Algebra and Graphs

Summary: Covers algebraic manipulation, linear and quadratic equations, simultaneous equations, inequalities, sequences, functions, coordinate geometry (straight line graphs, gradient, midpoint, distance, parallel and perpendicular lines), graphs of functions, and gradients of curves. This is the largest topic in IGCSE Mathematics 0580 and is heavily examined in both Core and Extended papers. Tags: igcse mathematics algebra graphs coordinate-geometry Created: 2026-07-16 Last Updated: 2026-07-16

1. Algebraic Manipulation

Collecting like terms: Terms with the same variable(s) raised to the same power(s) can be added/subtracted. Example: 3x + 5y − 2x + 7y = x + 12y.

Expanding brackets:

• Single bracket: a(b + c) = ab + ac
• Double brackets (FOIL): (a + b)(c + d) = ac + ad + bc + bd
• Difference of two squares: (a + b)(a − b) = a² − b²
• Perfect square: (a + b)² = a² + 2ab + b², (a − b)² = a² − 2ab + b²

Example: Expand (2x + 3)(x − 5) = 2x² − 10x + 3x − 15 = 2x² − 7x − 15.

Example: Expand (3x − 1)(x + 2)(x − 4) — multiply two brackets first, then the third.

Factorising:

• Common factor: 6x² + 9x = 3x(2x + 3)
• Difference of two squares: x² − 25 = (x + 5)(x − 5)
• Trinomial x² + bx + c: Find two numbers that multiply to c and add to b. Example: x² + 7x + 12 = (x + 3)(x + 4)
• Trinomial ax² + bx + c: Multiply a and c, find factors that sum to b, split the middle term, factorise by grouping.

Example: Factorise 6x² + 11x − 10. a×c = −60. Factors of −60 summing to 11: 15 and −4. = 6x² + 15x − 4x − 10 = 3x(2x + 5) − 2(2x + 5) = (3x − 2)(2x + 5).

Completing the square: x² + bx + c = (x + b/2)² − (b/2)² + c

Example: Write x² + 6x + 5 in the form (x + p)² + q. (x + 3)² − 9 + 5 = (x + 3)² − 4, so p = 3, q = −4.

2. Linear Equations

Solve by using inverse operations to isolate x, doing the same to both sides.

With fractions: Multiply every term by the LCM of denominators.

Example: Solve (x + 2)/3 + (x − 1)/2 = 5. Multiply by 6: 2(x + 2) + 3(x − 1) = 30 → 2x + 4 + 3x − 3 = 30 → 5x = 29 → x = 5.8.

Forming equations from word problems:

1. Identify the unknown and assign a variable.
2. Translate each part into algebra.
3. Solve and check in context.

Example: The sum of three consecutive integers is 57. Find them. Let integers be x, x+1, x+2. x + (x+1) + (x+2) = 57 → 3x = 54 → x = 18. Numbers: 18, 19, 20.

3. Simultaneous Equations

Elimination method:

1. Make coefficients of one variable equal (multiply equations as needed).
2. Add or subtract to eliminate that variable.
3. Solve for the remaining variable, then back-substitute.

Example: Solve 3x + 2y = 16 and 5x − 2y = 8. Add: 8x = 24 → x = 3. Substitute: 3(3) + 2y = 16 → 2y = 7 → y = 3.5.

Substitution method:

1. Rearrange one equation to make one variable the subject.
2. Substitute into the other equation and solve.

Example: Solve y = 2x + 1 and 3x + 4y = 23. 3x + 4(2x + 1) = 23 → 3x + 8x + 4 = 23 → 11x = 19 → x = 19/11, y = 49/11.

One linear, one quadratic (Extended): Substitute the linear into the quadratic, solve. This may yield two solution pairs.

Example: Solve y = 2x + 1 and y = x² + 1. x² + 1 = 2x + 1 → x² − 2x = 0 → x(x − 2) = 0 → x = 0 or x = 2. Solutions: (0, 1) and (2, 5).

