1. What Is Algebra? Basic Definitions

Algebra is the branch of mathematics that uses letters (variables) such as x, y and a to stand for unknown or changing numbers.

• Variable: a letter that represents a number (x, y, z).

• Constant: a fixed number (5, −3, π).

• Coefficient: the number multiplying a variable (in 7x, the coefficient is 7).

• Term: a single part of an expression (in 3x² + 5x − 2, the terms are 3x², 5x and −2).

• Expression: a combination of terms with no equals sign (3x + 4).

• Equation: a statement that two expressions are equal (3x + 4 = 19).

• Identity: an equation that is true for every value of the variable (a + b = b + a).

• Formula: an equation that relates quantities, such as Area = πr².

Equation vs. Identity vs. Formula

	Equation	Identity	Formula
True for	Only specific values	All values	Values that fit the real situation
Example	2x + 3 = 11 (only x = 4)	(a + b)² = a² + 2ab + b²	Speed = Distance ÷ Time
Goal	Find the unknown	Simplify, factorise, prove	Substitute and calculate

Remember: an equation is solved, an identity is proved. This difference is one of the most common exam questions.

2. Types of Algebraic Equations

1. Linear equation (degree 1): ax + b = 0. Example: 2x + 6 = 0, so x = −3.

2. Quadratic equation (degree 2): ax² + bx + c = 0. Example: x² − 5x + 6 = 0, so x = 2 or 3.

3. Cubic equation (degree 3): ax³ + bx² + cx + d = 0.

4. Simultaneous (system of) equations: two or more equations solved together.

5. Polynomial equations: equations of any degree.

6. Exponential, rational and radical equations: the variable sits in a power, a denominator or a root.

3. The Golden Rule for Solving Equations

Whatever you do to one side of an equation, do to the other side.

Think of an equation as a balanced weighing scale. If you add 5 kg to the left pan, you must add 5 kg to the right pan or the scale tips.

Step-by-step method: Solve 3x + 7 = 22

1. Subtract 7 from both sides: 3x = 15

2. Divide both sides by 3: x = 5

3. Check: 3(5) + 7 = 22 ✓

Always substitute your answer back into the original equation to verify it.

4. Essential Algebraic Identities (Algebra Formulas List)

These are the most important algebra formulas to memorise for school exams and competitions.

#	Identity	Name
1	(a + b)² = a² + 2ab + b²	Square of a sum
2	(a − b)² = a² − 2ab + b²	Square of a difference
3	a² − b² = (a + b)(a − b)	Difference of two squares
4	(x + a)(x + b) = x² + (a + b)x + ab	Product of two binomials
5	(a + b)³ = a³ + 3a²b + 3ab² + b³	Cube of a sum
6	(a − b)³ = a³ − 3a²b + 3ab² − b³	Cube of a difference
7	a³ + b³ = (a + b)(a² − ab + b²)	Sum of cubes
8	a³ − b³ = (a − b)(a² + ab + b²)	Difference of cubes
9	(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca	Square of a trinomial
10	a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca)	Three-cube identity

Bonus formulas

• Quadratic formula: x = [−b ± √(b² − 4ac)] / 2a

• Discriminant: D = b² − 4ac (D > 0: two real roots; D = 0: one repeated root; D < 0: no real roots)

• Sum of roots = −b/a; product of roots = c/a

• Binomial theorem: (a + b)ⁿ = Σ C(n,k) aⁿ⁻ᵏ bᵏ

5. How to Prove Algebraic Identities

There are several reliable methods. Learn them all, because exams ask for different ones.

Method 1: Algebraic expansion (the distributive law)

Prove (a + b)² = a² + 2ab + b²

(a + b)² = (a + b)(a + b)

= a(a + b) + b(a + b)

= a² + ab + ba + b²

= a² + 2ab + b² ∎

Prove (a − b)² = a² − 2ab + b²

(a − b)² = (a − b)(a − b) = a² − ab − ba + b² = a² − 2ab + b² ∎

Prove a² − b² = (a + b)(a − b)

(a + b)(a − b) = a² − ab + ba − b² = a² − b² ∎

Prove (a + b)³ = a³ + 3a²b + 3ab² + b³

(a + b)³ = (a + b)²(a + b)

= (a² + 2ab + b²)(a + b)

= a³ + a²b + 2a²b + 2ab² + ab² + b³

= a³ + 3a²b + 3ab² + b³ ∎

Prove a³ + b³ = (a + b)(a² − ab + b²)

(a + b)(a² − ab + b²) = a³ − a²b + ab² + a²b − ab² + b³ = a³ + b³ ∎

Method 2: Geometric (area) proof

This is the easiest way for students to see why an identity is true. Take a square with side (a + b) and split it into four regions: one square of area a², one square of area b², and two rectangles of area ab each. The total area is (a + b)², and it is also a² + ab + ab + b². So (a + b)² = a² + 2ab + b².

