Solve simultaneous equations in two unknowns by elimination or substitution, including equations that are linear and non-linear

IGCSE Additional Mathematics (0606) – Quick Reference Notes

Contents

1. Functions
2. Quadratic Functions & Equations
3. Polynomial Factorisation
4. Simultaneous Equations (Linear & Non‑linear)
5. Solving Inequalities
6. Logarithmic & Exponential Equations
7. Straight‑Line Graphs
8. Circles
9. Circular Measure (Radians)
10. Trigonometry
11. Permutations & Combinations
12. Series (Arithmetic & Geometric)
13. Vectors
14. Calculus (Differentiation & Integration)

1. Functions (Syllabus 1)

• Definition: A rule that assigns to each element of a domain exactly one element of a range. Notation: y = f(x).
• Domain & Range

  • Identify restrictions (division by zero, even roots, logarithms, etc.).
  • Example: for y = 1/(x‑2), domain = ℝ \ {2}, range = ℝ \ {0}.

• One‑to‑One (Injective) Functions

  • Pass the Horizontal Line Test.
  • Only one‑to‑one functions have inverses that are also functions.

• Inverse Function

  • Swap x and y and solve for y: y = f⁻¹(x).
  • Restrict the domain if necessary so that the function becomes one‑to‑one.
  • Graphically: reflection in the line y = x.

• Composition – (f ∘ g)(x) = f(g(x)). Order matters: f ∘ g ≠ g ∘ f in general.
• Common Function Types (with domain/range notes):

  • Linear: y = mx + c – domain & range = ℝ.
  • Quadratic: y = ax² + bx + c – domain = ℝ, range depends on a.
  • Power: y = xⁿ (n integer) – domain restrictions for even n.
  • Root: y = √(x‑k) – domain ≥ k.
  • Reciprocal: y = 1/(x‑k) – domain ≠ k.
  • Exponential: y = a·bˣ (b>0, b≠1) – domain = ℝ, range > 0.
  • Logarithmic: y = log_b(x) – domain > 0, range = ℝ.

Example – Finding an Inverse with Restriction

Given f(x) = √(x‑1), the domain is x ≥ 1. Interchange and square:

\$y = \sqrt{x-1}\;\Longrightarrow\;x = \sqrt{y-1}\;\Longrightarrow\;x^{2}=y-1\;\Longrightarrow\;y = x^{2}+1.\$

Thus f⁻¹(x) = x² + 1 with domain ≥ 0 (since x in the inverse is the original y).

2. Quadratic Functions & Equations (Syllabus 2)

• Standard form: y = ax² + bx + c (a ≠ 0).
• Vertex form: y = a(x‑h)² + k, where vertex = (h, k). Obtain by completing the square.
• Discriminant: Δ = b² – 4ac

  • Δ > 0 → two distinct real roots.
  • Δ = 0 → one real (double) root – the parabola is tangent to the x‑axis.
  • Δ < 0 → no real roots (complex pair).

• Factorisation

  • Find two numbers that multiply to ac and add to b.
  • If not obvious, use the quadratic formula or complete the square.

• Quadratic formula: x = [‑b ± √(b²‑4ac)] / (2a).
• Graph features

  • Axis of symmetry: x = –b/(2a).
  • Direction: opens upwards if a > 0, downwards if a < 0.

Example – Solving a Quadratic Equation

Solve 2x² – 5x – 3 = 0.

1. Δ = (‑5)² – 4·2·(‑3) = 25 + 24 = 49 > 0.
2. Roots: x = [5 ± √49] / 4 = (5 ± 7)/4 → x = 3 or x = –½.

3. Polynomial Factorisation (Syllabus 3)

• Common factor: Factor out the greatest common factor (GCF) first.
• Factor theorem: If f(r) = 0 then (x‑r) is a factor of f(x).
• Synthetic division – a quick way to divide by a linear factor (x‑r).
• Quadratic‑in‑form: Treat expressions like x⁴ – 5x² + 4 as a quadratic in x².

Example – Factorising a Cubic

Factor f(x) = x³ – 6x² + 11x – 6.

