A complete guide to algebra in three levels: notation, brackets and factorising, then equations, inequalities and quadratics, then sequences, graphs, functions, algebraic fractions, variation and polynomials.
Algebra is the language of patterns, relationships and unknown quantities. It can look like arithmetic with letters added, but its real purpose is more powerful: it lets you describe a general rule, solve for an unknown, model a real situation and reason about quantities that can change. Algebra is the foundation under equations, graphs, sequences, functions, coordinate geometry, variation, indices and surds. This guide builds the subject in three levels. Basic algebra covers notation, substitution, like terms and brackets. Intermediate algebra covers factorising, indices, surds, equations, inequalities, simultaneous equations and quadratics. Advanced algebra covers sequences, straight-line graphs, functions, algebraic fractions, variation and polynomials. It has 30 worked examples checked by computer, five diagrams, three tools, a practice drill, exam questions with answers, and a 20-question quiz.
Arithmetic asks you to calculate with known numbers. Algebra introduces variables so that one statement can stand for many possible values. In 3x + 5, the letter x represents a number whose value may be found later, may be restricted by a condition, or may be allowed to vary.
| Idea | Example | Meaning |
|---|---|---|
| Variable | x | A quantity whose value can vary or is unknown |
| Constant | 7 | A fixed number |
| Coefficient | 5 in 5x | The number multiplying a variable |
| Term | −3x² | One part of an expression, separated by + or − |
| Expression | 2x² + 3x − 1 | A combination of terms, with no equals sign |
| Equation | 2x + 1 = 9 | A statement that two expressions are equal |
| Inequality | 2x + 1 > 9 | A comparison that gives a range of values |
| Formula | A = ½bh | An equation that links real quantities |
“Five more than twice a number n” is 2n + 5. “Seven less than three times x” is 3x − 7. “The product of x and the number after x” is x(x + 1). Take care with order: “five less than x” is x − 5, not 5 − x.
Multiplication is written without a sign: 3 × x = 3x, and xy means x × y. A square means a number multiplied by itself, so x² = x × x. The expression 2x² means 2 × x², not (2x)². Substitution means replacing a variable with a known value. Put the whole value in brackets, especially when it is negative or a fraction.
If x = −3, evaluate 2x² − 5x + 1. Substitute: 2(−3)² − 5(−3) + 1 = 2(9) + 15 + 1 = 34. The brackets matter: (−3)² = 9, but −3² is read as −(3²) = −9.
In the formula A = ½bh, with b = 8 and h = 5: A = ½(8)(5) = 20 square units. Always state the unit when a formula describes a physical quantity.
Like terms have exactly the same variable part, including the same powers. So 3x and −7x are like terms, but 3x and 3x² are not. To simplify, add the coefficients and keep the variable part unchanged. This is different from multiplying: 3x + 5x = 8x, but 3x × 5x = 15x².
4x + 7 − 2x + 3 = (4x − 2x) + (7 + 3) = 2x + 10.
5a² − 3a + 2a² + 8a = 7a² + 5a.
3xy + 5x − 2xy − x = xy + 4x.
The expression x + x² cannot be written as 2x² or 2x, because x and x² are unlike terms. It is already as simple as it can be.
The distributive law says a(b + c) = ab + ac. Multiply every term inside the bracket by the factor outside, and watch the signs. For two brackets, multiply every term in the first by every term in the second.
3(2x − 5) = 6x − 15. −2(4x + 3) = −8x − 6.
(x + 4)(x + 3) = x² + 3x + 4x + 12 = x² + 7x + 12.
(2x − 1)(x + 5) = 2x² + 10x − x − 5 = 2x² + 9x − 5.
Three identities are worth learning because they save time:
(x + 3)² = x² + 6x + 9. (x − 5)² = x² − 10x + 25. (x + 7)(x − 7) = x² − 49.
Test it with x = 2: what do you get on each side?
The fix: (x + 3)² means (x + 3)(x + 3), so the middle term 2 × x × 3 = 6x is needed: x² + 6x + 9. With x = 2 the left side is 5² = 25, but x² + 9 gives only 13. A square of a sum is not the sum of the squares.
Factorising is the reverse of expanding: it writes an expression as a product. Always look first for a common factor, and take out the greatest one.
