Basic Algebra and Equations

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Algebra uses letters to represent unknown values. An equation states that two expressions are equal, and solving it means finding the value(s) of the unknown that make the statement true. The fundamental principle: whatever you do to one side of an equation, you must do to the other.

Essential Skills:

• Simplify algebraic expressions by collecting like terms
• Expand brackets: a(b + c) = ab + ac
• Factorise by finding common factors or recognising patterns
• Solve linear equations (unknown to power 1)
• Solve quadratic equations (unknown to power 2)
• Form and solve equations from word problems

Key Formulas:

• Linear: ax + b = c → x = (c − b)/a
• Quadratic: ax² + bx + c = 0 → x = (−b ± √(b² − 4ac)) / 2a
• Difference of squares: a² − b² = (a+b)(a−b)
• Perfect square: (a ± b)² = a² ± 2ab + b²

Linear Equation Solution: 3x + 7 = 16 → 3x = 9 → x = 3.

Quadratic Solution: x² − 5x + 6 = 0 → (x−2)(x−3) = 0 → x = 2 or x = 3.

⚡ Exam Tip: When a word problem gives you two conditions, form two equations in two unknowns. Solve using substitution or elimination. Always check your solutions in the original problem context (e.g., a negative length is not physically meaningful).

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🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Simplifying Algebraic Expressions

Collect like terms — terms with the same variable raised to the same power.

Example: 3x² + 5x − 2x² + 4 − x = (3x² − 2x²) + (5x − x) + 4 = x² + 4x + 4.

Expanding Brackets

Multiply every term inside the bracket by the term outside.

Example: 3(2x − 5) = 6x − 15. Example: −2x(3x² − x + 4) = −6x³ + 2x² − 8x.

For double brackets: (x + 3)(x − 2) = x² − 2x + 3x − 6 = x² + x − 6.

Factorisation

Factorisation is the reverse of expansion.

1. Common factor: 6x² + 9x = 3x(2x + 3).
2. Difference of two squares: x² − 16 = (x+4)(x−4).
3. Trinomial: x² + 5x + 6 = (x+2)(x+3) because 2+3=5 and 2×3=6.
4. Grouping: ax + ay + bx + by = (a+b)(x+y).

Solving Linear Equations

A linear equation has the unknown to power 1.

Example: 4(x − 3) = 2x + 8 → 4x − 12 = 2x + 8 → 4x − 2x = 8 + 12 → 2x = 20 → x = 10.

Solving Quadratic Equations

Three methods: factorisation, completing the square, or quadratic formula.

Factorisation: x² + 2x − 8 = 0 → (x+4)(x−2) = 0 → x = −4 or x = 2. Quadratic formula: Always works. For 2x² + 5x − 3 = 0: x = (−5 ± √(25 − 4(2)(−3))) / (2×2) = (−5 ± √(25+24)) / 4 = (−5 ± √49) / 4 = (−5 ± 7) / 4. So x = (−5+7)/4 = 2/4 = 0.5, or x = (−5−7)/4 = −12/4 = −3.

Simultaneous Equations — Two Variables

Two equations, two unknowns.

Substitution method: Equation 1: y = 2x + 1 Equation 2: 3x + y = 11 Substitute: 3x + (2x+1) = 11 → 5x = 10 → x = 2. Then y = 2(2)+1 = 5.

Elimination method: 3x + 2y = 16 … (1) 2x − y = 3 … (2) Multiply (2) by 2: 4x − 2y = 6 … (3) Add (1)+(3): 7x = 22 → x = 22/7 ≈ 3.14. From (2): 2(22/7) − y = 3 → y = 44/7 − 3 = 23/7 ≈ 3.29.

Forming Equations from Word Problems

Example: “A rectangle is 4 cm longer than it is wide. Its perimeter is 28 cm. Find its dimensions.” Let width = w cm. Length = w + 4. Perimeter = 2(length + width) = 2(w+4 + w) = 2(2w+4) = 4w + 8 = 28. 4w = 20 → w = 5 cm. Length = 9 cm. Check: 2(9+5) = 28 ✓.

Algebraic Fractions

Simplify: (x² − 9)/(x² + 5x + 6) ÷ (x−3)/(x+2) = [(x+3)(x−3)]/[(x+2)(x+3)] × (x+2)/(x−3) = 1. Domain: x ≠ −3, −2, 3.

