Use interval notation and number-line graphs, solve linear and quadratic inequalities, and work with the absolute value definition to solve equations and inequalities of the form |x| < k and |x| > k

What this dot point is asking

This dot point is about describing and reasoning with parts of the number line. Three skills sit together here. First, interval notation: naming a stretch of the real line and drawing it, being careful about which endpoints are included. Second, solving inequalities, both linear and quadratic, where the one idea that trips students up is that the direction of the sign can flip. Third, absolute value, which is best understood not as "make it positive" but as distance, because that single idea makes every $|x| < k$ and $|x| > k$ problem fall out without memorising templates blindly. NESA examines all three constantly, often hidden inside other questions (a domain is an interval, a "for what values" is an inequality), so the goal is to make the moves automatic and to never lose a mark to a flipped sign or an open-versus-closed slip.

The answer

Intervals and the number line

An interval is a connected piece of the real number line: every point between its ends is included. You describe one with an inequality and draw it on a number line.

• A closed interval contains its endpoints, written with $\le$ or $\ge$, for example $1 \le x \le 3$. On the number line each endpoint is a filled (closed) circle.
• An open interval does not contain its endpoints, written with $<$ or $>$, for example $-1 < x < 5$. Each endpoint is an empty (open) circle.
• An interval can be half-closed, including one end but not the other, for example $-2 \le x < 3$.
• An unbounded interval continues forever in one direction, for example $x \ge -5$ or $x < 2$. It has only one endpoint; the open direction is drawn with an arrow.

A common convention point: in Year 11 we write intervals with the inequality form $a < x < b$, not the round-bracket form $(a, b)$. The bracket form is held back to Year 12 to avoid confusing an interval on the number line with a point $(a, b)$ in the coordinate plane.

Solving linear inequalities

A linear inequality is solved almost exactly like a linear equation: do the same thing to both sides to keep it balanced. You may add or subtract any number from both sides freely, and you may multiply or divide both sides by any positive number with no change. There is exactly one extra rule, and it is the whole difficulty of the topic.

Why does the sign flip? Multiplying by a negative reflects every number through $0$ on the number line, which reverses their order: $2 < 5$ is true, but multiplying by $-1$ gives $-2$ and $-5$, and $-2 > -5$. The order has turned around, so the inequality symbol must turn around with it.

A safer habit in many cases is to keep the variable's coefficient positive by moving terms to whichever side avoids a negative coefficient, so you never need to flip at all. For example, rather than $4 - 3x \ge 13 \Rightarrow -3x \ge 9$, you could write $4 - 13 \ge 3x \Rightarrow -9 \ge 3x \Rightarrow -3 \ge x$, the same answer $x \le -3$ with no flip step. Either route is fine; just be deliberate.

If the inequality contains fractions, clear them first by multiplying every term by the lowest common denominator, exactly as with equations. Because the lowest common denominator is a positive number, this step never flips the sign.

Solving quadratic inequalities from the parabola

A quadratic inequality like $x^2 - x - 6 > 0$ asks "for which $x$ is this quadratic positive (or negative)?" You do not solve it like an equation and you do not simply divide. The reliable method is to look at the graph: a quadratic is a parabola, and the sign of the quadratic is just whether the parabola is above or below the $x$-axis.

The recipe is short:

1. Make one side zero (move everything to the left if it is not already).
2. Factor and find the zeros: these are the $x$-intercepts where the parabola crosses the axis.
3. Decide the shape: a positive leading coefficient opens upward (smile), a negative one opens downward (frown).
4. Read the sign off the sketch: an upward parabola is below the axis between its roots and above outside them; a downward one is the reverse.

So for $x^2 - x - 6 > 0$, the factors are $(x - 3)(x + 2)$ with roots $-2$ and $3$, the parabola opens upward, and it is above the axis (positive) outside the roots: $x < -2$ or $x > 3$. The stage-by-stage build below shows exactly this.

Read the sign off the parabola, stage by stage

The cleanest way to see a quadratic inequality is to sketch the parabola and shade where it sits on the wanted side of the axis. Below, $x^2 - x - 6 > 0$ is built up: axis and intercepts, then the curve, then the sign of each region, then the solution read onto a number line. The answer is $x < -2$ or $x > 3$.

Stage 1, mark the zeros on the axis. Factor $x^2 - x - 6 = (x - 3)(x + 2)$, so the parabola crosses the $x$-axis at $x = -2$ and $x = 3$. These two roots are the only places the sign can change, because the curve can only switch sides of the axis where it touches it. Plot them.