1. Test integer roots ±1, ±2, ±3, ±6. f(1)=0 ⇒ (x‑1) is a factor.
2. Synthetic division by 1 gives x² – 5x + 6.
3. Factor the quadratic: (x‑2)(x‑3).
4. Full factorisation: (x‑1)(x‑2)(x‑3).

4. Simultaneous Equations – Linear & Non‑linear (Syllabus 5)

4.1 General Strategy

1. Write each equation in a convenient form (standard, solved for a variable, or as a function).
2. Choose the method that minimises algebra:

   • Elimination – make coefficients of one variable opposites, add/subtract.
   • Substitution – isolate a variable in one equation, substitute into the other.

3. Solve the resulting single‑variable equation (linear, quadratic, cubic, absolute, etc.).
4. Back‑substitute to obtain the second variable.
5. Check every solution in both original equations (essential for non‑linear systems).

4.2 Linear Systems

Example A – Elimination (Two‑variable linear)

\$\$\begin{cases}

3x + 4y = 22\\

5x - 4y = 2

\end{cases}\$\$

1. Add the equations → 8x = 24 → x = 3.
2. Substitute into 3·3 + 4y = 22 → 9 + 4y = 22 → y = 13/4.

Solution: (3, 13/4).

Example B – Substitution (Linear)

\$\$\begin{cases}

2x - y = 5\\

x + 3y = 7

\end{cases}\$\$

1. From the first, y = 2x - 5.
2. Substitute: x + 3(2x‑5) = 7 → 7x‑15 = 7 → x = 22/7.
3. Back‑substitute: y = 2·22/7 - 5 = 9/7.

Solution: (22/7, 9/7).

4.3 Quadratic + Linear (or Quadratic + Quadratic)

1. Substitute the linear expression for y (or x) into the quadratic.
2. Solve the resulting quadratic (use discriminant to anticipate number of real solutions).
3. Find the corresponding second variable.
4. Check for extraneous roots (especially if you have squared both sides).

Example C – Quadratic + Linear

\$\$\begin{cases}

y = x^{2} - 3x + 2\\

2x + y = 5

\end{cases}\$\$

1. Substitute: 2x + (x²‑3x+2) = 5 → x² - x - 3 = 0.
2. Δ = 1 + 12 = 13 → two real roots: x = [1 ± √13]/2.
3. Corresponding y from the quadratic expression.
4. Both pairs satisfy the original system.

4.4 Cubic + Linear (or higher‑degree)

• Factor the higher‑degree equation first (factor theorem, synthetic division).
• Obtain all possible values of the isolated variable.
• For each value, compute the second variable from the linear equation.

Example D – Cubic + Linear

\$\$\begin{cases}

x^{3} - 4x^{2} + x = 0\\

x + 2y = 7

\end{cases}\$\$

1. Factor: x(x^{2}‑4x+1)=0. Roots: x=0 and x = 2 ± √3.
2. Use y = (7‑x)/2 for each x.

Solutions: (0, 3.5), (2+√3, (5‑√3)/2), (2‑√3, (5+√3)/2).

4.5 Absolute‑Value (Modulus) Systems

1. Isolate the absolute‑value expression.
2. Replace |A| = B with the two linear cases A = B and A = –B (remember B ≥ 0).
3. Solve each case together with the second equation.
4. Discard any pair that makes the original absolute‑value expression negative.

Example E – Modulus + Linear

\$\$\begin{cases}

|3x‑4| = y\\

2x + y = 10

\end{cases}\$\$

1. From the second, y = 10‑2x.
2. Substitute: |3x‑4| = 10‑2x (requires 10‑2x ≥ 0 → x ≤ 5).
3. Case 1: 3x‑4 = 10‑2x → 5x = 14 → x = 14/5 (valid, ≤5). y = 10‑28/5 = 22/5.
4. Case 2: ‑(3x‑4) = 10‑2x → -3x+4 = 10‑2x → -x = 6 → x = –6 (invalid because 10‑2x = 22 > 0 but |3x‑4| = 22 holds, so it is valid). y = 10‑2(‑6) = 22.

