Algebraic Expressions & Simple Equations

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Algebraic expressions combine numbers and variables using addition, subtraction, multiplication, and division. An equation states that two expressions are equal and can be solved to find the unknown variable’s value.

For linear equations in one variable, the goal is always to isolate the variable on one side. For 4x − 9 = 23: add 9 to both sides → 4x = 32; divide by 4 → x = 8. Always perform the same operation on both sides to maintain equality.

Key algebraic identities to memorise:

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

⚡ Exam tip: When solving word problems, define the unknown clearly first. If the problem says “three consecutive integers sum to 48,” let the numbers be n, n+1, n+2, giving 3n+3 = 48 → n = 15. The numbers are 15, 16, 17.

⚡ Exam tip: Always check your solution by substituting back into the original equation. This catches algebraic errors before they cost marks.

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🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Linear Equations in One Variable

A linear equation has the form ax + b = c, where a ≠ 0. The solution process involves inverse operations applied in reverse order of PEMDAS (reverse PEMDAS: subtract/add, then divide/multiply).

Solve 7(x − 3) = 2x + 12:

1. Expand: 7x − 21 = 2x + 12
2. Collect x terms: 7x − 2x = 12 + 21
3. Simplify: 5x = 33
4. Divide: x = 33/5 = 6.6

Systems of Linear Equations

Two equations with two unknowns can be solved by substitution or elimination.

Substitution method: From equation 1: y = 3x − 7. Substitute into equation 2: 2x + 3(3x − 7) = 8 → 2x + 9x − 21 = 8 → 11x = 29 → x = 29/11. Then y = 3(29/11) − 7 = 87/11 − 77/11 = 10/11.

Elimination method: Multiply equation 1 by 2 and equation 2 by 3, then subtract to eliminate x. The substitution method is often faster when one variable already has a simple coefficient.

Algebraic Expressions — Manipulation Skills

Factorisation is the reverse of expansion. Key techniques:

Factorising quadratic expressions where a = 1: x² + 5x + 6 = (x + 2)(x + 3) because 2 + 3 = 5 (coefficient of x) and 2 × 3 = 6 (constant).

Factorising by grouping: ax + ay + bx + by = a(x + y) + b(x + y) = (a + b)(x + y). For 2x + 2y + 3x + 3y: group as 2(x + y) + 3(x + y) = 5(x + y).

Common Mistakes to Avoid:

Mistake	Correct approach
Forgetting to multiply every term when expanding brackets	(x + 3)(x − 4) = x² − 4x + 3x − 12 = x² − x − 12
Cancelling terms instead of factors	Can only cancel factors: (x+2)/x+2 is NOT 1, but (x+2)/x+2 = 1 is wrong; must cancel the WHOLE numerator against denominator
Changing signs incorrectly when moving terms	Whatever you do to one side, you MUST do to the other
Not simplifying the final answer	x² − 4 can be factorised to (x+2)(x-2); always factorise if possible

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

The Fundamental Theorem of Algebra and Simple Equations

While the fundamental theorem states that a polynomial of degree n has exactly n complex roots, our focus is on practical solution methods for equations encountered in entrance exams.

Word Problem Strategies

Translating word problems into algebraic equations is a critical skill. Consider: “The sum of three times a number and 17 is 74. What is the number?” Translation: 3x + 17 = 74 → 3x = 57 → x = 19.

More complex example: “A father is four times as old as his son. In 20 years, he will be twice as old. Find their current ages.” Let son’s age = x, father’s age = 4x. In 20 years: 4x + 20 = 2(x + 20) → 4x + 20 = 2x + 40 → 2x = 20 → x = 20. Son is 20, father is 80.

Check: In 20 years, son is 40, father is 100. Father is 2.5 times, not twice. Wait — let me recalculate: 4x + 20 = 2(x + 20) → 4x + 20 = 2x + 40 → 2x = 20 → x = 10. Son is 10, father is 40. In 20 years: son 30, father 60. Yes, father is twice son’s age. Correct.

Partial Fractions

For rational expressions, partial fraction decomposition can simplify integration and equation solving. For 5x + 3 / (x − 1)(x + 2), write as A/(x − 1) + B/(x + 2). Multiply both sides by (x − 1)(x + 2): 5x + 3 = A(x + 2) + B(x − 1). Solve: let x = 1 → 8 = 3A → A = 8/3. Let x = −2 → −7 = −3B → B = 7/3.

Absolute Value Equations

Equations with absolute values require considering both cases. |2x − 3| = 7 gives two cases: 2x − 3 = 7 → 2x = 10 → x = 5, and 2x − 3 = −7 → 2x = −4 → x = −2.

For |x + 1| < 5, this means −5 < x + 1 < 5, so −6 < x < 4.

Historical Context

Algebra traces to the Babylonian mathematician al-Khwarizmi (c. 820 CE), whose treatise “Al-Jabr wa’l-Muqabala” gave us the word “algebra.” The Indian mathematician Brahmagupta (c. 628 CE) gave general solutions to quadratic equations. The symbolic notation we use today, with letters for unknowns, was developed by François Viète in 1591.

Exam Pattern Analysis

UI entrance exam questions frequently combine algebra with geometry or arithmetic. Common patterns include:

1. Age problems (usually solvable with one variable)
2. Work-rate problems (combined work rates add: if A takes 6 hours and B takes 3 hours, together they take 1/(1/6+1/3) = 2 hours)
3. Mixture problems (algebra with two unknowns)
4. Distance-speed-time (d = st, often combined with simultaneous equations)

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