A comprehensive guide to algebraic equations, unknowns, expressions, balancing, simplification, linear equations, brackets, fractions, powers, simultaneous equations, equations from word problems, checking solutions, common errors and advanced problem-solving strategies.
An algebraic equation is a mathematical statement that says two expressions are equal. Equations allow us to represent unknown quantities, describe relationships, model real situations and find values that satisfy given conditions. The central skill is not simply manipulating symbols: it is understanding why each step is valid, maintaining equality throughout the process, and checking that the final answer really satisfies the original equation.
An algebraic equation contains one or more variables, numerical values and mathematical operations, together with an equality sign.
For example, 3x + 5 = 20 is an equation. The letter x represents an unknown number. Solving the equation means finding the value or values of x that make the statement true.
| Term | Meaning | Example |
|---|---|---|
| Expression | A mathematical phrase without an equality statement | 3x + 5 |
| Equation | A statement that two expressions are equal | 3x + 5 = 20 |
| Identity | An equality true for every permitted value of the variable | 2(x + 3) = 2x + 6 |
Keeping these ideas separate prevents confusion. An expression is simplified or evaluated; an equation is solved; an identity remains true for all allowed values.
In 7x − 4, the variable is x, the coefficient of x is 7, and the constant is −4.
In 5a + 2b − 9, the terms are 5a, 2b and −9. The coefficients are 5 and 2, while −9 is a constant.
The symbol = does not mean “the answer comes next”. It means that the two sides have the same value.
For example, 7 + 5 = 12 is true because both sides have value 12. In an equation such as 2x + 1 = 9, we search for a value of x that makes both sides equal.
This balance interpretation is the foundation of equation solving.
A solution is a value that makes the equation true.
For 2x + 3 = 11, try x = 4.
Left side = 2(4) + 3 = 11.
Right side = 11.
Since both sides are equal, x = 4 is a solution.
Equation solving relies heavily on inverse operations, which undo one another.
| Operation | Inverse |
|---|---|
| Addition | Subtraction |
| Subtraction | Addition |
| Multiplication | Division |
| Division | Multiplication |
| Squaring | Square root, with attention to signs |
| Cubing | Cube root |
When solving an equation, the aim is to use inverse operations to isolate the unknown.
Solve x + 7 = 15.
Subtract 7 from both sides:
x + 7 − 7 = 15 − 7.
Therefore x = 8.
Check: 8 + 7 = 15, so the solution is correct.
Solve x − 9 = 4.
Add 9 to both sides:
x − 9 + 9 = 4 + 9.
Therefore x = 13.
Solve 6x = 42.
Divide both sides by 6:
x = 42 ÷ 6 = 7.
Solve x/5 = 9.
Multiply both sides by 5:
x = 9 × 5 = 45.
For an equation such as 3x + 4 = 19, remove the addition first and then undo the multiplication.
3x + 4 = 19
Subtract 4: 3x = 15.
Divide by 3: x = 5.
Check: 3(5) + 4 = 19.
A useful habit is to work systematically from the outside operations toward the variable.
Solve 4x + 3 = −9.
Subtract 3: 4x = −12.
Divide by 4: x = −3.
Check: 4(−3) + 3 = −12 + 3 = −9.
When working with negatives, keep the sign attached to the number and apply the operation to the entire side of the equation.
Before solving a complicated equation, it may be necessary to simplify each side. Like terms contain the same variables raised to the same powers.
3x + 5x − 2 = 8x − 2.
The terms 3x and 5x are like terms, so they combine to make 8x.
Terms such as 3x and 3x² are not like terms and cannot be combined directly.
Solve 5x + 2 = 2x + 17.
Subtract 2x from both sides: 3x + 2 = 17.
Subtract 2: 3x = 15.
Divide by 3: x = 5.
Moving terms across the equality sign is shorthand for performing the same operation on both sides. Thinking this way reduces sign errors.
The distributive property removes brackets:
Solve 3(x + 4) = 21.
Expand: 3x + 12 = 21.
Subtract 12: 3x = 9.
Divide by 3: x = 3.
Simplify −(x − 5).
Multiplying every term inside by −1 gives −x + 5.
A negative sign immediately before a bracket affects every term inside it. This is a frequent source of errors.
Solve 2(x + 3) + 4 = 18.
Expand: 2x + 6 + 4 = 18.
Combine: 2x + 10 = 18.
Subtract 10: 2x = 8.
Divide by 2: x = 4.
For longer equations, expand carefully, collect like terms, then isolate the variable.
Fractions can often be removed by multiplying every term by a common denominator.