Solutions: (14/5, 22/5) and (‑6, 22).

4.6 Systems Involving Inequalities

Replace the inequality by an equation to find the boundary, then test intervals (or use a sign chart) to determine where the inequality holds.

Example F – Linear + Quadratic Inequality

\$\$\begin{cases}

y \ge x^{2} - 4x + 3\\

y = 2x - 1

\end{cases}\$\$

1. Substitute: 2x‑1 ≥ x²‑4x+3 → 0 ≥ x²‑6x+4 → x²‑6x+4 ≤ 0.
2. Roots: x = [6 ± √(36‑16)]/2 = 3 ± √5.
3. Parabola opens upwards → inequality true between the roots.
4. Solution interval: 3‑√5 ≤ x ≤ 3+√5.

5. Solving Inequalities (Syllabus 4)

• Linear inequality – isolate x, reverse the sign when multiplying/dividing by a negative number.
• Quadratic inequality – bring to one side, factor (or use the quadratic formula), then draw a sign chart.
• Cubic or higher‑degree – factor into linear factors, list critical points, test intervals.
• Absolute‑value inequality – split into two linear inequalities: |A| < B → -B < A < B (with B > 0).
• Rational inequality – write as a single fraction, identify zeros of numerator and denominator, then use a sign chart (exclude points where denominator = 0).

Example – Rational Inequality

\$\frac{x+2}{x‑3} \le 0\$

1. Critical points: numerator = 0 → x = –2; denominator = 0 → x = 3 (excluded).
2. Sign chart:

   • (‑∞, –2): numerator < 0, denominator < 0 → quotient > 0.
   • (–2, 3): numerator > 0, denominator < 0 → quotient < 0.
   • (3, ∞): numerator > 0, denominator > 0 → quotient > 0.

3. Include the zero of the numerator (≤ 0) but exclude the denominator zero.
4. Solution: [-2, 3).

6. Logarithmic & Exponential Equations (Syllabus 4)

• Exponential form: a^{x} = b → x = log_{a}(b).
• Logarithmic form: log_{a}(x) = b → x = a^{b}.
• Use the laws of logarithms to combine or split terms:

  • log{a}(mn) = log{a}m + log_{a}n
  • log{a}(m/n) = log{a}m – log_{a}n
  • log{a}(m^{k}) = k·log{a}m

• When bases differ, change of base formula: log{a}b = log{c}b / log_{c}a.

Example – Solving an Exponential Equation

Solve 3^{2x‑1} = 27.

1. Write RHS as a power of 3: 27 = 3³.
2. Equate exponents: 2x‑1 = 3 → 2x = 4 → x = 2.

Example – Solving a Logarithmic Equation

Solve log{2}(x+3) – log{2}(x‑1) = 2.

1. Combine logs: log_{2}[(x+3)/(x‑1)] = 2.
2. Exponentiate: (x+3)/(x‑1) = 2² = 4.
3. Cross‑multiply: x+3 = 4x‑4 → 3x = 7 → x = 7/3.
4. Check domain: x‑1 > 0 → 7/3‑1 > 0 (valid).

7. Straight‑Line Graphs (Syllabus 4.1)

• General form: y = mx + c (gradient m, y‑intercept c).
• Slope formula: m = (y₂‑y₁)/(x₂‑x₁).
• Parallel lines: same gradient, different intercepts.
• Perpendicular lines: product of gradients = –1 (provided neither is vertical).
• Distance formula: d = √[(x₂‑x₁)² + (y₂‑y₁)²].
• Mid‑point formula: ((x₁+x₂)/2, (y₁+y₂)/2).

Example – Finding the Equation of a Line Through Two Points

Points A(1, 4) and B(5, ‑2).

1. Gradient: m = (‑2‑4)/(5‑1) = –6/4 = –3/2.
2. Use point‑slope form: y‑4 = (‑3/2)(x‑1) → y = (‑3/2)x + 11/2.

8. Circles (Syllabus 4.2)

• Standard form: (x‑h)² + (y‑k)² = r² where (h, k) is the centre and r the radius.
• Expand to general form: x² + y² + Dx + Ey + F = 0 (complete the square to find centre & radius).
• Diameter = 2r, circumference = 2πr, area = πr².

