Equations, Inequalities & Variation

Mathematics· 2.5, 2.6, E2.8· 25 min read

1. Core: Linear Equations & Changing the Subject of a Formula★★☆☆☆⏱ 6 min

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.

📘 Definition

Linear Equation

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

📐 Worked Example

Solve $3 x + 7 = 22$

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 $y = 4 x + 9$

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.

2. Core: Simultaneous Linear Equations & Representing Inequalities★★★☆☆⏱ 7 min

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: $2 x + y = 7$ and $x - 2 y = 6$

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.

3. Extended: Equations with Fractional Terms & Quadratic Equations★★★★☆Extended only⏱ 7 min

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

📘 Definition

Quadratic Equation

An equation of the form $a x^{2} + b x + c = 0$ 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 $2 x^{2} - 3 x - 7 = 0$ 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 $x = \frac{- b \pm b ^{2} - 4 a c}{2 a}$ :
$x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(2)(-7)}}{2(2)}$
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.

4. Extended: Mixed Simultaneous Equations, Inequality Regions & Proportion★★★★☆Extended only⏱ 5 min

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 $5 - 2 x \leq 13$ 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 $y = x^{2} - 3 x + 4$

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 $y = k x$ . 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.

5. Common Pitfalls

Wrong move:

Forgetting to flip the inequality sign when multiplying/dividing by a negative number

Why:

Violates basic inequality rules, giving the exact opposite solution to the correct one

Correct move:

Flip the direction of the inequality sign immediately whenever you multiply or divide both sides of an inequality by a negative value

Wrong move:

Core students attempting to solve quadratic equations algebraically

Why:

Quadratics are exclusively Extended tier content, so Core questions never require this skill, wasting time and leading to unnecessary errors

Correct move:

Core students only study linear content, skip Extended quadratic questions when practicing mixed tier papers

Wrong move:

Substituting the non-linear equation into the linear equation when solving mixed simultaneous pairs

Why:

Leads to unnecessarily complex algebra and a much higher risk of arithmetic errors

Correct move:

Always substitute the linear expression directly into the non-linear equation to simplify the problem

Wrong move:

Using open circles on number lines for $\leq$ or $\geq$ inequalities

Why:

Indicates the boundary value is not included in the solution set when it should be

Correct move:

Use closed circles for $\leq / \geq$ and open circles only for $< / >$

Wrong move:

Confusing direct and inverse proportion formulae, writing $y = k x$ for inverse proportion

Why:

Gives an incorrect constant of proportionality and wrong final answer

Correct move:

Remember inverse proportion uses $\frac{1}{x}$ , so $y = \frac{k}{x}$ , while direct proportion uses $y = k x$

6. Quick Reference Cheatsheet

Concept	Core Tier Requirement	Extended Tier Requirement
Linear equations	Construct & solve $a x + b = c$	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 $k$

7. Frequently Asked

Do Core tier students need to learn quadratic equations?

No, algebraic solving of quadratic equations is exclusively Extended tier content. Core students will never be asked to factorise or use the quadratic formula in their exams.

Do I have to flip the inequality sign if I multiply by a negative number?

Yes, always flip the direction of the inequality sign immediately whenever you multiply or divide both sides of an inequality by a negative value to keep the statement true.

Can I use the quadratic formula for all quadratic equation questions?

Yes, the quadratic formula works for all solvable quadratic equations, though factorisation is often faster for simpler expressions if you are able to do it quickly.

Going deeper

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.