Mathematics (Specification B) | Algebra

Welcome to Algebra: The Language of Mathematics!

Hello there! Don't worry if the word 'Algebra' makes your brain feel fuzzy. Algebra is simply the language we use to solve puzzles and describe relationships using letters (called variables) instead of just numbers. It’s like learning a secret code that unlocks the rest of high-level mathematics!

In this chapter, we will break down key algebraic techniques used in Specification B, making sure that every step, from simplifying basic expressions to solving complex quadratic equations, is crystal clear. Let’s dive in!

1. Algebraic Fundamentals: Simplifying and Substituting

1.1 Understanding Terms and Expressions

An expression is a mathematical phrase containing numbers, variables, and operators (like +, -, x, /). It does NOT have an equals sign (e.g., \(3x^2 - 5y + 1\)).

A term is a single part of an expression separated by a plus or minus sign (e.g., in \(3x^2 - 5y + 1\), the terms are \(3x^2\), \(-5y\), and \(1\)).

1.2 Simplifying Expressions (Collecting Like Terms)

You can only add or subtract terms that are like terms. This means they must have exactly the same letters (variables) and those letters must have the same power (index).

Analogy: Imagine 'x' terms are apples and 'y' terms are bananas. You can't add an apple to a banana!

Example: Simplify \(5a + 3b - 2a + 7\)

Step 1: Identify like terms.
(The 'a' terms are \(5a\) and \(-2a\)).
(The 'b' term is \(3b\)).
(The number term is \(7\)).

Step 2: Combine them.
\(5a - 2a = 3a\)
Result: \(3a + 3b + 7\) (This cannot be simplified further!)

1.3 Substitution

Substitution means replacing the variable (letter) with a given number. Always use brackets when substituting negative numbers to avoid sign errors.

Example: If \(x = 4\) and \(y = -2\), find the value of \(P = 3x - y^2\).

Step 1: Replace \(x\) with \(4\) and \(y\) with \(-2\).
\(P = 3(4) - (-2)^2\)

Step 2: Calculate using the order of operations (BIDMAS/BODMAS).
\(P = 12 - (4)\) (Remember: \((-2)^2 = (-2) \times (-2) = +4\)
\(P = 12 - 4\)
\(P = 8\)

Quick Takeaway: Algebra is organised counting! Only combine terms that match exactly (same letters, same powers). Substitution is just careful replacement.

2. Expanding Brackets (Multiplication)

Expanding means multiplying out the expression to remove the brackets. We use the Distributive Law: everything inside the bracket must be multiplied by everything outside.

2.1 Single Brackets

Multiply the term outside by every term inside. Pay careful attention to negative signs!

Example: Expand \(3(2x - 5)\)
\(3 \times 2x = 6x\)
\(3 \times (-5) = -15\)
Result: \(6x - 15\)

Example with Negatives: Expand \(-4(y - 3)\)
\(-4 \times y = -4y\)
\(-4 \times (-3) = +12\) (A negative times a negative is a positive!)
Result: \(-4y + 12\)

2.2 Double Brackets (Multiplying Two Binomials)

When you multiply two sets of brackets, you must ensure every term in the first bracket multiplies every term in the second.

Memory Aid: FOIL
This mnemonic helps you remember the four multiplications needed for brackets like \((a+b)(c+d)\):

• First terms
• Outer terms
• Inner terms
• Last terms

Example: Expand \((x + 2)(x - 5)\)

First: \(x \times x = x^2\)
Outer: \(x \times (-5) = -5x\)
Inner: \(2 \times x = +2x\)
Last: \(2 \times (-5) = -10\)

Combine the four results: \(x^2 - 5x + 2x - 10\)

Step 3: Simplify the like terms (\(-5x + 2x\)):
Result: \(x^2 - 3x - 10\)

2.3 Expanding Squared Brackets

If you see \((x+3)^2\), DO NOT write \(x^2 + 9\). This is a common mistake!

Remember that squaring means multiplying the term by itself:
\((x+3)^2 = (x+3)(x+3)\)

Then use FOIL to expand it correctly!

Quick Takeaway: Expanding is distribution. Use FOIL for double brackets and always collect the middle terms to simplify the final answer.

3. Factorisation

Factorising is the opposite of expanding. We are aiming to put brackets back into the expression.