Example – Finding Centre & Radius from General Form

Given x² + y² – 6x + 8y + 9 = 0.

1. Group and complete squares:

   • (x²‑6x) = (x‑3)² – 9
   • (y²+8y) = (y+4)² – 16

2. Rewrite: (x‑3)² + (y+4)² – 9 – 16 + 9 = 0 → (x‑3)² + (y+4)² = 16.
3. Centre = (3, ‑4), radius = 4.

9. Circular Measure (Radians) (Syllabus 4.3)

• Conversion: π rad = 180° → 1 rad = 180/π° ≈ 57.2958°.
• Arc length: s = rθ (θ in radians).
• Sector area: A = ½ r²θ.

Example – Arc Length

Find the length of an arc subtended by a central angle of 2π/3 radians in a circle of radius 5 cm.

s = rθ = 5 × 2π/3 = 10π/3 cm ≈ 10.47 cm.

10. Trigonometry (Syllabus 4.4)

• Primary ratios (right‑angled triangle): sin θ = opposite/hypotenuse, cos θ = adjacent/hypotenuse, tan θ = opposite/adjacent.
• Reciprocal ratios: csc θ = 1/sin θ, sec θ = 1/cos θ, cot θ = 1/tan θ.
• Pythagorean identities: sin²θ + cos²θ = 1, 1 + tan²θ = sec²θ, 1 + cot²θ = csc²θ.
• Angle of elevation/depression: use tan θ = opposite/adjacent.
• Trigonometric graphs: amplitude, period (2π), phase shift, vertical shift.
• Solving triangles (SAS, SSS, ASA, AAS): apply cosine rule, sine rule where needed.

Example – Solving a Right‑Angle Problem

A ladder leans against a wall, foot 4 m from the wall, makes an angle of 60° with the ground. Find the height reached.

Height = 4·tan 60° = 4·√3 ≈ 6.93 m.

11. Permutations & Combinations (Syllabus 5.1)

• Permutation (ordered): P(n, r) = n! / (n‑r)!.
• Combination (unordered): C(n, r) = n! / [r!(n‑r)!].
• When repetitions are allowed:

  • Permutations with repetition: n^{r}.
  • Combinations with repetition: C(n+r‑1, r).

Example – Committee Selection

From 8 boys and 6 girls, how many ways to form a committee of 4 with exactly 2 boys?

1. Choose boys: C(8,2) = 28.
2. Choose girls: C(6,2) = 15.
3. Total ways: 28 × 15 = 420.

12. Series (Arithmetic & Geometric) (Syllabus 5.2)

Arithmetic Series

• General term: aₙ = a₁ + (n‑1)d.
• Sum of first n terms: Sₙ = n/2 (a₁ + aₙ) or Sₙ = n/2 [2a₁ + (n‑1)d].

Geometric Series

• General term: aₙ = a₁·r^{n‑1} (r ≠ 1).
• Sum of first n terms: Sₙ = a₁ (1‑r^{n}) / (1‑r) (if |r| < 1, the infinite sum is a₁/(1‑r)).

Example – Sum of an Arithmetic Sequence

Find the sum of the first 20 terms of the sequence 5, 9, 13, …

• a₁ = 5, d = 4, n = 20.
• a₂₀ = 5 + 19·4 = 81.
• S₂₀ = 20/2 (5 + 81) = 10·86 = 860.

13. Vectors (Syllabus 5.3)

• Notation: →AB = (x₂‑x₁, y₂‑y₁) or ⟨a, b⟩.
• Magnitude: |→v| = √(a² + b²).
• Direction (angle θ): tan θ = b/a (adjust for quadrant).
• Vector addition: →u + →v = (a₁+a₂, b₁+b₂).
• Scalar multiplication: k·→v = (ka, kb).
• Dot product: →u·→v = a₁a₂ + b₁b₂ = |→u||→v|cos θ.

  • Perpendicular ⇔ dot product = 0.