Algebra Resources

Three connected algebra ideas: the area model for (x + 3)(x + 2) = x² + 5x + 6, solving 3x + 5 = 20 to get x = 5, and the lines y = 2x + 1 and y = −x + 7 meeting at (2, 5).

ALGEBRA • LEARN, PRACTISE, CONNECT

Find an algebra topic, understand the method, then choose a practice resource. This guide brings together an existing secondary-school resource library with original worked examples and a short diagnostic. Start with the explanations below if a method is unfamiliar, or jump to a topic in the resource directory when you know what you need.

The collection spans introductory algebra, GCSE-style work and some extension material. These are topic labels, not a guarantee that every linked task matches your exam board, tier or current specification. England’s national curriculum includes expressions, equations, sequences and graphs; individual courses organise and assess them differently.

Which algebra resource should I use?

Choose by the step you cannot yet explain. Being able to copy a procedure is different from recognising when it applies. Try one question without notes, inspect the first uncertain line, then practise that skill before moving on.

Letters feel unfamiliar: begin with forming expressions, substitution and function machines. Read 3x as three times x, not as the two-digit number “3x”.
You know the notation but lose signs: use simplifying, single brackets and common-factor factorising. Write the intermediate products rather than doing every step mentally.
You can simplify but cannot find an unknown: use linear equations, rearranging formulae and inequalities. Name the operation applied to both sides.
You need to connect patterns and pictures: use linear sequences, linear graphs and simultaneous equations. Explain what each coordinate or term means.
You are ready for more demanding work: choose quadratics, algebraic fractions, functions or proof. Check prerequisites and restrictions first.

For a smaller practice collection, use the GCSE algebra worksheets guide. The broader algebra worksheet library provides another topic-based route. Work independently first; a calculator can check a result, but it cannot replace the written reasoning your teacher or examination requires.

Ten worked examples that connect the main ideas

1. Form an expression, then substitute carefully

A notebook costs p pounds. Three notebooks and a £2 delivery charge cost 3p + 2 pounds. The delivery charge is added once. The expression 3(p + 2) would charge delivery three times.

If p = 4.50, the total is 3 × 4.50 + 2 = £15.50. Keep the units in the interpretation: p represents a price measured in pounds, while the coefficient 3 counts notebooks.

For a negative input, brackets protect the sign. At x = −2, x² − 3x = (−2)² − 3(−2) = 4 + 6 = 10. The expression −x² means −(x²), so it would equal −4 at this input.

2. Collect like terms and distribute a negative

Simplify 4x + 3 − 2(x − 5). First multiply both terms inside the bracket by −2: −2(x − 5) = −2x + 10. Then 4x + 3 − 2x + 10 = 2x + 13.

Terms can be combined by addition when their variable parts match. Thus 4x − 2x = 2x, but 4x + 3 cannot become 7x. Likewise x + x = 2x, whereas x × x = x². Addition and multiplication use different rules.

3. Expand two brackets using area

For (x + 3)(x + 2), multiply each term in the first bracket by each term in the second: x² + 2x + 3x + 6 = x² + 5x + 6. The two middle products both matter.

Area model for (x plus 3)(x plus 2), split into x squared, 3x, 2x and 6, which combine to x squared plus 5x plus 6. — Four smaller areas account for all four products. The geometric picture uses positive lengths; the expanded algebraic identity holds for every real x.

Factorising reverses this process: x² + 5x + 6 = (x + 2)(x + 3), because 2 + 3 = 5 and 2 × 3 = 6. Always expand your proposed factors to check them. For 6x² + 9x, take the common factor 3x to obtain 3x(2x + 3).

4. Solve an equation by preserving equality

Solve 3(x − 2) + 4 = 2x + 9. Expanding gives 3x − 6 + 4 = 2x + 9, so 3x − 2 = 2x + 9. Subtract 2x from both sides: x − 2 = 9. Add 2 to both sides: x = 11.

Check in the original equation: 3(11 − 2) + 4 = 31 and 2(11) + 9 = 31. Matching sides confirm this candidate solution. A step such as “move it across and change the sign” is shorthand for a valid operation on both sides, not a separate algebraic law.

Some equations have no solution or infinitely many solutions. For example, 2(x + 1) = 2x + 3 reduces to 2 = 3, so it has no solution. The equation 2(x + 1) = 2x + 2 is true for every real x. Do not divide by a variable expression without considering whether it can be zero.

5. Rearrange a formula and keep restrictions

Make h the subject of A = ½bh. Multiply by 2, giving 2A = bh. Divide by b to get h = 2A/b, provided b ≠ 0. For a triangle with A = 24 cm² and b = 8 cm, h = 48/8 = 6 cm.