3.1 Factorising by finding the Highest Common Factor (HCF)

Look for the biggest number and/or the highest power of the variable that divides into all terms. This HCF goes outside the bracket.

Example: Factorise \(6x^2 + 15x\)

Step 1: Find HCF of 6 and 15. HCF is 3.
Step 2: Find the common power of x. The lowest power is \(x^1\).
The total HCF is \(3x\).

Step 3: Divide each original term by the HCF:
\(6x^2 \div 3x = 2x\)
\(15x \div 3x = 5\)

Result: \(3x(2x + 5)\)

3.2 Factorising Simple Quadratic Expressions (\(x^2 + bx + c\))

We are looking for two numbers that:

1. Multiply together to make the constant term (\(c\)).
2. Add/Subtract together to make the middle coefficient (\(b\)).

Example: Factorise \(x^2 + 6x + 8\)
We need two numbers that multiply to 8 and add to 6.

Factors of 8: (1, 8), (2, 4).
Which pair adds to 6? 2 and 4.

Result: \((x + 2)(x + 4)\)

3.3 Factorising the Difference of Two Squares (DOTS)

This is a special, fast method. If you have a squared term minus another squared term, the factors always follow a specific pattern:

Rule: \(a^2 - b^2 = (a - b)(a + b)\)

Example: Factorise \(x^2 - 49\)
This is \(x^2 - 7^2\).
Result: \((x - 7)(x + 7)\)

Quick Takeaway: Always look for the HCF first. For quadratics, the signs are critical: if the last term (c) is negative, your brackets will have one plus and one minus.

4. Solving Equations and Inequalities

The goal when solving an equation (which contains an equals sign, \(=\)) is to isolate the variable on one side.

4.1 Solving Linear Equations

Use the 'balancing' method. Whatever you do to one side of the equation, you must do to the other side. You are effectively "undoing" the operations to leave the variable alone.

Example: Solve \(5x - 3 = 17\)

1. Add 3 to both sides (to undo the -3):
   \(5x = 17 + 3\) -> \(5x = 20\)
2. Divide both sides by 5 (to undo the multiplication):
   \(x = 20 / 5\)
3. Result: \(x = 4\)

4.2 Solving Quadratic Equations

A quadratic equation has \(x^2\) as the highest power and must be set to zero: \(ax^2 + bx + c = 0\). You will usually get two solutions for \(x\).

Method A: Factorisation

If the equation can be factorised, set each resulting bracket equal to zero.

Example: Solve \(x^2 - 3x - 10 = 0\)

1. Factorise the quadratic (two numbers that multiply to -10 and add to -3: -5 and +2):
   \((x - 5)(x + 2) = 0\)
2. Set each bracket equal to zero and solve:
   \(x - 5 = 0\) OR \(x + 2 = 0\)
3. Results: \(x = 5\) OR \(x = -2\)

Method B: The Quadratic Formula

If you cannot factorise or you are asked for answers to a specific degree of accuracy (e.g., 3 significant figures), use the formula.

The Formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] Don't worry! This formula is usually provided in the exam. Just be careful plugging in the values of \(a\), \(b\), and \(c\).

4.3 Simultaneous Equations (Solving Together)

We use two equations with two unknown variables (like \(x\) and \(y\)) to find the unique solution that works for both.

Method A: Elimination (Often easier)

Multiply one or both equations so that the coefficients of one variable (\(x\) or \(y\)) are the same, then add or subtract the equations to eliminate that variable.

Example setup: \begin{align*} 3x + y &= 7 \quad (1) \\ x + y &= 3 \quad (2) \end{align*} Since the \(y\) terms are the same, subtract (2) from (1):
\((3x - x) + (y - y) = (7 - 3)\)
\(2x = 4\)
\(x = 2\)
Substitute \(x=2\) back into (2): \(2 + y = 3\), so \(y = 1\).
Solution: \(x=2\), \(y=1\).

4.4 Solving Linear Inequalities

Inequalities (\(<, >, \le, \ge\)) are solved exactly like equations, with one critical exception.

The Golden Rule: If you multiply or divide both sides of an inequality by a negative number, you must reverse the inequality sign.

