Year 11 Methods (Unit 1 & 2) Reviewing Linear Equations

Solving Linear Inequalities

Q: What do open and closed circles mean on a number line?

An open circle marks a strict endpoint, less than or greater than, that is not included. A closed circle marks less than or equal to or greater than or equal to, which is included.

Q: Does adding a negative number flip the sign?

No. Only multiplying or dividing by a negative flips the sign. Adding or subtracting never does.

Q: How do you solve a worded inequality?

Define a variable, translate at most as less than or equal to and at least as greater than or equal to, solve, then apply any context limits such as a distance being non-negative.

20 practice questions 1 video lesson Theory + worked examples

Solving linear inequalities is a foundational skill assumed for Queensland Year 11 Mathematical Methods (QCAA). An inequality has a whole range of solutions rather than a single value, describing every number that satisfies the condition.

You will solve inequalities step by step, apply the crucial rule of reversing the sign when you multiply or divide by a negative, and graph the solution on a number line.

Practice 20 questions

Practice questions

Every question with a fully worked solution.

Start practising

Watch 1 video(s)

Solving Linear Inequalities - Video - Solving linear inequalities Watch

Theory

In Year 11 Mathematical Methods (QCAA), a linear inequality such as $3x-5<7$ has a whole range of solutions, not a single value. This page reviews how to solve linear inequalities in one variable, the crucial rule of flipping the sign when you multiply or divide by a negative, how to show a solution on a number line, and how to write it in interval notation.

A linear inequality compares two linear expressions with one of the signs $<$, $>$, $\le$ or $\ge$, for example $3x-5<7$. Its solution is the set of all values of the variable that make the statement true — usually a range such as $x<4$.

You solve an inequality almost exactly like an equation, using inverse operations. The one extra rule is decisive: when you multiply or divide both sides by a negative number, reverse the inequality sign.

On a number line, an open circle marks a strict endpoint ($<$ or $>$) that is not included, and a closed circle marks $\le$ or $\ge$, which is included; the shaded ray shows the direction of all the solutions.

Flip when you divide by a negative. Solving \(-2x\ge6\) gives \(x\le-3\) — the sign turns around because both sides were divided by \(-2\).

$x<4$: an open circle at $4$ (not included), shaded to the left.

$-3\le x<2$: closed at $-3$ (included), open at $2$ (not included).

The solving rules are the same as for equations, with one exception:

\[a>b\;\Longrightarrow\;a+c>b+c\quad(\text{add/subtract: sign unchanged})\]

a + c > b + c

\[a>b,\ k>0\;\Longrightarrow\;ka>kb\quad(\text{positive multiple: sign unchanged})\]

k a > k b

\[a>b,\ k<0\;\Longrightarrow\;ka

k a < k b

Interval notation writes the solution set compactly, using a square bracket for an included endpoint and a round bracket for one that is excluded:

\[x<4\;\equiv\;(-\infty,\,4),\qquad -3\le x<2\;\equiv\;[-3,\,2)\]

[- 3, 2)

Only a negative multiplier flips the sign. Adding or subtracting a negative, or multiplying by a positive, leaves the inequality sign as it is.

How to solve a linear inequality

Simplify each side: expand brackets and clear fractions by multiplying every term by the (positive) lowest common denominator.
Collect the variable on one side and the numbers on the other with inverse $+/-$.
Divide by the coefficient — and flip the sign if that coefficient is negative.
Show the solution on a number line and, if asked, in interval notation.

Example 1 — A two-step inequality

Solve $3x-5<7$ and show the solution on a number line.

Solution

Add $5$ to both sides:

$3x-5$	$<$	$7$
$3x$	$<$	$12$

Divide both sides by $3$ (positive, so the sign stays):

$\dfrac{3x}{3}$	$<$	$\dfrac{12}{3}$
$x$	$<$	$4$

Solution: $x<4$, or $(-\infty,\,4)$.

x < 4

Example 2 — Flipping the sign

Solve $4-2x\ge10$.

