Topic 2 of 4Pearson EdExcel

Algebra & Equations

Master solving linear, simultaneous, and quadratic equations and applying them to real-world scenarios.

### Introduction to Algebra & Equations

Algebra is the language of mathematics that uses letters (like x and y) to represent unknown values. It allows us to build mathematical models of real-world situations, from calculating a shop's profit to planning a satellite's orbit. This topic focuses on solving different types of equations, which are the fundamental tools for finding these unknown values.

### 1. Linear Equations and Inequalities

A linear equation involves one or more variables raised to the power of 1. The goal is to find the value of the unknown that makes the equation true.

Solving a Linear Equation (e.g., 5(x - 2) = 2x + 8):

Expand Brackets: Distribute any numbers outside the brackets.

`5x - 10 = 2x + 8`

Collect Variable Terms: Move all terms with the variable (x) to one side of the equation. It's often easiest to move the smaller term to avoid negatives.

`5x - 2x - 10 = 8`

`3x - 10 = 8`

Collect Constant Terms: Move all constant numbers to the other side.

`3x = 8 + 10`

`3x = 18`

Isolate the Variable: Divide by the coefficient of the variable.

`x = 18 / 3`

`x = 6`

Linear Inequalities work similarly, but use symbols like < (less than), > (greater than), ≤ (less than or equal to), and ≥ (greater than or equal to).

Key Rule for Inequalities: The process for solving is the same as for equations, with one critical exception. If you multiply or divide both sides of an inequality by a negative number, you must reverse the direction of the inequality sign.

Example: `-2x < 10` → `x > 10 / -2` → `x > -5`

Solutions are often represented on a number line. Use an open circle (○) for < and > to show the endpoint is not included, and a closed circle (●) for ≤ and ≥ to show it is included.

### 2. Rearranging Formulae

This skill involves changing the subject of a formula. It uses the same principles as solving equations: apply inverse operations to both sides to isolate the desired variable.

Example: Make 'u' the subject of `v = u + at`.

Subtract `at` from both sides: `v - at = u`

Therefore, `u = v - at`.

### 3. Simultaneous Linear Equations

These are pairs of equations with two unknowns (e.g., x and y). The solution is the pair of values that satisfies *both* equations at the same time. Graphically, this is the point where two straight lines intersect.

Method 1: Elimination

Use this when the coefficients of one variable are the same or multiples of each other.

Equations:

(1) `3x + 2y = 18`

(2) `5x - 2y = 14`

Eliminate: The coefficients of y are `+2` and `-2`. Adding the equations will eliminate `y`.

`(3x + 5x) + (2y - 2y) = (18 + 14)`

`8x = 32` → `x = 4`

Substitute: Put `x = 4` into either original equation (e.g., eq 1).

`3(4) + 2y = 18` → `12 + 2y = 18` → `2y = 6` → `y = 3`

Solution: `x = 4, y = 3`

Method 2: Substitution

Use this when one equation can be easily rearranged to make one variable the subject.

Equations:

(1) `y = 2x - 1`

(2) `4x + 3y = 23`

Substitute: Substitute the expression for `y` from (1) into (2).

`4x + 3(2x - 1) = 23`

Solve: Now it's a linear equation with one variable.

`4x + 6x - 3 = 23` → `10x = 26` → `x = 2.6`

Back-substitute: Put `x = 2.6` into the rearranged equation (1).

`y = 2(2.6) - 1` → `y = 5.2 - 1` → `y = 4.2`

Common Exam Trap: When subtracting equations in the elimination method, remember to subtract *every* term, paying close attention to negative signs. For example, `(4y) - (-y)` becomes `5y`.

### 4. Quadratic Equations (ax² + bx + c = 0)

A quadratic equation is an equation of degree 2. It can have up to two distinct real solutions (roots).

Method 1: Factorising

This is the quickest method if the quadratic expression can be factored.

Example: `x² - 5x + 6 = 0`

We need two numbers that multiply to `+6` and add to `-5`. These are `-2` and `-3`.

`(x - 2)(x - 3) = 0`

For the product to be zero, one of the brackets must be zero.

`x - 2 = 0` or `x - 3 = 0`

Solutions: `x = 2` or `x = 3`

Method 2: Completing the Square

This method converts one side of the equation into a perfect square trinomial.

Example: `x² + 6x - 5 = 0`

Move constant: `x² + 6x = 5`

Half the x-coefficient, square it, and add to both sides: `(6/2)² = 3² = 9`.

`x² + 6x + 9 = 5 + 9`

Factor the perfect square: `(x + 3)² = 14`

Square root both sides: `x + 3 = ±√14`

Isolate x: `x = -3 ± √14`

Method 3: The Quadratic Formula

This formula solves any quadratic equation. It is essential to memorise.

For `ax² + bx + c = 0`, the solution is: `x = (−b ± √(b² − 4ac)) / 2a`

The Discriminant (b² - 4ac)

The part inside the square root, `b² - 4ac`, is the discriminant. It tells us about the nature of the roots without fully solving:

If `b² - 4ac > 0`, there are two distinct real roots (the parabola crosses the x-axis twice).

If `b² - 4ac = 0`, there is one repeated real root (the parabola touches the x-axis at one point).

If `b² - 4ac < 0`, there are no real roots (the parabola never crosses the x-axis).

### Real-World Applications

Algebraic equations are used to model problems in finance, engineering, and daily life. The key is to translate a word problem into a mathematical equation.

Example: A mobile phone plan in Pakistan costs PKR 500 per month plus PKR 2 for every MB of data over the included limit. If the total bill was PKR 850, how much data was used over the limit?

Let `d` be the data used over the limit (in MB).

Equation: `500 + 2d = 850`

Solve: `2d = 350` → `d = 175 MB`.

Key Points to Remember

1FOIL: (a+b)(c+d) = ac + ad + bc + bd
2Factorising quadratics: find two numbers that multiply to c and add to b
3Quadratic formula works when factorising is difficult
4Flip inequality sign when multiplying/dividing by a negative

Pakistan Example

Profit Maximisation — A Karachi Shopkeeper's Algebra

A Saddar shopkeeper sells umbrellas. Each costs Rs. 200 and sells for Rs. 200 + p. Daily sales are (300 − p) umbrellas. Profit = p(300 − p) = 300p − p². To maximise: complete the square or use vertex formula → p = 150. Optimal profit = 150 × 150 = Rs. 22,500/day. Forming and solving quadratic equations solves real business problems every day in Karachi.

Quick Revision Infographic

Mathematics — Quick Revision