6x² + 9x = 3x(2x + 3).
xy + 2x + 3y + 6 = x(y + 2) + 3(y + 2) = (x + 3)(y + 2).
For a quadratic x² + bx + c, find two numbers with product c and sum b. For ax² + bx + c with a ≠ 1, multiply a × c, find two numbers with that product that add to b, and use them to split the middle term.
(a) x² + 7x + 12: numbers with product 12 and sum 7 are 3 and 4, so (x + 3)(x + 4).
(b) x² − x − 12: product −12 and sum −1 gives −4 and 3, so (x − 4)(x + 3).
(c) 2x² + 7x + 3: a × c = 6, and the numbers 6 and 1 add to 7. So 2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3).
(d) 3x² − 10x + 8: a × c = 24, and −4 and −6 add to −10. So 3x² − 4x − 6x + 8 = x(3x − 4) − 2(3x − 4) = (3x − 4)(x − 2).
a² − b² = (a − b)(a + b). So 9x² − 25 = (3x − 5)(3x + 5), and 2x² − 50 = 2(x² − 25) = 2(x − 5)(x + 5). Always take out a common factor first.
For the same non-zero base: aᵐ × aⁿ = aᵐ⁺ⁿ, aᵐ ÷ aⁿ = aᵐ⁻ⁿ and (aᵐ)ⁿ = aᵐⁿ. Also a⁰ = 1 and a⁻ⁿ = 1/aⁿ. Fractional indices link powers and roots: a^(1/2) = √a and a^(1/3) is the cube root of a.
x³ × x⁵ = x⁸. y⁹ ÷ y⁴ = y⁵. (2a³)² = 4a⁶. 16x⁻² = 16/x² for x ≠ 0. 8^(1/3) = 2, and 8^(2/3) = 2² = 4.
A surd is an exact root that is not a rational number, such as √2. To simplify, take out square factors: √72 = √(36 × 2) = 6√2. Surds add only when the simplified root is the same: 3√2 + 5√2 = 8√2, but √2 + √3 cannot be combined.
(a) √50 = √(25 × 2) = 5√2.
(b) Rationalise 1/√3: multiply the top and bottom by √3 to get √3/3.
(c) Rationalise 2/(2 + √3): multiply the top and bottom by the conjugate (2 − √3). The bottom becomes (2 + √3)(2 − √3) = 4 − 3 = 1, so the result is 2(2 − √3) = 4 − 2√3.
To solve an equation, keep it balanced: whatever you do to one side, do to the other. The full method, with brackets, fractions and special cases, is in our algebraic equations guide.
(a) 5x − 7 = 18. Add 7: 5x = 25. Divide by 5: x = 5.
(b) x/3 + 2 = 7. Subtract 2: x/3 = 5. Multiply by 3: x = 15.
Inequalities follow the same steps with one critical exception: when you multiply or divide by a negative number, reverse the inequality sign.
(a) −2x > 8. Divide by −2 and reverse the sign: x < −4.
(b) 3 − 2x < 9. Subtract 3: −2x < 6. Divide by −2 and reverse: x > −3. Check x = 0: 3 − 0 = 3, which is less than 9. ✓
(c) 2 < x + 3 ≤ 7. Subtract 3 from every part: −1 < x ≤ 4.
Test x = −3 and x = −5 in the original inequality.
The fix: x = −5 works (−2 × −5 = 10 > 8) but x = −3 does not (−2 × −3 = 6, which is not above 8). So the solutions are x < −4. Dividing by a negative number reverses the inequality sign.
Two equations with two unknowns are solved together by elimination, substitution or a graph, where the crossing point gives the solution.
Solve 2x + y = 11 and 3x − y = 9. Add the equations: 5x = 20, so x = 4. Substitute: 8 + y = 11, so y = 3. Check: 3(4) − 3 = 9. ✓
Solve 4x + 3y = 17 and y = 2x − 1. Substitute y: 4x + 3(2x − 1) = 17, so 10x − 3 = 17 and x = 2. Then y = 2(2) − 1 = 3. Check: 4(2) + 3(3) = 17. ✓
A quadratic equation has the form ax² + bx + c = 0 with a ≠ 0. It can have two, one or no real solutions. When it factorises, use the zero-product rule: if AB = 0, then A = 0 or B = 0.
x² − 5x + 6 = 0 gives (x − 2)(x − 3) = 0, so x = 2 or x = 3.
