Linear inequalities and absolute values – Linear and Quadratic Equations and Inequalities (Senior 4 Mathematics) | Ganzaa

Linear and Quadratic Equations and Inequalities

Lessons

Senior 4 · Mathematics

Linear and Quadratic Equations and Inequalities

5 lessons

Lesson 1 of 5

Linear inequalities and absolute values

25 min

How big can the error be?

A technician in a Kigali workshop measures the diameter of a bolt with a vernier caliper and reads $12.4\,\mathrm{mm}$ . The caliper is good to $0.1\,\mathrm{mm}$ , so the true diameter $d$ is not known exactly: all we can say is that $d$ is within $0.1\,\mathrm{mm}$ of $12.4\,\mathrm{mm}$ . In symbols,

|d - 12.4| \le 0.1 \quad\Longleftrightarrow\quad 12.3 \le d \le 12.5 .

Statements like this, a quantity that may lie anywhere in a range, are inequalities, and the absolute value is the natural way to write "within so much of". This lesson revises linear equations and inequalities, solves inequalities whose expressions are fractions of linear factors, and then solves equations and inequalities that contain absolute values.

Linear equations and inequalities in one unknown

A linear equation in $x$ can always be rearranged to $a x = b$ with real numbers $a$ and $b$ . Three cases can happen:

Case	Solutions of $a x = b$	Example
$a \ne 0$	exactly one: $x = \frac{b}{a}$	$3x = -6 \Rightarrow x = -2$
$a = 0,\ b = 0$	every real number	$2 (x + 1) - 2 x = 2$ becomes $0 x = 0$
$a = 0,\ b \ne 0$	none (the equation is impossible)	$4 x - 4 (x - 1) = 7$ becomes $0 x = 3$

A linear inequality such as $5 - 2x \ge 3x + 20$ is solved like an equation, with one extra rule: multiplying or dividing both sides by a negative number reverses the sign. Here $-5x \ge 15$ , and dividing by $- 5$ gives $x \le -3$ , the interval $(-\infty, -3]$ . (REB's books write open ends with reversed brackets, $]{-}\infty, -3]$ ; both notations mean the same set.)

Definition

Solution set of an inequality in $x$ : the set of all real numbers that make it true, usually written as an interval or a union of intervals.

Example: the solution set of $2x - 1 \lt 5$ is $\{x \in \mathbb{R} : x \lt 3\} = (-\infty, 3)$ .

Products and quotients of linear factors

To solve $\frac{2x + 1}{x - 3} \le 1$ you may not multiply both sides by $x - 3$ : you do not know whether it is positive or negative. Instead, bring everything to one side, write it as one fraction, and study the sign of each linear factor in a sign table. A linear factor $a x + b$ (with $a \ne 0$ ) is zero at $x = -\frac{b}{a}$ , has the sign of $a$ to the right of that root, and the opposite sign to the left.

Worked example 1: a quotient inequality

Solve $\frac{2x + 1}{x - 3} \le 1$ .

Solution

The fraction is undefined at $x = 3$ . Subtract 1 and use a common denominator:

\frac{2x + 1}{x - 3} - \frac{x - 3}{x - 3} \le 0 \quad\Longleftrightarrow\quad \frac{x + 4}{x - 3} \le 0 .

The critical values are $x = - 4$ (numerator zero) and $x = 3$ (denominator zero).

Sign table for (x + 4) ÷ (x − 3); ‖ marks the value where the fraction is undefined.
$x$		$- 4$		$3$
$x + 4$	−	0	+		+
$x - 3$	−		−	0	+
$\frac{x + 4}{x - 3}$	+	0	−	‖	+

The fraction is negative between the roots and zero at $x = - 4$ ; $x = 3$ is excluded. So the solution set is $[- 4, 3)$ .

Check with $x = 0$ : $\frac{1}{-3} \le 1$ ✓. With $x = 4$ : $\frac{9}{1} = 9 \le 1$ is false ✓ (4 is outside the set).

Warning

Multiplying $\frac{2x + 1}{x - 3} \le 1$ by $x - 3$ gives $2x + 1 \le x - 3$ , so $x \le -4$ , which is wrong: it keeps $x = - 10$ , where the fraction equals $\frac{19}{13} \gt 1$ . Multiply by an expression only when you know its sign.