Title: Diagram - Description: Geometric proof of (a+b) squared

Similarly, (a + b)³ can be pictured as a large cube made of an a³ cube, a b³ cube, three a²b slabs and three ab² slabs. Counting the volumes gives the identity directly.

Method 3: Numerical verification (a quick sanity check, not a formal proof)

Substitute simple numbers to check that you have not made an error. Let a = 3 and b = 2: LHS: (3 + 2)² = 25. RHS: 9 + 12 + 4 = 25 ✓. A numerical check alone does not prove an identity, because one example could be a coincidence. Use it only to catch mistakes.

Method 4: Proof by factorising or transforming one side

Start from the more complicated side and simplify it until it matches the other side. This is how the three-cube identity is proved: expanding (a + b + c)(a² + b² + c² − ab − bc − ca) term by term, the cross terms cancel in pairs, leaving a³ + b³ + c³ − 3abc ∎

Proof of the Quadratic Formula (completing the square)

Start with ax² + bx + c = 0.

1. Divide by a: x² + (b/a)x + c/a = 0

2. Move the constant: x² + (b/a)x = −c/a

3. Add (b/2a)² to both sides: x² + (b/a)x + b²/4a² = b²/4a² − c/a

4. Factor the left side: (x + b/2a)² = (b² − 4ac)/4a²

5. Take square roots: x + b/2a = ±√(b² − 4ac)/2a

6. Therefore x = [−b ± √(b² − 4ac)] / 2a ∎

6. Worked Examples (Step by Step)

Example 1: Using identities for mental maths

• 102² = (100 + 2)² = 10000 + 400 + 4 = 10,404

• 98² = (100 − 2)² = 10000 − 400 + 4 = 9,604

• 53 × 47 = (50 + 3)(50 − 3) = 50² − 3² = 2500 − 9 = 2,491

Example 2: Factorisation

• 4x² − 25 = (2x)² − 5² = (2x + 5)(2x − 5)

• x² + 10x + 25 = x² + 2(x)(5) + 5² = (x + 5)²

• 8a³ + 27b³ = (2a)³ + (3b)³ = (2a + 3b)(4a² − 6ab + 9b²)

Example 3: Solving a quadratic

Solve x² − 7x + 12 = 0. Find two numbers that multiply to 12 and add to −7: −3 and −4. So (x − 3)(x − 4) = 0, giving x = 3 or x = 4.

Example 4: Simultaneous equations

Solve x + y = 10 and x − y = 4. Add the equations: 2x = 14, so x = 7. Then y = 3. Check: 7 + 3 = 10 ✓ and 7 − 3 = 4 ✓.

Example 5: Using an identity to find a hidden value

If a + b = 7 and ab = 12, find a² + b². a² + b² = (a + b)² − 2ab = 49 − 24 = 25. This is one of the most frequent exam patterns, so practise it.

7. Real-Life Examples of Algebraic Equations

Students learn faster when they can connect algebra to daily life. Here are real situations.

1. Shopping and discounts (linear equation). A jacket costs $80 and is on sale for 25% off. The price you pay is P = 80 − 0.25(80) = $60. Reverse the question: if you paid $45 after a 25% discount, the original price x satisfies 0.75x = 45, so x = $60.

2. Mobile phone plans (linear equation). Plan A costs $10 plus $0.05 per minute and Plan B costs $0.15 per minute. They cost the same when 10 + 0.05m = 0.15m, so m = 100 minutes. Algebra tells you which plan is cheaper for your usage.

3. Travel time (formula). Distance = Speed × Time. A train travelling at 80 km/h takes t = 320/80 = 4 hours to cover 320 km.

4. Cooking and recipes (proportion). A recipe for 4 people needs 300 g of rice. For 10 people, x/10 = 300/4, so x = 750 g.

5. Area of a garden (identity). A square garden has a side of (a + b) metres. Its area is (a + b)², which is the sum of a flower bed (a²), a vegetable bed (b²) and two paths (ab each). This is exactly the geometric proof above.

