Algebra & Equations

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):

1. Expand Brackets: Distribute any numbers outside the brackets.

    `5x - 10 = 2x + 8`

1. 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`

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

    `3x = 8 + 10`
    `3x = 18`

1. 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.

1. Subtract at from both sides: v - at = u
2. 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.

1. Equations:

    (1) `3x + 2y = 18`
    (2) `5x - 2y = 14`

1. 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`

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

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

1. Solution: x = 4, y = 3

Method 2: Substitution

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

1. Equations:

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

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

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

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

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

1. 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

1. Move constant: x² + 6x = 5
2. Half the x-coefficient, square it, and add to both sides: (6/2)² = 3² = 9.

        `x² + 6x + 9 = 5 + 9`
    3.  Factor the perfect square: `(x + 3)² = 14`
    4.  Square root both sides: `x + 3 = ±√14`
    5.  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.