Rearranging changes the subject while keeping the relationship. If the formula contains a square, square-rooting may introduce two algebraic possibilities: x² = 25 gives x = ±5. A positive-length context can then rule out the negative value.

6. Reverse an inequality when dividing by a negative

Solve 5 − 2x < 11. Subtract 5: −2x < 6. Dividing both sides by −2 reverses the order, so x > −3. The boundary −3 is excluded because the original inequality is strict: it gives 11 < 11, which is false.

Test one included value, x = 0: 5 < 11 is true. On a number line, use an open circle at −3 and shade to the right. For a non-strict inequality, the boundary would be included and shown with a filled circle.

7. Find a sequence rule, then test it

The arithmetic sequence 7, 11, 15, 19, … increases by 4. With the first term at n = 1, its nth term is 7 + 4(n − 1) = 4n + 3. The tenth term is 43. To test whether 50 occurs, solve 4n + 3 = 50: n = 47/4 = 11.75, which is not a positive integer. Therefore 50 is not a term of this stated arithmetic sequence.

A finite list alone does not determine a unique infinite sequence. The answer above uses the stated constant-difference rule. For quadratic-sequence tasks with equally spaced n, constant second differences suggest a quadratic rule; further checks are still needed. Use the nth-term calculator as a check after you state the pattern assumption.

8. Connect simultaneous equations to an intersection

Solve y = 2x + 1 and y = −x + 7 together. At an intersection the y-values are equal, so 2x + 1 = −x + 7. Hence 3x = 6, x = 2 and y = 5. The ordered pair is (2, 5); it satisfies both original equations.

The lines y equals 2x plus 1 and y equals minus x plus 7 cross at the labelled point (2, 5). — A simultaneous solution lies on both graphs. Reading a graph gives an estimate; substitution confirms the exact coordinates.

The gradients are 2 and −1, while the y-intercepts are 1 and 7. Parallel distinct lines would have no intersection; two equations describing the same line would have infinitely many shared points.

9. Factorise a quadratic and respect a fraction’s domain

Solve x² − x − 12 = 0. Numbers −4 and 3 multiply to −12 and add to −1, so (x − 4)(x + 3) = 0. At least one factor must be zero: x = 4 or x = −3. This zero-product reasoning applies because the product equals zero.

Now simplify (x² − 9)/(x − 3). Factor the numerator as (x − 3)(x + 3), then cancel the common factor to obtain x + 3 for x ≠ 3. The original fraction remains undefined at x = 3. Cancelling a factor does not restore an excluded input, and individual terms cannot be cancelled across addition.

10. Compose functions and give a general proof

Let f(x) = 2x + 1 and g(x) = x². Then f(g(3)) = f(9) = 19, while g(f(3)) = g(7) = 49. Read composition from the inside out. The order matters, and f(x) means the output of f at x, not f multiplied by x.

For a short proof, take any three consecutive integers n, n + 1 and n + 2, where n is an integer. Their sum is 3n + 3 = 3(n + 1), an integer multiple of 3. This proves the claim for every such triple. Checking 4 + 5 + 6 = 15 illustrates the claim but does not prove all cases.

A 12-question algebra check, with explained answers

Try each question on paper before opening its answer. If your result differs, compare the first line of your method with the explanation. Questions 1–6 focus on foundations; 7–12 connect sequences, graphs and more advanced algebra.

Write the cost of five tickets at t pounds each plus one £3 booking fee.
Show answer 1
5t + 3 pounds. The fee is paid once. 5(t + 3) would add five fees.
Evaluate 2a² − b when a = −3 and b = 4.
Show answer 2
14. 2(−3)² − 4 = 2 × 9 − 4 = 14. Square the negative number before multiplying by 2.
Simplify 7x − 3 + 2x + 8.
Show answer 3
9x + 5. Combine the x-terms and the constants separately.
Expand −3(2x − 5).
Show answer 4
−6x + 15. Multiply both terms by −3; a negative multiplied by a negative is positive.
Factorise 8x² − 12x fully.
Show answer 5
4x(2x − 3). Both terms contain 4x. Expanding checks the result.
Solve 4x + 7 = 2x − 5.
Show answer 6
x = −6. Subtract 2x and 7 to get 2x = −12, then divide by 2. Both original sides equal −17.
Find the nth term of the arithmetic sequence 12, 9, 6, 3, …, starting at n = 1.
Show answer 7
15 − 3n. Use 12 − 3(n − 1). The first term is 12 and each increase in n decreases the value by 3.
Solve 2 − 3x ≤ 14.
Show answer 8
x ≥ −4. Subtract 2, then divide by −3 and reverse the inequality. The boundary is included.
Find the intersection of y = x + 2 and y = 8 − 2x.
Show answer 9
(2, 4). Equate x + 2 = 8 − 2x to obtain 3x = 6. Substitute x = 2 into either line.
Solve x² + 2x − 15 = 0.
Show answer 10
x = 3 or x = −5. Factorise as (x + 5)(x − 3) = 0, then set each factor equal to zero.
Simplify (x² − 16)/(x − 4), including the restriction.
Show answer 11
x + 4, with x ≠ 4. The numerator is (x − 4)(x + 4). The original denominator cannot be zero.
Prove that the sum of any two consecutive odd integers is divisible by 4.
Show answer 12
Write them as 2n + 1 and 2n + 3 for an integer n. Their sum is 4n + 4 = 4(n + 1), so it is divisible by 4. The reasoning includes negative odd integers.