When a quadratic does not factorise easily, use the quadratic formula:
For 2x² + 3x − 2 = 0: a = 2, b = 3, c = −2. Then b² − 4ac = 9 + 16 = 25 and x = (−3 ± 5) ÷ 4. So x = 1/2 or x = −2.
Completing the square rewrites a quadratic as a squared bracket plus a number. It shows the turning point and can solve equations.
x² + 6x + 5 = (x + 3)² − 9 + 5 = (x + 3)² − 4. Since a square is never negative, the lowest value of the expression is −4, at x = −3. To solve x² + 6x + 5 = 0: (x + 3)² = 4, so x + 3 = ±2 and x = −1 or x = −5.
The part under the square root, b² − 4ac, is the discriminant. If it is positive there are two different real roots, if it is zero there is one repeated root, and if it is negative there are no real roots.
(a) Show that 2x² − 3x + 5 = 0 has no real roots: b² − 4ac = 9 − 40 = −31, which is negative.
(b) Find k so that x² + kx + 4 = 0 has a repeated root: k² − 16 = 0, so k = 4 or k = −4. For example, k = 4 gives x² + 4x + 4 = (x + 2)².
Rearranging a formula uses the same balance rules as solving an equation.
(a) A = πr². Divide by π: A/π = r². Take the positive square root: r = √(A/π).
(b) v = u + at. Subtract u: v − u = at. Divide by a: t = (v − u)/a.
(c) y = (x + 1)/(x − 2), where x appears twice. Multiply by (x − 2): y(x − 2) = x + 1, so yx − 2y = x + 1. Collect the x terms: yx − x = 2y + 1, so x(y − 1) = 2y + 1 and x = (2y + 1)/(y − 1). Check with x = 3: y = 4/1 = 4, and (2 × 4 + 1)/(4 − 1) = 3. ✓
A sequence is an ordered list that follows a rule. In an arithmetic sequence the difference between terms is constant, so the nth term is a + (n − 1)d, where a is the first term and d the common difference.
5, 8, 11, 14, … has d = 3 and first term 5, so the nth term is 5 + 3(n − 1) = 3n + 2. The 20th term is 3(20) + 2 = 62.
The sum of the first n terms is Sₙ = n/2 × [2a + (n − 1)d]. For 20 terms: S₂₀ = 10 × (10 + 57) = 670. Equivalently, 20/2 × (first + last) = 10 × (5 + 62) = 670.
In a quadratic sequence the second differences are constant, and if the second difference is 2a, the nth term starts with an².
2, 7, 14, 23, … has first differences 5, 7, 9 and constant second difference 2, so the nth term starts with n². Subtract n² (1, 4, 9, 16) from the terms: 1, 3, 5, 7, which is 2n − 1. So the nth term is n² + 2n − 1. Check n = 4: 16 + 8 − 1 = 23. ✓
In a geometric sequence each term is multiplied by a constant ratio r, so the nth term is a × rⁿ⁻¹.
3, 6, 12, 24, … has a = 3 and r = 2. The nth term is 3 × 2ⁿ⁻¹, so the 10th term is 3 × 512 = 1536. The sum of n terms is a(rⁿ − 1)/(r − 1), so the sum of the first 10 terms is 3(1024 − 1) = 3069.
The equation y = mx + c describes a straight line with gradient m and y-intercept c. The gradient between two points is m = (y₂ − y₁) ÷ (x₂ − x₁). Parallel lines have the same gradient. Perpendicular lines have gradients whose product is −1.
A line passes through (2, 5) and (6, 13). The gradient is (13 − 5) ÷ (6 − 2) = 8 ÷ 4 = 2. Put the point (2, 5) into y = 2x + c: 5 = 4 + c, so c = 1. The line is y = 2x + 1. Check the other point: 2(6) + 1 = 13. ✓
The midpoint of the two points is ((2 + 6) ÷ 2, (5 + 13) ÷ 2) = (4, 9). A line perpendicular to y = 2x + 1 has gradient −1/2, so y = −x/2 + 3 is perpendicular to it and crosses the y-axis at 3.