6. Throwing a ball (quadratic equation). The height of a ball is h = −5t² + 20t. It lands when h = 0, so −5t(t − 4) = 0 and t = 4 seconds. Sports scientists, engineers and game designers use this kind of model.

7. Bank interest (formula). Compound interest is A = P(1 + r/n)^(nt). With $1,000 at 5% for 2 years, A = 1000(1.05)² = $1,102.50. The expansion (1 + r)² uses the (a + b)² identity.

8. Temperature conversion (linear formula). F = (9/5)C + 32. For 25 °C, F = 45 + 32 = 77 °F.

9. Computer science and cryptography. Algebraic identities and polynomial equations sit behind data encryption, error-correcting codes and computer graphics.

10. Medicine. Doctors use formulas such as Dosage = Body weight × mg per kg to calculate medicine amounts.

8. Common Mistakes Students Make

1. Writing (a + b)² = a² + b². This is wrong. You must include the middle term 2ab.

2. Sign errors: −(x − 3) = −x + 3, not −x − 3.

3. Forgetting to apply an operation to both sides of an equation.

4. Dividing by a variable that could be zero, which loses solutions.

5. Mixing up a³ + b³ and (a + b)³. They are different expressions.

6. Not checking the answer by substitution.

9. Quick Memory Tricks

• “First squared, last squared, twice the product” gives a² + 2ab + b².

• SOAP for cubes: in a³ ± b³ = (a ± b)(a² ∓ ab + b²), the signs go Same, Opposite, Always Positive.

• Difference of squares: “conjugates multiply to a difference.”

• Discriminant: positive means two answers, zero means one, negative means none (in real numbers).

10. Practice Questions and Answers (Beginner to Intermediate)

Q1. Solve 5x − 9 = 26.

Answer: 5x = 35, so x = 7.

Q2. Expand (2x + 3y)².

Answer: = 4x² + 12xy + 9y².

Q3. Factorise x² − 49.

Answer: = (x + 7)(x − 7).

Q4. Evaluate 99² using an identity.

Answer: (100 − 1)² = 10000 − 200 + 1 = 9,801.

Q5. If x + 1/x = 5, find x² + 1/x².

Answer: x² + 1/x² = (x + 1/x)² − 2 = 25 − 2 = 23.

Q6. Solve 2x² − 8x = 0.

Answer: 2x(x − 4) = 0, so x = 0 or x = 4.

Q7. Solve x² + 6x + 5 = 0.

Answer: (x + 1)(x + 5) = 0, so x = −1 or −5.

Q8. Find the discriminant and nature of roots of 3x² − 4x + 2 = 0.

Answer: D = 16 − 24 = −8 < 0, so no real roots.

Q9. Simplify (a + b)² − (a − b)².

Answer: = 4ab. (Expand both and subtract.)

Q10. Solve 2x + y = 11 and x − y = 1.

Answer: Add them: 3x = 12, so x = 4 and y = 3.

Q11. Expand (x + 2)³.

Answer: = x³ + 6x² + 12x + 8.

Q12. A rectangle has length (x + 5) and width (x − 2). Its area is 40. Find x.

Answer: (x + 5)(x − 2) = 40, so x² + 3x − 50 = 0. Then x = [−3 + √209]/2 ≈ 5.73 (the negative root is rejected because a length cannot be negative).

11. Global Competition Questions and Answers

These questions are in the style of international contests such as AMC 8/10/12, UKMT, Singapore Math Olympiad, IMO, Canadian Math Olympiad and JEE/NTSE foundation. They are written in that style, not copied from past papers.

C1. (Olympiad style) If a + b = 5 and ab = 3, find a³ + b³.

Answer: a³ + b³ = (a + b)³ − 3ab(a + b) = 125 − 45 = 80.

C2. If x + y = 10 and x² + y² = 58, find xy.

Answer: (x + y)² = x² + y² + 2xy, so 100 = 58 + 2xy and xy = 21.

C3. If a + b + c = 0, prove that a³ + b³ + c³ = 3abc.

Answer: Using the three-cube identity, a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca). Since a + b + c = 0, the right side is 0. So a³ + b³ + c³ = 3abc ∎

C4. Evaluate 2026² − 2025².

Answer: Difference of squares: (2026 + 2025)(2026 − 2025) = 4051 × 1 = 4,051.

C5. Evaluate 1000³ − 999³ − 3 × 1000 × 999.