How to turn a worksheet into useful practice

Choose a short set with a clear purpose: fluent substitution, careful negative signs, selecting a method or explaining why a claim is true. Attempt it before consulting answers. Mark errors by cause, such as an incorrect expansion or a lost restriction, then write a corrected example without copying the answer line.

After success on one familiar format, change the form of the task. Explain an error, compare two methods, match an equation to a graph or create a question with a specified solution. Return to a few questions later without notes. One high score immediately after a worked example does not establish lasting understanding.

Teachers should preview each external resource, confirm its licence and check its answers before classroom use. Some links lead to a landing page, an account requirement or paid material. Titles containing “new GCSE”, an old grade label or an older exam year are historical resource titles, not a claim about today’s qualification.

For checking symbolic steps after an independent attempt, try the algebra calculator. Keep exact fractions where appropriate, explain any rounding, and use your course’s permitted calculator rules.

Jump to a resource topic

The existing collection is retained below in Sources & References, grouped by topic. Each jump takes you to the original resource links and credits. Use your browser’s Find command to locate a named author or a phrase such as “answers”.

Sources & References

Resource credit. The retained external library follows Jo Morgan’s Resourceaholic Algebra collection. Its original first-person credits have been clarified as Jo Morgan/Resourceaholic credits, while the named resource creators remain credited beside their links. External resources are linked at the URLs recorded in the original library; availability and access conditions may change. Inclusion does not transfer copyright or imply endorsement.

Jo Morgan, Resourceaholic: Algebra resource library — provenance of the retained collection and its author credits
Resourceaholic: About Jo Morgan — authorship context
Department for Education: mathematics programmes of study for England — curriculum context, not a match guarantee for every resource
OpenStax: a general strategy for solving linear equations — equality-preserving operations and checking solutions

Access reviewed 4 October 2026 (UTC). Original destinations are retained, including links with access limitations. A working web page or file endpoint does not establish that an entire worksheet, answer key or download has been checked. Notices identify unavailable responses and pages/viewers that may require further steps. A provider may require registration, payment or permission. Do not enter credentials on an unexpected destination; return to the provider’s official site. The worked examples and twelve answers above are original HeLovesMath teaching material.

Which algebra resource should I use?

Ten worked examples that connect the main ideas

1. Form an expression, then substitute carefully

2. Collect like terms and distribute a negative

3. Expand two brackets using area

4. Solve an equation by preserving equality

5. Rearrange a formula and keep restrictions

6. Reverse an inequality when dividing by a negative

7. Find a sequence rule, then test it

8. Connect simultaneous equations to an intersection

9. Factorise a quadratic and respect a fraction’s domain

10. Compose functions and give a general proof

A 12-question algebra check, with explained answers

How to turn a worksheet into useful practice

Jump to a resource topic

Sources & References

Forming Expressions

Function Machines

Simplifying Expressions

Expanding Single Brackets

Expanding Double Brackets

Expanding Cubics

Factorising (Single Brackets)

Rearranging Formulae

Substitution

Linear Sequences

Quadratic Sequences

Other Sequences

Linear Equations

Linear and Graphical Inequalities

Quadratic Inequalities

Linear Graphs

Quadratics

Factorising

Quadratic graphs – plotting, sketching and recognising

Forming and solving quadratic equations

Quadratic Formula

Completing the Square

Mixed

Iteration

Simultaneous Equations

Algebraic Fractions

Functions

Graphing Functions

Graph Transformations

Algebraic Proof and Identities

Equation of a Circle & Tangent to a Circle

Gradient of a Curve

Area under a Graph

Algebra Mixed/Revision

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