# Equations, Inequalities & Variation

> Mathematics · CIE IGCSE 0580 2025-2027
> Source: https://www.owlsprep.com/study/cie-0580-u2-equations-inequalities-variation/

This guide covers all Core and Extended content for equations, inequalities and variation in CIE IGCSE Maths 0580, including step-by-step solutions for common exam problems and tier-specific guidance.

**Prerequisites:** [Basic algebraic manipulation](https://www.owlsprep.com/study/cie-0580-u2-basic-algebraic-manipulation/); [Number line fundamentals](https://www.owlsprep.com/study/cie-0580-u1-number-systems/)

## Learning objectives

- Construct and solve linear equations (Core tier)
- Solve pairs of simultaneous linear equations in two variables (Core + Extended)
- Rearrange simple formulae to change the subject with one step (Core)
- Represent and interpret inequalities on a number line (Core); construct and solve linear inequalities (Extended)
- Solve equations with fractional/algebraic terms (Extended)
- Solve quadratic equations via factorisation, completing the square and quadratic formula, including surd solutions (Extended)
- Solve pairs of one linear + one non-linear simultaneous equation (Extended)
- Represent inequalities as regions on 2D coordinate grids (Extended)
- Solve direct and inverse proportion problems, calculating the constant of proportionality k (Extended)

## Core: Linear Equations & Changing the Subject of a Formula

Linear equations are algebraic expressions where the highest power of the unknown variable is 1. You will be asked to both construct linear equations from word problems and solve them for the unknown value.

**Linear Equation** — An equation of the form $ax + b = c$ where $a, b, c$ are constants and $a \neq 0$, with no exponents higher than 1 on the variable.

**Worked example:** Solve $3x + 7 = 22$

1. Subtract 7 from both sides to isolate the term with $x$:

   $$3x = 22 - 7 = 15$$
2. Divide both sides by 3 to solve for $x$:

   $$x = \frac{15}{3} = 5$$
3. Check your answer by substituting back into the original equation: $3(5) +7 = 22$, which is correct.

Changing the subject of a formula for Core tier only requires one rearrangement step, with no nested operations or multiple rearrangements needed.

**Worked example:** Make $x$ the subject of the formula $y = 4x + 9$

1. Subtract 9 from both sides to isolate the term with $x$:

   $$y - 9 = 4x$$
2. Divide both sides by 4 to make $x$ the subject:

   $$x = \frac{y - 9}{4}$$

> **Exam tip:** Always show every rearrangement step for full marks, even if you can solve the equation mentally.

## Core: Simultaneous Linear Equations & Representing Inequalities

Simultaneous linear equations are pairs of equations with two unknowns, which you solve to find values that satisfy both equations. You can use either elimination or substitution methods, whichever you find easier for the given question.

**Worked example:** Solve the pair of simultaneous equations: $2x + y =7$ and $x - 2y =6$

1. Multiply the first equation by 2 to align the coefficients of $y$ for elimination:

   $$4x + 2y = 14$$
2. Add the adjusted first equation to the second equation to eliminate $y$:

   $$5x = 20 \implies x = 4$$
3. Substitute $x=4$ back into the first original equation to find $y$:

   $$2(4) + y =7 \implies y = -1$$
4. Check the solution in the second equation: $4 - 2(-1) = 6$, which is correct. The solution is $x=4, y=-1$.

Linear inequalities compare two expressions using $<, >, \leq, \geq$. For the Core tier you need to represent and interpret a given inequality on a number line: use open circles for strict inequalities ($<$ or $>$) and closed circles for inclusive inequalities ($\leq$ or $\geq$), then shade the values that satisfy it. (Constructing and solving linear inequalities is Extended content.)

**Worked example:** Represent $-2 < x \leq 3$ on a number line and list the integer values of $x$ that satisfy it.

1. Place an open circle at $-2$ (strict $<$, not included) and a closed circle at $3$ (inclusive $\leq$, included).
2. Shade the segment between the two circles to show all values with $-2 < x \leq 3$.
3. Read off the integers in this range:

   $$-1, 0, 1, 2, 3$$

> **Exam tip:** Always test your solution for simultaneous equations in both original equations to catch arithmetic errors before moving on.

## Extended: Equations with Fractional Terms & Quadratic Equations

Extended tier requires solving equations with fractional or algebraic denominators, plus quadratic equations using three methods: factorisation, completing the square, and the quadratic formula.

**Quadratic Equation** — An equation of the form $ax^2 + bx + c = 0$ where $a \neq 0$, with the highest power of $x$ equal to 2.

**Worked example:** Solve the equation with fractional terms: $\frac{x+2}{3} + \frac{x-1}{2} = 4$

1. Multiply all terms by 6 (the lowest common multiple of 3 and 2) to eliminate denominators:

   $$2(x+2) + 3(x-1) = 24$$
2. Expand and simplify the left-hand side:

   $$2x +4 + 3x -3 = 5x +1 = 24$$
3. Rearrange and solve for $x$:

   $$5x = 23 \implies x = 4.6$$

**Worked example:** Solve $2x^2 -3x -7 = 0$ using the quadratic formula, giving answers to 2 decimal places.

1. Identify the values $a=2$, $b=-3$, $c=-7$ from the standard quadratic form.
2. Substitute into the quadratic formula $x = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}$:

   $$x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(2)(-7)}}{2(2)}$$
3. Calculate the discriminant and simplify:

   $$x = \frac{3 \pm \sqrt{9 + 56}}{4} = \frac{3 \pm \sqrt{65}}{4}$$
4. Evaluate both solutions: $x \approx 2.77$ or $x \approx -1.27$

> **Exam tip:** When asked for exact solutions to quadratic equations, leave surds in your final answer instead of evaluating to decimal form unless explicitly told to do so.