Answer: Let a = 1000 and b = 999. Then a³ − b³ = (a − b)³ + 3ab(a − b). With a − b = 1 this is 1 + 3ab, so the expression equals 1 + 3ab − 3ab = 1.

C6. If x − 1/x = 4, find x³ − 1/x³.

Answer: x³ − 1/x³ = (x − 1/x)³ + 3(x − 1/x) = 64 + 12 = 76.

C7. (AMC-style) Find the sum of the squares of the roots of x² − 8x + 5 = 0.

Answer: r + s = 8 and rs = 5, so r² + s² = 64 − 10 = 54.

C8. (UKMT-style) Solve for real x: (x² − 5x + 5)^(x² − 9x + 20) = 1.

Answer: The expression equals 1 when: (i) the exponent is 0: x² − 9x + 20 = 0, so x = 4 or 5 (base is 1 and 5, valid); (ii) the base is 1: x² − 5x + 4 = 0, so x = 1 or 4; (iii) the base is −1 with an even exponent: x² − 5x + 6 = 0, so x = 2 or 3, and the exponents are 6 and 2 (both even). Solutions: x = 1, 2, 3, 4, 5.

C9. (Olympiad style) Prove that n⁵ − n is divisible by 30 for every integer n.

Answer: n⁵ − n = n(n − 1)(n + 1)(n² + 1). The factors n − 1, n, n + 1 are three consecutive integers, so the product is divisible by 2 and 3. For 5, check n modulo 5: if n ≡ 0, ±1 then one of n, n − 1, n + 1 is divisible by 5; if n ≡ ±2 then n² + 1 ≡ 5 ≡ 0 (mod 5). So the expression is divisible by 2, 3 and 5, hence by 30 ∎

C10. (Challenge) If a² + b² + c² = ab + bc + ca for real a, b, c, prove a = b = c.

Answer: Multiply by 2 and rearrange: (a − b)² + (b − c)² + (c − a)² = 0. A sum of squares is zero only if every square is zero, so a = b = c ∎

C11. Prove that for positive reals, (a + b)/2 ≥ √(ab).

Answer: (√a − √b)² ≥ 0 gives a − 2√(ab) + b ≥ 0, so a + b ≥ 2√(ab) ∎ (This is the AM–GM inequality, built on the (a − b)² identity.)

C12. Find all integer solutions of x² − y² = 15.

Answer: (x − y)(x + y) = 15. The factor pairs are (1, 15), (3, 5), (−1, −15), (−3, −5) and, with the order swapped, (15, 1), (5, 3), (−15, −1), (−5, −3). Solving x = (sum of factors)/2 and y = (second − first)/2 gives (x, y) = (8, 7), (4, 1), (−8, −7), (−4, −1), (8, −7), (4, −1), (−8, 7), (−4, 1): 8 solutions in total.

12. Frequently Asked Questions (FAQ)

What is the difference between an algebraic equation and an algebraic expression?

An expression has no equals sign and cannot be solved (3x + 2). An equation has an equals sign and can be solved (3x + 2 = 11).

What are the 3 most important algebraic identities?

(a + b)², (a − b)² and a² − b². Almost every other identity builds on these.

How many algebraic identities should I learn for exams?

Learn the 8 to 10 listed in Section 4. They cover nearly all school-level exams.

Why do we need to prove an identity?

A proof shows an identity holds for all values, not just the ones you tested.

What is the easiest way to remember algebra formulas?

Use the geometric area pictures, say the formula aloud as words, and practise 10 questions per formula.

Is algebra useful in real life?

Yes. Budgeting, programming, engineering, medicine, finance and science all use algebraic equations.

How do I get better at algebra quickly?

Master the fundamentals (signs, fractions, brackets), learn identities with their proofs, practise 20 to 30 minutes daily, and always check your answers.

13. Global-Level Competition Question & Answer

International math competitions love testing symmetric polynomial manipulation and clever substitutions. Here is a challenge problem characteristic of regional mathematical Olympiads.

