.

Why? $2 < 5$ , but $-2 > -5$ . Multiplying by $-1$ flips the numbers to the other side of 0, so their order swaps, and the sign must turn round too. Adding or subtracting only slides the numbers along, so it never changes the sign.

More: solving linear inequalities

Clearing fractions

Multiply every term on both sides by the LCM of the denominators (the smallest number they all divide into). The LCM is positive, so the sign stays the same.

Worked example · WAEC 2019

WAEC 2019 · Paper 2 · Q1 (a)

Solve the inequality $\dfrac{1 + 4x}{2} - \dfrac{5 + 2x}{7} < x - 2$ .

Clear the fractions
Multiply every term by 14:
$\begin{aligned} &7(1 + 4x) - 2(5 + 2x) \\ &< 14x - 28 \end{aligned}$
Think first. What is the LCM of 2 and 7? Multiply every term by it.
Expand
- Expand: ${7 + 28x - 10 - 4x < 14x - 28}$ .
- Collect: ${24x - 3 < 14x - 28}$ .
Think first. Watch the sign in front of the 2.
Collect and solve
- Take $14x$ from both sides and add 3: ${10x < -25}$ .
- Divide by 10: ${x < -2\frac12}$ .

More: clearing fractions

Double inequalities

An inequality with three parts, such as $3x - 2 < 10 + x < 2 + 5x$ , is really two inequalities. Split it, solve each, and put the answers together.

Worked example · WAEC 2018

WAEC 2018 · Paper 2 · Q8 (b)

Find the range of values of $x$ which satisfies the inequality $3x - 2 < 10 + x < 2 + 5x$ .

Split it
$3x - 2 < 10 + x$ and $10 + x < 2 + 5x$ .
Think first. What are the two inequalities?
Solve each
- First: ${2x < 12}$ , so ${x < 6}$ .
- Second: ${8 < 4x}$ , so ${x > 2}$ .
Put them together
$x$ is more than 2 and less than 6: $2 < x < 6$ .
Think first. Which values satisfy both?

When only the middle part has $x$ , as in $-3 < 2x + 1 \le 7$ , you can do the same to all three parts at once. Here, take 1 from all three parts, then divide all three by 2.

Two inequalities together

Two inequalities togetherChange the signs and the numbers

−1 ≤ x < 4both together{−1, 0, 1, 2, 3}whole numbers in the range

first number = −1second number = 4

Top line: x ≥ −1. Middle line: x < 4. The bottom line is where they overlap: −1 ≤ x < 4. An open circle means the end number is left out (< or >); a filled circle means it's included (≤ or ≥). The whole numbers in the range are −1, 0, 1, 2, 3.

The values that satisfy both inequalities are where the two ranges overlap. If they don’t overlap, no value works.

More: double inequalities and whole numbers

Harder cases: x underneath, and distances

When $x$ is in a denominator, you can only multiply through by it if you know its sign. Take $\frac1x > 2$ :

the left side is more than 2, so it is positive, and $x$ must be positive too;
multiplying by $x$ (positive) gives $1 > 2x$ ;
so the answer is $0 < x < \frac12$ .

If the sign isn’t clear, bring everything to one side as a single fraction. Then ask when that fraction is positive or negative.

A modulus is a distance. $|x - a|$ is the distance from $x$ to $a$ on the number line. So $|x - a| < b$ means $x$ is within $b$ of $a$ :

|x - a| < b \quad\Longleftrightarrow\quad a - b < x < a + b

|x − 1| < 4Within 4 of 1: −3 < x < 5

More: x underneath, and modulus

Your turn

WAEC 2015 · Paper 2 · Q2 (a)

(a)
Solve the inequality $4 + \frac34(x + 2) \le \frac38x + 1$ .
Show the answer
$x \le -12$

Worked solution (try it first)

(a)

The denominators are 4 and 8, so multiply every term by 8: $32 + 6(x + 2) \le 3x + 8$ .
Expand: $32 + 6x + 12 \le 3x + 8$ , so $6x + 44 \le 3x + 8$ .
Collect: $3x \le -36$ , so $x \le -12$ .

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