4. Quadratic Equations

Solving by factorising: If (x + p)(x + q) = 0, then x = −p or x = −q.

Example: Solve x² − 5x + 6 = 0. (x − 2)(x − 3) = 0 → x = 2 or x = 3.

Quadratic formula (Extended): x = [−b ± √(b² − 4ac)] / 2a

Example: Solve 2x² − 3x − 5 = 0. a = 2, b = −3, c = −5. x = [3 ± √(9 + 40)] / 4 = [3 ± 7] / 4 → x = 2.5 or x = −1.

Completing the square method: Rewrite as (x + p)² = q, then x = −p ± √q.

Example: Solve x² + 8x + 3 = 0. (x + 4)² − 16 + 3 = 0 → (x + 4)² = 13 → x = −4 ± √13.

The Discriminant: Δ = b² − 4ac

• Δ > 0 → two distinct real roots
• Δ = 0 → one repeated real root (equal roots)
• Δ < 0 → no real roots

Example: How many real roots does 3x² − 2x + 1 = 0 have? Δ = (−2)² − 4(3)(1) = 4 − 12 = −8 → no real roots.

5. Inequalities

Solve linear inequalities the same as equations, except: when multiplying or dividing by a negative number, reverse the inequality sign.

Number line representation: Open circle ○ for < and >. Closed circle ● for ≤ and ≥.

Example: Solve 5 − 2x > 3. −2x > −2 → x < 1 (sign reversed). Number line: open circle at 1, arrow pointing left.

Compound inequalities: Solve each part, find the overlap.

Example: Solve −3 < 2x + 1 ≤ 7. −4 < 2x ≤ 6 → −2 < x ≤ 3.

Quadratic inequalities (Extended): Solve the related equation, sketch the graph, determine the region where the inequality holds.

Example: Solve x² − x − 6 < 0. Roots: (x − 3)(x + 2) = 0 → x = −2, 3. Parabola opens upward (∪). y < 0 between the roots: −2 < x < 3.

6. Sequences

Linear (arithmetic) sequences: nth term = a + (n − 1)d, where a = first term, d = common difference.

Example: Find the nth term of 7, 12, 17, 22, … d = 5, a = 7. nth term = 7 + (n − 1)×5 = 5n + 2.

Quadratic sequences (Extended): nth term = an² + bn + c.

1. Find the second difference (constant). Then 2a = second difference.
2. Subtract an² from each term to get a linear sequence.
3. Find b and c from the linear part.

Example: Find the nth term of 3, 10, 21, 36, 55, … First differences: 7, 11, 15, 19. Second difference = 4 → 2a = 4 → a = 2. Subtract 2n²: 3−2=1, 10−8=2, 21−18=3, 36−32=4, 55−50=5 → nth term = n. Original nth term = 2n² + n.

Special sequences:

• Square numbers: 1, 4, 9, 16, 25, … nth term = n²
• Triangular numbers: 1, 3, 6, 10, 15, … nth term = n(n+1)/2
• Fibonacci: 1, 1, 2, 3, 5, 8, 13, … (each term is sum of previous two)
• Cubic numbers: 1, 8, 27, 64, … nth term = n³

7. Functions

Function notation: f(x) means “f of x”. f(x) = 2x + 3 means take x, double it, add 3.

Domain: The set of all valid input values (x-values). Range: The set of all possible output values (y-values).

Composite functions (Extended): fg(x) means f(g(x)) — apply g first, then f.

Example: f(x) = 3x + 1, g(x) = x². Find fg(2) and gf(x). fg(2) = f(g(2)) = f(4) = 3(4) + 1 = 13. gf(x) = g(f(x)) = g(3x + 1) = (3x + 1)² = 9x² + 6x + 1.

Inverse functions (Extended): f⁻¹(x) reverses the effect of f(x). Method: write y = f(x), swap x and y, solve for y.

Example: Find f⁻¹(x) for f(x) = (2x + 1)/3. y = (2x + 1)/3 → x = (2y + 1)/3 → 3x = 2y + 1 → y = (3x − 1)/2. So f⁻¹(x) = (3x − 1)/2.

