Solving linear inequalities, representing solutions on a number line, solving quadratic inequalities at Higher tier, and graphing inequality regions.

What this dot point is asking

AQA wants you to solve linear inequalities and show solutions on a number line, and at Higher tier to solve quadratic inequalities and shade regions defined by linear inequalities on a graph. Inequalities behave almost exactly like equations, with one crucial difference around negative multiplication, and the topic links closely to solving equations and straight line graphs.

Solving linear inequalities

An inequality such as $3x + 2 > 11$ is solved exactly like an equation: subtract $2$ to get $3x > 9$, then divide by $3$ to get $x > 3$. The solution is a range of values rather than a single number.

You can avoid the sign reversal entirely by keeping the unknown positive. For $-2x < 6$, add $2x$ to both sides and subtract $6$ to get $-6 < 2x$, then divide by the positive $2$ to get $-3 < x$, which is the same as $x > -3$.

Double inequalities are solved by operating on all three parts at once. For $-1 \le 2x + 3 < 9$, subtract $3$ throughout to get $-4 \le 2x < 6$, then divide by $2$ to get $-2 \le x < 3$.

Representing solutions on a number line

A solution is shown as a circle on the value with an arrow indicating the direction. An open (unfilled) circle means the endpoint is not included ($<$ or $>$); a closed (filled) circle means it is included ($\le$ or $\ge$). For $x > 3$ draw an open circle at $3$ with an arrow to the right. For a double inequality like $-2 \le x < 3$, mark a filled circle at $-2$, an open circle at $3$, and shade the segment between them.

Integer solutions are sometimes asked for separately: the integers satisfying $-2 \le x < 3$ are $-2, -1, 0, 1, 2$ (note $3$ is excluded).

Solving quadratic inequalities at Higher tier

A quadratic inequality is solved by finding where the quadratic equals zero, then deciding which side of those critical values satisfies the inequality.

Graphing inequality regions

A linear inequality such as $y \le 2x + 1$ defines a region of the plane. Draw the boundary line $y = 2x + 1$ (solid for $\le$ or $\ge$, dashed for $<$ or $>$), then shade the side that satisfies the inequality. Test a point not on the line, often the origin: for $y \le 2x + 1$ at $(0,0)$ we get $0 \le 1$, which is true, so shade the side containing the origin. With several inequalities, the required region is where all the shaded areas overlap.

Integer solutions and listing

A common question asks for the integer values that satisfy an inequality or a pair of inequalities. After solving, list the whole numbers in range carefully, watching whether each endpoint is included. For $-3 < x \le 2$ the integers are $-2, -1, 0, 1, 2$, since $-3$ is excluded but $2$ is included. When two inequalities are combined, solve each, then take the overlap: $x > 1$ and $x \le 5$ gives $1 < x \le 5$, so the integers are $2, 3, 4, 5$. Errors here almost always come from mishandling a strict versus inclusive endpoint, so checking each boundary explicitly is worth the moment it takes.