Solution

Subtract $4$ from both sides:

$4-2x$	$\ge$	$10$
$-2x$	$\ge$	$6$

Divide both sides by $-2$ — reverse the sign:

$\dfrac{-2x}{-2}$	$\le$	$\dfrac{6}{-2}$
$x$	$\le$	$-3$

Check — test $x=-4$ (which satisfies $x\le-3$):

$4-2(-4)$

$=$

$4+8=12\ge10\;\checkmark$

Solution: $x\le-3$, or $(-\infty,\,-3]$.

x \leq - 3

Example 3 — Brackets and variable on both sides

Solve $2(x-3)>5x+9$.

Solution

Expand the bracket:

$2(x-3)$	$>$	$5x+9$
$2x-6$	$>$	$5x+9$

Collect the variable — subtract $5x$ from both sides:

$2x-6-5x$	$>$	$9$
$-3x-6$	$>$	$9$

Add $6$ to both sides:

$-3x$

$>$

$15$

Divide by $-3$ — reverse the sign:

$x$

$<$

$-5$

Solution: $x<-5$, or $(-\infty,\,-5)$.

x < - 5

Example 4 — A worded inequality

A taxi charges a $\$4$ flagfall plus $\$2$ per kilometre. Ravi has at most $\$30$ to spend. How far can he travel?

Solution

Define — let $d$ be the distance in km. ‘At most $\$30$’ means $\le30$:

$4+2d$

$\le$

$30$

Subtract $4$, then divide by $2$ (positive):

$2d$	$\le$	$26$
$d$	$\le$	$13$

A distance cannot be negative, so $0\le d\le13$.

Ravi can travel up to $13$ km, i.e. $0\le d\le13$.

d \leq 13

Common pitfalls

Forgetting to flip the sign. Dividing or multiplying both sides by a negative number reverses the inequality: $-3x>15$ gives $x<-5$, not $x>-5$.

Flipping when you should not. Only a negative multiplier or divisor flips the sign. Adding or subtracting a negative does not.

Open vs closed circle. Use an open circle for $<$ or $>$ and a closed circle for $\le$ or $\ge$; getting this wrong changes whether the endpoint is included.

Ignoring context limits. A distance or a count cannot be negative, so a worded answer may need $0\le d\le13$ rather than just $d\le13$.

Frequently asked questions

How is solving an inequality different from solving an equation?

You use the same inverse operations, but there is one extra rule: reverse the inequality sign whenever you multiply or divide both sides by a negative number.

When do you flip the inequality sign?

Only when you multiply or divide both sides by a negative number. Solving $-2x\ge6$ gives $x\le-3$.

What do open and closed circles mean on a number line?

An open circle marks a strict endpoint ($<$ or $>$) that is not included; a closed circle marks $\le$ or $\ge$, which is included.

How do you write an inequality in interval notation?

Use a round bracket for an excluded endpoint and a square bracket for an included one: $x<4$ is $(-\infty,\,4)$, and $-3\le x<2$ is $[-3,\,2)$.

Does adding a negative number flip the sign?

No. Only multiplying or dividing by a negative flips the sign; adding or subtracting never does.

How do you solve a worded inequality?

Define a variable, translate ‘at most’ as $\le$ and ‘at least’ as $\ge$, solve, then apply any context limits such as a distance being non-negative.

← Previous subtopic

Constructing Simultaneous Linear Equations

Next subtopic →

Using And Transposing Formulas

\(3x-5\)	\(<\)	\(7\)
\(3x\)	\(<\)	\(12\)

\(\dfrac{3x}{3}\)	\(<\)	\(\dfrac{12}{3}\)
\(x\)	\(<\)	\(4\)

\(4-2x\)	\(\ge\)	\(10\)
\(-2x\)	\(\ge\)	\(6\)

\(\dfrac{-2x}{-2}\)	\(\le\)	\(\dfrac{6}{-2}\)
\(x\)	\(\le\)	\(-3\)

\(2(x-3)\)	\(>\)	\(5x+9\)
\(2x-6\)	\(>\)	\(5x+9\)

\(2x-6-5x\)	\(>\)	\(9\)
\(-3x-6\)	\(>\)	\(9\)

\(2d\)	\(\le\)	\(26\)
\(d\)	\(\le\)	\(13\)

Solving Linear Inequalities

📖 Theory

How to solve a linear inequality

Common pitfalls

Frequently asked questions

Theory