Definition

Absolute value of a real number $x$ :

|x| = \begin{cases} x & \text{if } x \ge 0 \\ -x & \text{if } x \lt 0 \end{cases}

It is the distance from $x$ to 0 on the number line, so $∣ x - a ∣$ is the distance between $x$ and $a$ .

Example: $∣ - 7 ∣ = 7$ , and $∣ x - 2 ∣ = 5$ means " $x$ is 5 units from 2", so $x = 7$ or $x = - 3$ .

Important

Rules for absolute values (from the distance meaning). Let $X$ be any expression and $c \gt 0$ :

$|X| = c \iff X = c \ \text{or}\ X = -c$
$|X| \lt c \iff -c \lt X \lt c$ (within $c$ of 0)
$|X| \gt c \iff X \lt -c \ \text{or}\ X \gt c$ (more than $c$ from 0)
$|X| = |Y| \iff X = Y \ \text{or}\ X = -Y$

If $c \lt 0$ : $|X| \lt c$ has no solution and $|X| \gt c$ holds for every $x$ where $X$ is defined, because $|X| \ge 0$ always.

Worked example 2: "within" and "beyond"

Solve (a) $|2x - 3| \le 5$ (b) $|3 - x| \gt 2$ .

Solution

$-5 \le 2x - 3 \le 5$ . Add 3: $-2 \le 2x \le 8$ . Divide by 2: $-1 \le x \le 4$ . Solution set $[- 1, 4]$ .
$3 - x \lt -2$ or $3 - x \gt 2$ , that is $x \gt 5$ or $x \lt 1$ . Solution set $(-\infty, 1) \cup (5, +\infty)$ .

A closed circle means the endpoint is included; an open circle means it is not. Part (a) is one interval; part (b) is two rays, because it describes the points far from 3.

Worked example 3: two absolute values equal

Solve $∣ x - 1 ∣ = ∣ 2 x + 4 ∣$ .

Solution

Either $x - 1 = 2 x + 4$ , giving $x = - 5$ , or $x - 1 = - (2 x + 4)$ , giving $3 x = - 3$ , so $x = - 1$ .

Check: $x = - 5$ : $∣ - 6 ∣ = ∣ - 6 ∣$ ✓. $x = - 1$ : $∣ - 2 ∣ = ∣ 2 ∣$ ✓. Solution set ${- 5, - 1}$ .

Worked example 4: a sum of absolute values

Solve $|x + 2| + |x - 3| \le 7$ .

Solution

The expressions inside the bars change sign at $x = - 2$ and $x = 3$ . On each interval, remove the bars with the right sign:

Interval	$∣ x + 2 ∣ + ∣ x - 3 ∣$	Inequality	Solutions in the interval
$x \lt -2$	$- (x + 2) - (x - 3) = - 2 x + 1$	$-2x + 1 \le 7 \Rightarrow x \ge -3$	$[- 3, - 2)$
$-2 \le x \le 3$	$(x + 2) - (x - 3) = 5$	$5 \le 7$ , always true	$[- 2, 3]$
$x \gt 3$	$(x + 2) + (x - 3) = 2 x - 1$	$2x - 1 \le 7 \Rightarrow x \le 4$	$(3, 4]$

The union is $[- 3, 4]$ . Geometrically, $∣ x + 2 ∣ + ∣ x - 3 ∣$ is the sum of the distances from $x$ to $- 2$ and to $3$ . Between them the sum is always the gap, 5; outside, it grows by 2 for every unit you move away, so it reaches 7 one unit beyond each end.

The graph lies on or below the line $y = 7$ exactly for $-3 \le x \le 4$ (the green segment on the x-axis), the solution of Worked example 4.

Worked example 5: an unknown on both sides

Solve $|2x - 1| \lt x + 1$ .

Solution

The rule $|X| \lt Y \iff -Y \lt X \lt Y$ holds for any expression $Y$ (if $Y \le 0$ neither side can be true). So

-(x + 1) \lt 2x - 1 \lt x + 1 .

The left part gives $0 \lt 3x$ , so $x \gt 0$ ; the right part gives $x \lt 2$ . Solution set $(0, 2)$ .

Check: $x = 1$ : $|1| = 1 \lt 2$ ✓. $x = 3$ : $|5| = 5 \lt 4$ is false ✓.