Problem-Solving Strategies:

• In simultaneous equation word problems, identify what two things are unknown and form two equations relating them
• For quadratic word problems, form the equation, solve it, then check both solutions in context — sometimes one is extraneous (e.g., a negative age)
• When the equation has a squared term, check whether both positive and negative roots make sense

Common Mistakes:

• Forgetting to multiply every term inside brackets when expanding
• Making sign errors when moving terms across the equals sign
• In simultaneous equations, forgetting to multiply the correct equation
• Cancelling terms that are added rather than multiplied: (x+3)/(x+5) cannot cancel to 3/5

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Quadratic Formula — Derivation

Starting from ax² + bx + c = 0 (a ≠ 0):

1. ax² + bx = −c
2. Divide by a: x² + (b/a)x = −c/a
3. Complete square: x² + (b/a)x + (b/2a)² = −c/a + (b/2a)²
4. Left = (x + b/2a)². Right = (b² − 4ac) / 4a²
5. (x + b/2a) = ±√(b²−4ac) / 2a
6. x = (−b ± √(b²−4ac)) / 2a ✓

Nature of Quadratic Roots

The discriminant Δ = b² − 4ac determines the roots:

• Δ > 0: two distinct real roots
• Δ = 0: two equal real roots (repeated root)
• Δ < 0: no real roots (complex conjugates)

This is useful for determining whether a quadratic equation is solvable without actually solving it.

Sum and Product of Roots

For ax² + bx + c = 0 with roots α and β: α + β = −b/a αβ = c/a

Example: Find the sum and product of roots of 2x² − 7x + 3 = 0. Sum = −(−7)/2 = 7/2 = 3.5. Product = 3/2 = 1.5. Check by solving: discriminant = 49 − 24 = 25. Roots = (7±5)/4 = 3 or 0.5. Sum = 3.5, product = 1.5 ✓.

Word Problems Leading to Quadratic Equations

Example: “A garden measuring 12 m by 5 m is surrounded by a path of uniform width. The area of the path is 24 m². Find the width of the path.” Let path width = x metres. Outer dimensions: (12+2x) by (5+2x). Area of outer rectangle = (12+2x)(5+2x) = 60 + 24x + 10x + 4x² = 4x² + 34x + 60. Area of path = outer − inner = 4x² + 34x + 60 − 60 = 4x² + 34x = 24. 4x² + 34x − 24 = 0 → 2x² + 17x − 12 = 0. x = (−17 ± √(289 + 96)) / 4 = (−17 ± √385) / 4 ≈ (−17 ± 19.62) / 4. Positive root: (2.62)/4 ≈ 0.655 m. Negative root: (−36.62)/4 ≈ −9.15 (reject). So path width ≈ 0.655 m ≈ 65.5 cm.

Functions — Basics

A function f maps each input to exactly one output. f(x) = 2x + 3. Domain: all valid inputs. Range: all valid outputs. Composite function: f(g(x)) means apply g first, then f. Inverse function f⁻¹: the function that undoes f. If y = 2x + 3, then x = (y − 3)/2, so f⁻¹(x) = (x − 3)/2.

Linear Inequalities

Solve 3x − 5 < 7: 3x < 12 → x < 4. For inequalities with negative coefficient of x, reverse the inequality when dividing. −2x > 8 → x < −4 (divide by −2, reverse > to <).

For quadratic inequalities, find the roots and test intervals. x² − x − 6 < 0 → (x−3)(x+2) < 0. Roots at x = 3 and x = −2. Test x = 0: (0−3)(0+2) = (−3)(2) = −6 < 0 ✓. So solution is −2 < x < 3.

UI Entrance Exam Patterns

Algebra questions in the UI Academic Potential test typically include:

1. Simplifying algebraic expressions
2. Solving linear equations (including those arising from word problems)
3. Solving quadratic equations by factorisation or formula
4. Simultaneous equations
5. Evaluating expressions by substitution

⚡ Exam Strategy: When a quadratic equation looks difficult to factor, go straight to the quadratic formula. The discriminant calculation is straightforward and reliable. For word problems, write down what the variable represents before forming the equation.

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