A function gives one output for each input. The notation f(x) means “the value of f when the input is x”, not f multiplied by x. A composite function fg(x) means f(g(x)): apply g first, then f. An inverse function f⁻¹ undoes f.
If f(x) = 3x − 4, then f(5) = 3(5) − 4 = 11. To solve f(x) = 8, set 3x − 4 = 8 so x = 4.
To find the inverse, write y = 3x − 4, swap x and y to get x = 3y − 4, and solve for y: f⁻¹(x) = (x + 4)/3. Check: f⁻¹(11) = 15 ÷ 3 = 5, the original input. ✓
Let f(x) = 2x + 1 and g(x) = x². Then fg(3) = f(g(3)) = f(9) = 19, but gf(3) = g(f(3)) = g(7) = 49. The order matters.
Work out each side.
The fix: f(3) = 7 and f(2) = 5, so f(3) + f(2) = 12. But f(5) = 11. Function notation is not multiplication, so f(a) + f(b) is not f(a + b) in general.
Algebraic fractions follow the rules of numerical fractions. Factorise first, then cancel factors, never separate terms joined by + or −. Note any values that make a denominator zero.
(a) Simplify (x² − 9)/(x² + 5x + 6) = [(x − 3)(x + 3)] / [(x + 2)(x + 3)] = (x − 3)/(x + 2), with x ≠ −3 and x ≠ −2.
(b) 1/x + 1/(x + 2) = [(x + 2) + x] / [x(x + 2)] = (2x + 2)/[x(x + 2)], with x ≠ 0 and x ≠ −2.
(c) Solve 3/(x − 1) = 2/(x + 1). Cross-multiply: 3(x + 1) = 2(x − 1), so 3x + 3 = 2x − 2 and x = −5. Check: 3/(−6) = −0.5 and 2/(−4) = −0.5. ✓
Try x = 4.
The fix: with x = 4, (4 + 2)/4 = 1.5, not 2. You can cancel only a factor of the whole top and the whole bottom. The x in x + 2 is a term, not a factor, so (x + 2)/x cannot be simplified this way. By contrast, (2x + 4)/(x + 2) = 2(x + 2)/(x + 2) = 2, because (x + 2) is a factor of the whole top.
If y is directly proportional to x, then y = kx. If y is inversely proportional to x, then y = k/x. Powers can appear, such as y = kx². Find k from one pair of values, then use the formula.
(a) y ∝ x² and y = 20 when x = 2. Then 20 = k(4), so k = 5 and y = 5x². When x = 3, y = 45.
(b) y is inversely proportional to x, and y = 6 when x = 4. Then k = 6 × 4 = 24, so y = 24/x. When x = 8, y = 3.
A polynomial such as x³ − 2x² − 5x + 6 has whole-number powers of x. The remainder theorem says that when a polynomial f(x) is divided by (x − a), the remainder is f(a). The factor theorem says that (x − a) is a factor exactly when f(a) = 0.
(a) The remainder when x² + 3x + 1 is divided by (x − 2) is f(2) = 4 + 6 + 1 = 11.
(b) Factorise f(x) = x³ − 2x² − 5x + 6. Try f(1) = 1 − 2 − 5 + 6 = 0, so (x − 1) is a factor. Dividing gives x² − x − 6 = (x − 3)(x + 2). So f(x) = (x − 1)(x − 3)(x + 2), and the roots are 1, 3 and −2. Check: f(3) = 27 − 18 − 15 + 6 = 0 and f(−2) = −8 − 8 + 10 + 6 = 0. ✓
Many problems hide an equation. Define the variables, translate each condition, solve, and then interpret the answer. A correct solution to the equation can still be wrong for the situation, for example a negative length.