Competition Challenge

Question: If real numbers $x, y,$ and $z$ satisfy the system of equations:

$x + y + z = 2$
$x^2 + y^2 + z^2 = 6$
$x^3 + y^3 + z^3 = 8$

Find the exact value of the product: xyz

Expert Solution Strategy:
1. Recall the algebraic identity linking sums of powers with elementary symmetric sums:
  $(x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx)$
2. Substitute the known values ( $x + y + z = 2$ and $x^2 + y^2 + z^2 = 6$ ):
  $(2)^2 = 6 + 2(xy + yz + zx)$
  $4 = 6 + 2(xy + yz + zx)$
  $-2 = 2(xy + yz + zx) \implies xy + yz + zx = -1$
3. Now, recall the expanded algebraic identity for the sum of cubes:
  $x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - (xy + yz + zx))$
4. Substitute all our known values into this master identity:
  - $x^3 + y^3 + z^3 = 8$
  - $x + y + z = 2$
  - $x^2 + y^2 + z^2 = 6$
  - $xy + yz + zx = -1$
  Plugging these in:
  $8 - 3xyz = (2) \cdot [6 - (-1)]$
  $8 - 3xyz = 2 \cdot [6 + 1]$
  $8 - 3xyz = 2 \cdot 7$
  $8 - 3xyz = 14$
5. Isolate $3xyz$ :
  $-3xyz = 14 - 8$
  $-3xyz = 6$
  $xyz = \frac{6}{-3} = -2$
Final Answer: -2

14. Comprehensive Practice Q&A for Students

Level 1: Beginner (Evaluating Expressions & Basic Equations)

Question 1: Solve the linear equation for $x$ : $5(x - 3) + 2 = 3(x + 4) - 1$ .

Solution:
1. Distribute the numbers outside the parentheses:
  $5x - 15 + 2 = 3x + 12 - 1$
2. Simplify both sides by combining constants:
  $5x - 13 = 3x + 11$
3. Subtract $3x$ from both sides to gather variables on the left:
  $2x - 13 = 11$
4. Add $13$ to both sides:
  $2x = 24$
5. Divide by $2$ :
  $x = 12$
Answer: $x = 12$ .

Level 2: Intermediate (Applying Algebraic Identities)

Question 2: Simplify the expression using algebraic identities: $(3x + 4)^2 - (3x - 4)^2$ .

Solution:
- Method A (Expanding individually):
  $(9x^2 + 24x + 16) - (9x^2 - 24x + 16)$
  $= 9x^2 + 24x + 16 - 9x^2 + 24x - 16$
  $= 48x$
- Method B (Using the Difference of Squares Identity $A^2 - B^2 = (A-B)(A+B)$ ):
  Let $A = (3x + 4)$ and $B = (3x - 4)$ .
  $[(3x + 4) - (3x - 4)] \cdot [(3x + 4) + (3x - 4)]$
  $= [3x + 4 - 3x + 4] \cdot [3x + 4 + 3x - 4]$
  $= [8] \cdot [6x] = 48x$
Answer: $48x$ .

Level 3: Advanced Word Problem

Question 3: A train travels 300 kilometres at a certain speed. If the train's speed had been 10 km/h faster, the trip would have taken 1 hour less. Find the original speed of the train.

Solution:
1. Let the original speed of the train be $r$ km/h.
2. Original time taken = $\frac{300}{r}$ hours.
3. New speed = $r + 10$ km/h, and new time taken = $\frac{300}{r + 10}$ hours.
4. According to the problem, the new time is 1 hour less than the original time:
  $\frac{300}{r} - \frac{300}{r + 10} = 1$
5. Multiply the entire equation by $r(r + 10)$ to clear the denominators:
  $300(r + 10) - 300r = r(r + 10)$
6. Expand and simplify:
  $300r + 3000 - 300r = r^2 + 10r$
  $3000 = r^2 + 10r$
  $r^2 + 10r - 3000 = 0$
7. Factor the quadratic equation (find two numbers that multiply to $-3000$ and add to $+10$ ):
  $(r + 60)(r - 50) = 0$
8. This yields $r = -60$ or $r = 50$ . Since speed cannot be negative, $r = 50$ .
Answer: The original speed of the train was 50 km/h.

15. Study Plan (7 Days)

Day	Focus
1	Variables, expressions, solving linear equations
2	Squares identities (1–3) with geometric proofs
3	Factorisation using identities
4	Cubic identities and the trinomial identity
5	Quadratic equations and the quadratic formula
6	Simultaneous equations and word problems
7	Competition questions (Section 11) and a timed mock test

16. Key Takeaways

• An equation is solved for specific values; an identity is true for all values and must be proved.

• Memorise the core identities and understand them through expansion, geometry and substitution checks.

• Always keep the equation balanced and check your answer.

• Algebra models real life: shopping, travel, sport, finance, science and technology.

• Competition problems reward spotting patterns: difference of squares, sum and product, completing the square.

Algebraic Equations & Identities: Formulas, Proofs, Examples