Topic 7: Algebraic Expressions and Equations

🟢 Lite — Quick Review (1h–1d)

Algebraic expressions and equations form the language through which quantitative relationships are described and analysed. An algebraic expression—such as 3x² + 2y - 7—combines numbers (constants), letters (variables), and operation symbols without an equals sign. An equation, by contrast, asserts equality between two expressions, as in 3x + 5 = 14. The fundamental skill in algebra involves manipulating these expressions according to established rules to extract information about unknown quantities—the variables we seek to determine.

Understanding terminology is essential. A term is a single algebraic expression separated from others by addition or subtraction (in 4x² - 3x + 7, the terms are 4x², -3x, and 7). The coefficient is the numerical factor of a term (in 5xy², the coefficient is 5). Like terms are terms containing the same variables raised to the same powers—they can be combined by adding or subtracting their coefficients. The degree of a term is the sum of the exponents of its variables (4x³y² has degree 5). The degree of a polynomial is the highest degree of any term.

Key Facts:

• Combining like terms: add/subtract coefficients, keep variables unchanged
• Expanding brackets: multiply each term inside by the factor outside
• Factorisation: expressing as a product of factors (reverse of expanding)
• The zero-product property: if AB = 0, then A = 0 or B = 0 (or both)
• A linear expression has degree 1; a quadratic has degree 2; a cubic has degree 3
• Substituting values: replace each variable with its given value and evaluate

⚡ Exam Tip: In the WASSCE Objective paper, always check your final answer by substituting it back into the original equation. If it satisfies the equation, your answer is correct.

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🟡 Standard — Regular Study (2d–2mo)

Simplifying Algebraic Expressions

Simplify: 3(2x - 4) - 2(x + 5)

Step 1 — Expand brackets: 3(2x - 4) = 6x - 12; -2(x + 5) = -2x - 10 Step 2 — Rewrite: 6x - 12 - 2x - 10 Step 3 — Group like terms: (6x - 2x) + (-12 - 10) = 4x - 22

Factorising Expressions

Common factor method: Find the highest common factor (HCF) of all terms.

Factorise: 12x² + 8x

• HCF of 12 and 8 is 4
• HCF of x² and x is x
• HCF = 4x
• 12x² + 8x = 4x(3x + 2)

Difference of two squares: a² - b² = (a + b)(a - b)

Factorise: x² - 16 = (x + 4)(x - 4)

Trinomial factorisation: For ax² + bx + c, find two numbers that multiply to ac and add to b.

Factorise: x² + 7x + 12

• Two numbers multiplying to 12 and adding to 7 are 3 and 4
• x² + 7x + 12 = (x + 3)(x + 4)

Comparison Table: Factorisation Methods

Expression Type	Method	Example
Common factor	Take HCF outside brackets	6x² + 9x = 3x(2x + 3)
Difference of squares	(a+b)(a-b)	x² - 25 = (x+5)(x-5)
Perfect square trinomial	(a±b)²	x² + 6x + 9 = (x+3)²
General trinomial	Find pair summing to b	x² + 5x + 6 = (x+2)(x+3)
Grouping	Group terms to reveal common factor	x² + 3x + 2x + 6 = (x+2)(x+3)

Evaluating Expressions

If x = -3, y = 4, evaluate: 2x² - 3y + 5

2(-3)² - 3(4) + 5 = 2(9) - 12 + 5 = 18 - 12 + 5 = 11

Note: (-3)² = 9, not -9. Squaring a negative number always yields a positive result.

Formulating Expressions from Word Problems

“The cost of a notebook is GH₵ x and a pen is GH₵ y. Express the cost of 4 notebooks and 3 pens.”

Total cost = 4(x) + 3(y) = 4x + 3y cedis

Common Mistakes to Avoid:

1. Forgetting to multiply every term inside brackets when expanding
2. Confusing 3(x + 2) with x + 3 × 2
3. Failing to change signs when subtracting expressions
4. Forgetting that subtraction of a negative becomes addition
5. Incorrectly identifying like terms (e.g., x² and x are NOT like terms)

Problem-Solving Strategy:

1. Read the problem carefully and identify what is unknown
2. Translate words into algebraic symbols
3. Form the equation or expression
4. Simplify both sides before attempting to solve
5. Solve for the variable
6. Check your answer by substitution

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🔴 Extended — Deep Study (3mo+)

Algebraic Proof Techniques

Algebra provides powerful tools for proving general statements. Consider proving that the difference between the squares of two consecutive integers is always odd.

Let n and n+1 be consecutive integers. (n+1)² - n² = (n² + 2n + 1) - n² = 2n + 1

Since 2n is always even, 2n + 1 is always odd. Q.E.D.

The Binomial Theorem (Basic)

For non-negative integer n: (a + b)ⁿ = Σ C(n,r) aⁿ⁻ʳ bʳ where C(n,r) = n!/[r!(n-r)!]

Key binomial expansions:

• (a + b)² = a² + 2ab + b²
• (a - b)² = a² - 2ab + b²
• (a + b)³ = a³ + 3a²b + 3ab² + b³
• (a - b)³ = a³ - 3a²b + 3ab² - b³

Advanced Factorisation: Quadratic Trinomials

For factorising expressions where the coefficient of x² is not 1:

Factorise: 2x² + 7x + 3

1. Multiply coefficient of x² (2) by constant (3) = 6
2. Find two numbers multiplying to 6 and adding to 7 → 6 and 1
3. Rewrite 7x as 6x + x: 2x² + 6x + x + 3
4. Group: (2x² + 6x) + (x + 3) = 2x(x + 3) + 1(x + 3)
5. Factor: (2x + 1)(x + 3)

Remainder and Factor Theorems

Remainder Theorem: When polynomial P(x) is divided by (x - a), the remainder is P(a).

If P(x) = x³ - 4x² + 2x + 1, find the remainder when divided by (x - 2): P(2) = 8 - 16 + 4 + 1 = -3

Factor Theorem: (x - a) is a factor of P(x) if and only if P(a) = 0.

Partial Fractions

Expressing a rational expression as a sum of simpler fractions:

Express (5x + 3)/(x² - x - 2) in partial fractions.

x² - x - 2 = (x - 2)(x + 1)

Let (5x + 3)/(x² - x - 2) = A/(x-2) + B/(x+1)

5x + 3 = A(x + 1) + B(x - 2) Setting x = 2: 5(2) + 3 = 13 = A(3) → A = 13/3 Setting x = -1: 5(-1) + 3 = -2 = B(-3) → B = 2/3

Thus: (5x + 3)/(x² - x - 2) = (13/3)/(x-2) + (2/3)/(x+1)

WASSCE Examination Patterns:

WASSCE questions in this topic typically require:

1. Simplifying algebraic expressions (Objective)
2. Factorising given expressions (Objective and Theory)
3. Evaluating expressions for given values (Objective)
4. Solving linear and quadratic equations (Theory)
5. Formulating and solving equations from word problems (Theory)

Algebraic Identities to Memorise:

• (a + b)² = a² + 2ab + b²
• (a - b)² = a² - 2ab + b²
• (a + b)(a - b) = a² - b²
• (a + b)³ = a³ + 3a²b + 3ab² + b³
• a³ + b³ = (a + b)(a² - ab + b²)
• a³ - b³ = (a - b)(a² + ab + b²)

⚡ Pro Exam Tip: In WASSCE Theory paper, always show your working for equation-solving questions. Partial credit is awarded for correct methodology even if arithmetic errors occur. When factorising quadratics, always check by expanding your answer mentally.

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