A rectangular garden is x m wide and (x + 4) m long, with area 60 m². Then x(x + 4) = 60, so x² + 4x − 60 = 0 and (x + 10)(x − 6) = 0. So x = −10 or x = 6. A width cannot be negative, so x = 6. The garden is 6 m by 10 m, and 6 × 10 = 60. ✓
Solve x² − 4x − 5 = 0. It factorises: (x − 5)(x + 1) = 0, so x = 5 or x = −1. The formula would give the same answer (b² − 4ac = 16 + 20 = 36), but factorising is quicker. If the factors are not obvious, check the discriminant first: if it is not a perfect square, the roots are not rational and the formula is the right tool.
| If you see… | Think about… | Typical action |
|---|---|---|
| Several like terms | Simplifying | Collect the coefficients |
| Brackets | Expanding or factorising | Distribute, or take out a factor |
| x² with a product and sum pattern | A quadratic | Factorise, complete the square or use the formula |
| Two unknowns | Simultaneous equations | Eliminate or substitute |
| < or > | An inequality | Solve, and reverse the sign when dividing by a negative |
| y = mx + c | A straight line | Use the gradient and intercept |
| f(x) | A function | Substitute, compose or invert |
| Constant differences | A sequence | Find the nth term |
| “proportional to” | Variation | Introduce a constant k |
| A fraction containing x | An algebraic fraction | Factorise, then use a common denominator |
The progression is deliberate: language, then manipulation, then equations, then functions and graphs, then modelling. If the early skills are secure, advanced questions become combinations of familiar ideas.
Enter a, b and c for ax² + bx + c = 0. The tool shows the discriminant, the roots, the factorised form, the completed-square form and the turning point. Try 1, 6, 5 to match Figure 3.
Enter two points. You get the gradient, the equation of the line, the intercepts, the midpoint, the distance, and the gradients of parallel and perpendicular lines.
Choose arithmetic or geometric, then enter the first term, the common difference (or ratio) and n. You get the first terms, the nth term and the sum of the first n terms.
Tap the correct answer for each question. Your best score is saved on this device.
1. (x − 3)(x − 3) = x² − 3x − 3x + 9 = x² − 6x + 9.
2. Add 3: x/2 = 8. Multiply by 2: x = 16.
3. The common difference is −3, so the nth term starts with −3n. When n = 1 the term is 12, so we add 15: −3n + 15 (or 15 − 3n).
1. (a) 2x² + 10x − x − 5 = 2x² + 9x − 5. (b) 3x(2x + 3).
2. (a) 3x − 6 ≤ 2x + 5, so x ≤ 11. Check x = 11: 27 ≤ 27. ✓ (b) Subtract 3 from every part: −1 < x ≤ 4.
3. (a) (2x − 1)(x + 2) = 0, so x = 1/2 or x = −2. (b) x² + 6x + 5 = (x + 3)² − 9 + 5 = (x + 3)² − 4. The minimum value is −4, at x = −3.
4. From 2x − y = 1, y = 2x − 1. Substitute: 4x + 3(2x − 1) = 17, so 10x − 3 = 17 and x = 2. Then y = 3. Check: 8 + 9 = 17 and 4 − 3 = 1.
5. (a) The difference is 4 and the first term is 5, so the nth term is 4n + 1. (b) 4(50) + 1 = 201. (c) 4n + 1 = 100 gives n = 24.75, which is not a whole number, so 100 is not a term.
6. (a) f(4) = 8 − 3 = 5. (b) fg(x) = f(x² + 1) = 2(x² + 1) − 3 = 2x² − 1. (c) y = 2x − 3, swap: x = 2y − 3, so y = (x + 3)/2. f⁻¹(x) = (x + 3)/2.
7. (a) y = k/x and k = 6 × 4 = 24, so y = 24/x. (b) y = 24/8 = 3. (c) 2 = 24/x, so x = 12.
Algebra is easier when learners see what the symbols mean. Use area models, like the one for (x + 4)(x + 3), to show why expanding and factorising are opposites. Ask learners to test an algebraic claim with a number before they trust it: if (x + 3)² = x² + 9, try x = 2. Use function machines for equations and inverse functions, and ask learners to describe each step in words. The Quadratic Explorer, Straight-Line Calculator and Sequence Calculator on this page let learners check their own working, but they should always show the algebra first. Encourage a final check by substitution every time.
This article is original EDUSAMBAM writing. Examination requirements differ between boards, so compare your practice with the specification of your own board.
20 questions covering basic, intermediate and advanced mathematics algebra. Answer every question, then submit to see your score instantly.