Algebra: Equations, Inequalities and Simultaneous Equations

This lesson covers solving quadratic and simultaneous equations, including the case of one linear and one quadratic, and representing inequality solutions on graphs.

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Solving Quadratic Equations

By Factorising

Set equal to zero, factorise, then solve each factor.

x² − 7x + 10 = 0 → (x − 2)(x − 5) = 0 → x = 2 or x = 5

Using the Quadratic Formula

x = (−b ± √(b² − 4ac)) / 2a (see previous lesson).

By Completing the Square

Convert to (x + p)² = q form, then square root both sides (see previous lesson).

Solving Equations Reducible to Quadratics

Look for a substitution. For x⁴ − 5x² + 4 = 0, let u = x²: u² − 5u + 4 = 0 → (u − 1)(u − 4) = 0 → u = 1 or u = 4 → x = ±1 or x = ±2

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Linear Simultaneous Equations

Elimination Method

Example: 3x + 2y = 12 and 5x − 2y = 4 Add the equations: 8x = 16 → x = 2 → y = 3

Example: 2x + 5y = 11 and 3x − 2y = 7 Multiply so that coefficients match, then eliminate. Here multiply the first by 2 and the second by 5: 4x + 10y = 22; 15x − 10y = 35 → add: 19x = 57 → x = 3; then y = 1.

Substitution Method

Make one variable the subject and substitute into the other equation.

Example: y = 2x − 1 and 3x + y = 9 Substitute: 3x + (2x − 1) = 9 → 5x = 10 → x = 2; y = 3

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Linear and Quadratic Simultaneous Equations [H]

Substitute the linear equation into the quadratic.

Example: y = x + 3 and x² + y² = 29 Substitute: x² + (x + 3)² = 29 → x² + x² + 6x + 9 = 29 2x² + 6x − 20 = 0 → x² + 3x − 10 = 0 → (x + 5)(x − 2) = 0 x = −5 → y = −2; x = 2 → y = 5

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Linear Inequalities

Solve like equations, but reverse the inequality sign when multiplying or dividing by a negative number.

Example: 3 − 2x > 7 → −2x > 4 → x < −2

Double-sided: −1 ≤ 3x + 2 < 11 → subtract 2: −3 ≤ 3x < 9 → −1 ≤ x < 3

Integer Solutions

List all integers satisfying the inequality. −1 ≤ x < 3: integers are {−1, 0, 1, 2}

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