The graph of y = f⁻¹(x) is the reflection of y = f(x) in the line y = x.

8. Coordinate Geometry: Straight Line Graphs

The equation of a straight line: y = mx + c

• m = gradient (slope)
• c = y-intercept (where the line crosses the y-axis, when x = 0)

Finding gradient from two points (x₁, y₁) and (x₂, y₂): m = (y₂ − y₁) / (x₂ − x₁)

Midpoint of two points: ((x₁ + x₂)/2, (y₁ + y₂)/2)

Distance between two points: √[(x₂ − x₁)² + (y₂ − y₁)²]

Equation of a line given gradient and a point: y − y₁ = m(x − x₁)

Example: Find the equation of the line through (2, 5) with gradient 3. y − 5 = 3(x − 2) → y − 5 = 3x − 6 → y = 3x − 1.

Example: Find the equation of the line through (2, 5) and (6, 13). m = (13 − 5)/(6 − 2) = 2. y − 5 = 2(x − 2) → y = 2x + 1.

Parallel lines: Have the same gradient (m₁ = m₂).

Perpendicular lines: The product of their gradients is −1 (m₁ × m₂ = −1), so m₂ = −1/m₁.

Example: Find the equation of the line perpendicular to y = 3x + 2, passing through (6, 4). m_perp = −1/3. y − 4 = (−1/3)(x − 6) → y = −x/3 + 6.

9. Graphs of Functions

Quadratic graphs (y = ax² + bx + c):

• Parabola: ∪ if a > 0 (minimum), ∩ if a < 0 (maximum).
• y-intercept: set x = 0.
• x-intercepts (roots): solve ax² + bx + c = 0.
• Turning point: x = −b/(2a).

Example: Sketch y = x² − 4x + 3. x-intercepts: (x − 1)(x − 3) = 0 → x = 1, 3. y-intercept: (0, 3). Turning point: x = 4/2 = 2, y = 4 − 8 + 3 = −1. Minimum at (2, −1).

Cubic graphs (Extended): y = ax³ + bx² + cx + d. Find x and y intercepts, general shape determined by the sign of a.

Reciprocal graphs (Extended): y = k/x. Hyperbola with asymptotes at x = 0 and y = 0. If k > 0, graph is in quadrants I and III. If k < 0, quadrants II and IV.

Exponential graphs (Extended): y = aˣ.

• a > 1: growth curve, increases slowly then rapidly. Asymptote: y = 0 (x → −∞).
• 0 < a < 1: decay curve. Asymptote: y = 0 (x → +∞).

10. Gradient of a Curve (Extended)

The gradient of a curve at a specific point is found by drawing the tangent at that point.

Method:

1. Plot the curve or use the given graph.
2. At the required x-value, draw a tangent — a straight line that just touches the curve at that point.
3. Extend the tangent to form a right-angled triangle.
4. Gradient = (change in y) / (change in x) using the triangle.

Example: On the graph of y = x², estimate the gradient at x = 2. Draw tangent at (2, 4). Tangent passes approximately through (0, −4) and (4, 12). Gradient ≈ (12 − (−4))/(4 − 0) = 16/4 = 4.

Sources

• BBC Bitesize GCSE Mathematics — Algebraic Expressions and Equations, BBC (free educational resource)
• OpenStax Math — Algebra and Functions, Rice University (free, CC BY 4.0)
• Cambridge IGCSE Mathematics 0580 — Algebra and Graphs, Cambridge Assessment International Education
• CK-12 Mathematics — Algebraic Concepts, CK-12 Foundation (free, CC BY-NC 3.0)

Related Notes

• Number — Fractions, surds, and algebraic fractions
• Geometry — Angle properties, circle theorems used in coordinate problems
• Mensuration — Area/volume formulas used in algebraic word problems
• Trigonometry — Gradient and tan θ relationship
• IGCSE-Maths-Index — Full IGCSE Mathematics index

Common Misconceptions