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Cambridge IGCSE Mathematics — 0580 Extended

Topic 2.6: Algebra and graphs — Inequalities

Number lines

An inequality compares two expressions. On a number line, the boundary is an open circle for \(<\) or \(>\) (value not included) and a closed circle for \(\leq\) or \(\geq\) (value included). Shade the values that satisfy the inequality.

Method

  1. Mark each endpoint. Closed circle if the inequality includes equals; open if it does not.
  2. Shade every number that makes the inequality true — between the endpoints for a compound statement, or out to an arrow for a single inequality.

Show \(-3 \leq x < 1\) on a number line.

Number line with closed circle at negative 3, open circle at 1, interval shaded between
Closed at \(-3\), open at \(1\), shade between.

Solving linear inequalities

Solve an inequality with the same steps as an equation, with one extra rule: if you multiply or divide by a negative number, reverse the inequality sign.

Which method?

You multiply or divide by a positive number

Keep the inequality sign the same.

You multiply or divide by a negative number

Reverse the inequality sign (\( < \) becomes \( > \), \(\leq\) becomes \(\geq\)).

Solve \(3x < 2x + 4\).

Worked solution of 3x less than 2x plus 4 giving x less than 4
Subtract \(2x\): \(x < 4\).

Solve \(-3 \leq 3x - 2 < 7\).

Worked compound inequality negative 3 less than or equal to 3x minus 2 less than 7 giving negative one third less than or equal to x less than 3
Add \(2\), then divide by \(3\): \(-\dfrac{1}{3} \leq x < 3\).

Paper 2 (non-calculator)

If the question asks for integer values of \(x\) satisfying \(-\dfrac{1}{3} \leq x < 3\), list \(x = 0, 1, 2\). Do not include \(3\) (\(x < 3\)) and do not include \(-1\) (because \(-1 < -\dfrac{1}{3}\)).

Two-variable graphs

A linear inequality in \(x\) and \(y\) is a half-plane. Draw the boundary line \(ax + by = c\). Use a broken line for \(<\) or \(>\) and a solid line for \(\leq\) or \(\geq\).

Cambridge papers shade the unwanted region unless the question says otherwise. Linear programming is not in this syllabus.

Method

  1. Draw each boundary. Solid if equals is allowed; broken if not.
  2. Test a point (often the origin) in each inequality to see which side is wanted.
  3. Shade the unwanted side of every line. The unshaded region, including solid boundaries, is the solution.

Show the region \(x \geq 0\), \(y \geq 0\), \(x + y \leq 4\).

Coordinate diagram of the triangle x plus y at most 4 in the first quadrant with unwanted region hatched
Solid on \(x + y = 4\). Shade away from the origin as unwanted. Wanted is the triangle including the axes.

Listing inequalities from a region

Read each boundary, then decide the direction from the unshaded (wanted) side. A broken line is strict; a solid line includes equality.

Write the three inequalities that define the unshaded region.

Region with dashed line x equals 2, solid y equals x and solid y equals 6, unwanted hatched, wanted x greater than 2, y at least x, y at most 6
Broken \(x = 2\) so \(x > 2\); solid \(y = x\) with wanted above so \(y \geq x\); solid \(y = 6\) with wanted below so \(y \leq 6\).

Try this

Solve \(5 - 2x \geq 11\).

Show answer
Answer

  1. Subtract 5 from both sides

    \[ -2x \geq 6 \]

  2. Divide by \(-2\) and reverse the sign

    \[ x \leq -3 \]

  3. Final answer

    \[ x \leq -3 \]

Exam Traps

  • Reverse the inequality sign when you multiply or divide by a negative. Forgetting this is the usual mark-loss on \(5 - 2x \geq 11\).
  • CIE shades unwanted regions unless the question tells you to shade the required region. Do not assume “shade the answer”.

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