Example: Solve \(-2x + 1 > 7\)

1. Subtract 1 from both sides:
   \(-2x > 6\)
2. Divide by -2 and reverse the sign:
   \(x < 6 / -2\)
3. Result: \(x < -3\)

Quick Takeaway: Equations use balancing (or the Quadratic Formula). Inequalities use balancing, but remember to flip the sign if you multiply/divide by a negative.

5. Algebraic Fractions

Working with algebraic fractions uses the exact same rules as working with number fractions.

5.1 Simplifying Algebraic Fractions

To simplify, you must factorise the numerator (top) and the denominator (bottom) first, and then cancel any common factors. You cannot cancel terms that are separated by a plus or minus sign!

Example: Simplify \(\frac{x^2 + 2x}{x^2 - 4}\)

1. Factorise the Numerator (HCF): \(x(x + 2)\)
2. Factorise the Denominator (DOTS): \((x - 2)(x + 2)\)
3. Rewrite the fraction: \(\frac{x(x + 2)}{(x - 2)(x + 2)}\)
4. Cancel the common factor \((x + 2)\):
5. Result: \(\frac{x}{x - 2}\)

5.2 Adding and Subtracting Algebraic Fractions

Just like with numbers, you need a Common Denominator.

Example: Simplify \(\frac{2}{x} + \frac{3}{y}\)

Step 1: The common denominator is \(xy\).

Step 2: Adjust the fractions:
\(\frac{2 \times y}{x \times y} + \frac{3 \times x}{y \times x}\)
\(= \frac{2y}{xy} + \frac{3x}{xy}\)

Step 3: Combine the numerators:
Result: \(\frac{2y + 3x}{xy}\)

Did you know? Algebraic manipulation techniques are essential for physics and engineering, where complex real-world relationships are modeled using fractions and variables!

Quick Takeaway: Factorise before you simplify! When adding/subtracting, find the common denominator and then combine the numerators.

6. Functions and Sequences

6.1 Understanding Function Notation

A function is a rule that links an input value (\(x\)) to an output value. We often write this as \(f(x)\) or \(g(x)\).

Analogy: A function is a machine. You put the input (x) in, the machine follows the rule, and the output \(f(x)\) comes out.

Example: If the function is defined as \(f(x) = 5x - 3\), find \(f(4)\).

We substitute \(x=4\) into the rule:
\(f(4) = 5(4) - 3\)
\(f(4) = 20 - 3\)
Result: \(f(4) = 17\)

6.2 Finding the Inverse Function (\(f^{-1}(x)\))

The inverse function undoes what the original function did.

Step-by-Step for finding \(f^{-1}(x)\):

1. Replace \(f(x)\) with \(y\). (e.g., \(y = 5x - 3\))
2. Swap the \(x\) and \(y\) variables. (e.g., \(x = 5y - 3\))
3. Rearrange the new equation to make \(y\) the subject (isolate \(y\)).
   \(x + 3 = 5y\)
   \(y = \frac{x + 3}{5}\)
4. Replace \(y\) with \(f^{-1}(x)\).
   Result: \(f^{-1}(x) = \frac{x + 3}{5}\)

6.3 Sequences and the \(n^{th}\) Term (Linear)

A sequence is a list of numbers that follow a specific pattern or rule. In Specification B, you need to find the rule for linear sequences (where the difference between terms is constant). This rule is called the \(n^{th}\) term.

Rule structure: \(n^{th} \text{ term} = dn + (a - d)\)
Where \(d\) is the common difference, and \(a\) is the first term.

Example: Find the \(n^{th}\) term for the sequence: 5, 8, 11, 14...

1. Find the Common Difference (\(d\)): The sequence goes up by 3 each time. So, \(d = 3\).
2. The rule starts with \(3n\). (The sequence \(3n\) is 3, 6, 9, 12...)
3. Compare \(3n\) to the original sequence:
   Term 1: 5. Rule \(3(1) = 3\). Difference: \(5 - 3 = 2\).
4. Add the difference to the rule: \(3n + 2\).
5. Result: \(n^{th} \text{ term} = 3n + 2\)

Quick Takeaway: Functions are input-output rules. The inverse function reverses the input and output. For sequences, the \(n^{th}\) term helps you find any term in the sequence.

You've mastered the fundamentals of Algebra! Keep practicing these techniques, especially factorisation and solving simultaneous equations, as they form the backbone of your Specification B exams. You've got this!