Stage 2, draw the parabola. The coefficient of $x^2$ is $+1$, so the parabola opens upward. Its vertex sits halfway between the roots at $x = \tfrac{-2 + 3}{2} = 0.5$, where $y = (0.5)^2 - 0.5 - 6 = -6.25$. Draw a smooth upward parabola through the two roots with its lowest point at $(0.5, -6.25)$.

Stage 3, sign each region. The two roots split the line into three regions. Where the curve is above the axis the quadratic is positive ($+$); where it dips below the axis it is negative ($-$). Reading the curve: positive for $x < -2$, negative for $-2 < x < 3$, positive for $x > 3$. (Quick test points confirm it: at $x = -3$, value $9 > 0$; at $x = 0$, value $-6 < 0$; at $x = 4$, value $9 > 0$.)

Stage 4, read the solution onto a number line. We want $x^2 - x - 6 > 0$, that is, the positive regions, and the inequality is strict ($>$) so the roots themselves are excluded (hollow circles). The solution is $x < -2$ or $x > 3$, drawn as two outward rays.

Absolute value as distance

The absolute value $|x|$ is most usefully defined as the distance of $x$ from $0$ on the number line. So $|5| = 5$ and $|{-5}| = 5$: both are $5$ units from the origin. Distance is never negative, so $|x| \ge 0$ for every real $x$, and $|{-x}| = |x|$. There are two equivalent ways to write it down:

• As cases: $|x| = x$ when $x \ge 0$, and $|x| = -x$ when $x < 0$. (When $x$ is negative, $-x$ is the positive version of it.)
• As a square root: $|x| = \sqrt{x^2}$, since squaring kills the sign and the square-root symbol returns the non-negative root.

The distance idea extends: $|x - a|$ is the distance from $x$ to $a$ on the number line. So $|x - 4| = 6$ says "$x$ is $6$ away from $4$", giving $x = 4 - 6 = -2$ or $x = 4 + 6 = 10$ at once.

Solving $|x| < k$ and $|x| > k$

Reading absolute value as distance makes the two inequality shapes obvious, and they are opposites of each other. Let $k > 0$.

• $|x| < k$ means "$x$ is less than $k$ from $0$", that is, $x$ lies in the band between $-k$ and $k$. So $|x| < k \iff -k < x < k$, a single interval.
• $|x| > k$ means "$x$ is more than $k$ from $0$", that is, $x$ lies outside the band. So $|x| > k \iff x < -k$ or $x > k$, two rays.

The same templates work with an expression inside: replace $x$ by $ax + b$. For example $|2x - 3| < 5 \iff -5 < 2x - 3 < 5$, and $|x - 4| > 6 \iff x - 4 < -6$ or $x - 4 > 6$.

Here are the two solved on number lines. First $|2x - 3| < 5$, which gives the single band $-1 < x < 4$ (hollow endpoints, strict inequality).

Now $|x - 4| > 6$, which gives the two outward rays $x < -2$ or $x > 10$ (more than $6$ from the centre $4$).

How exam questions ask about intervals, inequalities and absolute value

These ideas are nearly always the working inside a larger question, so learn the cue words:

• "Solve the inequality" or "find the values of $x$ for which..." wants a solution set, written as an inequality (and sometimes graphed). For a linear one, isolate $x$ and watch for the negative-divide flip; for a quadratic one, make one side zero, factor, sketch the parabola and read off the sign.
• "Sketch on a number line" or "graph the solution" wants the right circles (filled for $\le$ / $\ge$, hollow for $<$ / $>$) and the correct rays or band; the marks are for the endpoints and the shading, so do not be sloppy with them.
• "For what values is the parabola below the $x$-axis" is a quadratic inequality $f(x) < 0$ in disguise: it is the region between the roots for an upward parabola.
• Absolute value: "solve $|\dots| = k$" has two answers (or none if $k < 0$); "solve $|\dots| < k$" is the single band $-k < \dots < k$; "solve $|\dots| > k$" is the two pieces. Reading them as distance stops you from guessing the wrong template.
• "Write the interval" wants inequality form such as $2 < x \le 7$ in Year 11 (not round/square brackets), with the correct strict-versus-inclusive signs.
• "Hence" tells you to reuse a result, for example a factorisation you found in a previous part, rather than starting the quadratic inequality from scratch.