## Extended: Mixed Simultaneous Equations, Inequality Regions & Proportion

Extended students construct and solve linear inequalities (remembering to flip the inequality sign when multiplying or dividing by a negative number), solve pairs of simultaneous equations where one is linear and one is non-linear (e.g. quadratic), represent inequalities as shaded regions on a coordinate grid, and solve direct and inverse proportion problems.

**Worked example:** Solve $5 - 2x \leq 13$ and represent the solution on a number line.

1. Subtract 5 from both sides:

   $$-2x \leq 8$$
2. Divide both sides by -2, flipping the inequality sign:

   $$x \geq -4$$
3. Draw a closed circle at -4 on the number line, and shade all values to the right of -4 to represent the solution set.

**Worked example:** Solve the simultaneous equations: $y = x + 1$ and $y = x^2 -3x +4$

1. Substitute the linear expression for $y$ directly into the quadratic equation:

   $$x + 1 = x^2 -3x +4$$
2. Rearrange to standard quadratic form:

   $$x^2 -4x +3 = 0$$
3. Factorise and solve for $x$:

   $$(x-1)(x-3) = 0 \implies x=1 \text{ or } x=3$$
4. Find corresponding $y$ values by substituting back into the linear equation: solutions are $(1,2)$ and $(3,4)$.

Direct proportion means one quantity increases as the other increases, written $y \propto x$ so $y = kx$. Inverse proportion means one quantity decreases as the other increases, written $y \propto \frac{1}{x}$ so $y = \frac{k}{x}$, where $k$ is the constant of proportionality.

**Worked example:** $y$ is inversely proportional to $x$. When $y=6$, $x=4$. Find the value of $y$ when $x=8$.

1. Write the inverse proportion formula:

   $$y = \frac{k}{x}$$
2. Substitute the known values to solve for $k$:

   $$6 = \frac{k}{4} \implies k = 24$$
3. Substitute $x=8$ and the value of $k$ to find $y$:

   $$y = \frac{24}{8} = 3$$

> **Exam tip:** When shading inequality regions on a coordinate grid, always explicitly label the required region to avoid losing marks for shading the wrong side of the line.

## Common pitfalls

- **Wrong:** Forgetting to flip the inequality sign when multiplying/dividing by a negative number
  - Why it fails: Violates basic inequality rules, giving the exact opposite solution to the correct one
  - Correct: Flip the direction of the inequality sign immediately whenever you multiply or divide both sides of an inequality by a negative value
- **Wrong:** Core students attempting to solve quadratic equations algebraically
  - Why it fails: Quadratics are exclusively Extended tier content, so Core questions never require this skill, wasting time and leading to unnecessary errors
  - Correct: Core students only study linear content, skip Extended quadratic questions when practicing mixed tier papers
- **Wrong:** Substituting the non-linear equation into the linear equation when solving mixed simultaneous pairs
  - Why it fails: Leads to unnecessarily complex algebra and a much higher risk of arithmetic errors
  - Correct: Always substitute the linear expression directly into the non-linear equation to simplify the problem
- **Wrong:** Using open circles on number lines for $\leq$ or $\geq$ inequalities
  - Why it fails: Indicates the boundary value is not included in the solution set when it should be
  - Correct: Use closed circles for $\leq/\geq$ and open circles only for $</>$
- **Wrong:** Confusing direct and inverse proportion formulae, writing $y=kx$ for inverse proportion
  - Why it fails: Gives an incorrect constant of proportionality and wrong final answer
  - Correct: Remember inverse proportion uses $\frac{1}{x}$, so $y = \frac{k}{x}$, while direct proportion uses $y=kx$

## Cheatsheet

| Concept | Core Tier Requirement | Extended Tier Requirement |
| --- | --- | --- |
| Linear equations | Construct & solve $ax + b = c$ | Solve with fractional/algebraic denominators |
| Simultaneous equations | Solve 2 linear equations | Solve 1 linear + 1 non-linear pair |
| Formula rearrangement | 1-step rearrangement only | Multi-step rearrangement (separate topic) |
| Inequalities | Represent & interpret on a number line | Construct & solve linear inequalities; represent as 2D coordinate regions |
| Quadratic equations | Not required | Solve via factorisation, completing the square, quadratic formula, give surd solutions |
| Proportion | Not required | Solve direct & inverse proportion, calculate constant $k$ |

## What's next

Now that you have mastered equations, inequalities and variation for CIE IGCSE Maths 0580, you can move on to more advanced algebraic topics. For Core students, next steps include learning to plot and interpret linear graphs, which builds directly on your linear equation skills. For Extended students, you can apply your quadratic equation knowledge to graphing quadratic functions, as well as tackling more complex algebraic manipulation challenges. Practice with tier-specific past paper questions to reinforce your understanding and identify any gaps before your exam.

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