Warning

Never write $|X| \gt c$ as " $-c \gt X \gt c$ ": no number is both less than $- c$ and greater than $c$ . "Greater than" gives two separate pieces joined by or; "less than" gives one piece between $- c$ and $c$ .

Tip

Read $|x - a| \lt r$ aloud as " $x$ is within $r$ of $a$ ". Then the answer $a - r \lt x \lt a + r$ is immediate, and you can sketch it on a number line before doing any algebra.

Khan Academy: Absolute value inequalities (about 13 min). Watch how each inequality is turned into a compound inequality, and how the solution is drawn on a number line before it is written.

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Practice

Show your working; questions 7 and 8 are harder.

Solve $5 - 2x \ge 3x + 20$ and write the solution as an interval.
Solve $∣ x + 3 ∣ = 7$ .
Solve $|4x - 1| \lt 7$ .
Solve $|5 - 2x| \ge 3$ .
Solve $\frac{x + 2}{x - 1} \ge 2$ .
Solve $∣ x - 1 ∣ = ∣ 3 x + 5 ∣$ .
Solve $|x| + |x - 4| \gt 6$ .
A cold room must keep its temperature $T$ (in °C) so that $|T - 18| \le 3$ . Write this as a double inequality, then find the allowed range in degrees Fahrenheit, using $F = 1.8 T + 32$ .

Show answersHide answers

$-5x \ge 15$ , so $x \le -3$ : $(-\infty, -3]$ .
$x + 3 = 7$ or $x + 3 = - 7$ : $x = 4$ or $x = - 10$ .
$-7 \lt 4x - 1 \lt 7 \Rightarrow -6 \lt 4x \lt 8 \Rightarrow -1.5 \lt x \lt 2$ .
$5 - 2x \le -3$ or $5 - 2x \ge 3$ : $x \ge 4$ or $x \le 1$ , that is $(-\infty, 1] \cup [4, +\infty)$ .
$\frac{x + 2 - 2(x - 1)}{x - 1} \ge 0 \iff \frac{4 - x}{x - 1} \ge 0$ . Sign table: positive for $1 \lt x \lt 4$ , zero at 4, undefined at 1. Solution $(1, 4]$ .
$x - 1 = 3x + 5 \Rightarrow x = -3$ ; $x - 1 = -3x - 5 \Rightarrow x = -1$ . Both check: $∣ - 4 ∣ = ∣ - 4 ∣$ , $∣ - 2 ∣ = ∣ 2 ∣$ .
$x \lt 0$ : $4 - 2x \gt 6 \Rightarrow x \lt -1$ . $0 \le x \le 4$ : $4 \gt 6$ is false. $x \gt 4$ : $2x - 4 \gt 6 \Rightarrow x \gt 5$ . Solution $(-\infty, -1) \cup (5, +\infty)$ .
$15 \le T \le 21$ . Then $1.8 (15) + 32 = 59$ and $1.8 (21) + 32 = 69.8$ , and $F$ increases with $T$ , so $59 \le F \le 69.8$ .

Quick check

Which set is the solution of |x − 4| < 3?

Key points

$|x - a| \lt r \iff a - r \lt x \lt a + r$ ; $|x - a| \gt r \iff x \lt a - r$ or $x \gt a + r$ .
For a quotient or product of linear factors, move everything to one side and use a sign table; never multiply by an expression of unknown sign.
With several absolute values, split the number line at the zeros of the expressions inside the bars, solve on each interval, and take the union.
Check one value inside and one outside your solution set.

Sources and credits

Written by Ganzaa: the explanations, the diagrams, the worked examples, the practice items and the quiz questions. The unit follows the topic list of REB's S4 Core Mathematics (2026–2027); no REB text is reproduced.
Photo: Scale on vernier caliper, Peter Southwood, Wikimedia Commons, CC BY-SA 3.0.
Video: Absolute value inequalities, Khan Academy, embedded from YouTube.
This lesson adapts CC BY-SA material and is shared under CC BY-SA 4.0.

1How big can the error be?

2Linear equations and inequalities in one unknown

3Products and quotients of linear factors

How big can the error be?

Linear equations and inequalities in one unknown

Products and quotients of linear factors