Solve x/4 + 3 = 8.
Subtract 3: x/4 = 5.
Multiply by 4: x = 20.
Solve x/3 + x/6 = 9.
The lowest common denominator is 6. Multiply every term by 6:
2x + x = 54.
3x = 54.
x = 18.
The multiplication must apply to the entire equation, not just one term.
Solve 0.4x + 1.2 = 5.2.
Subtract 1.2: 0.4x = 4.
Divide by 0.4: x = 10.
Alternatively, multiply the original equation by 10 to remove the decimal places before solving.
Some equations contain a variable raised to a power. For a simple square equation, the square root operation can undo the square, but both positive and negative roots may need to be considered.
Solve x² = 49.
Taking square roots gives x = ±7.
Both values work because 7² = 49 and (−7)² = 49.
This is different from an equation such as x² = −4, which has no real-number solution.
| Type | Typical feature | Example |
|---|---|---|
| Linear | Variable has highest power 1 | 3x + 5 = 17 |
| Quadratic | Variable can have power 2 | x² − 5x + 6 = 0 |
| Cubic | Variable can have power 3 | x³ − 8 = 0 |
| Rational | Variable may appear in a denominator | 1/x = 4 |
Different equation types require different solving methods.
A quadratic equation often has the form ax² + bx + c = 0. When factorisation is possible, the zero-product property can be used.
Solve x² − 5x + 6 = 0.
Factorise: (x − 2)(x − 3) = 0.
Therefore x − 2 = 0 or x − 3 = 0.
Solutions: x = 2 or x = 3.
If two factors multiply to give zero, at least one factor must be zero:
This principle is what allows a factorised quadratic to produce separate solutions.
When a quadratic cannot easily be factorised, the quadratic formula can be used:
For an equation in the form ax² + bx + c = 0, identify a, b and c carefully before substituting.
For x² − 3x − 4 = 0, a = 1, b = −3 and c = −4.
x = [3 ± √(9 + 16)] ÷ 2 = [3 ± 5] ÷ 2.
Therefore x = 4 or x = −1.
The expression b² − 4ac is called the discriminant. It provides information about the roots of a quadratic.
| Discriminant | Real solutions |
|---|---|
| Positive | Two distinct real solutions |
| Zero | One repeated real solution |
| Negative | No real solutions |
A pair of simultaneous equations contains two unknowns whose values must satisfy both equations at the same time.
Solve:
x + y = 11
x − y = 3
Add the equations: 2x = 14, so x = 7.
Substitute into x + y = 11: 7 + y = 11, so y = 4.
Solve:
y = 2x + 1
x + y = 10
Substitute y = 2x + 1 into the second equation:
x + 2x + 1 = 10.
3x = 9, so x = 3.
Then y = 2(3) + 1 = 7.
Algebra becomes especially powerful when a real situation is translated into an equation.
One number is 4 greater than another. Their sum is 30. Let the smaller number be x.
Larger number = x + 4.
Equation: x + (x + 4) = 30.
2x + 4 = 30.
2x = 26, so x = 13.
The two numbers are 13 and 17.
| Words | Algebraic form |
|---|---|
| A number increased by 7 | x + 7 |
| A number decreased by 7 | x − 7 |
| Three times a number | 3x |
| A number divided by 4 | x/4 |
| Five less than twice a number | 2x − 5 |
| The sum of x and y | x + y |
| The difference between x and y | x − y, depending on the stated order |
Read phrases carefully. “Five less than twice a number” means 2x − 5, not 5 − 2x.
Many ratio problems can be expressed as equations. If two quantities are proportional, an unknown can be found by forming equivalent ratios.
If 4 items cost 12 units of currency, let x be the cost of 10 items.
4/12 = 10/x.
Cross-multiply: 4x = 120.
x = 30.
What number has 20% equal to 18?
0.20x = 18.
x = 18 ÷ 0.20 = 90.
Writing the percentage as a decimal or fraction makes the algebraic relationship explicit.
A formula is an equation that describes a relationship between quantities. You may need to rearrange it to make a different variable the subject.
Given A = lw, make w the subject.
Divide both sides by l:
w = A/l, provided l ≠ 0.
The same balance principle applies when rearranging formulas. Treat the desired variable as the unknown and use inverse operations.
Given v = u + at, make t the subject.
Subtract u: v − u = at.
Divide by a: t = (v − u)/a, provided a ≠ 0.
Checking is one of the most reliable ways to catch algebraic errors.
Suppose the solution to 5x − 7 = 18 is x = 5.
Substitute: 5(5) − 7 = 25 − 7 = 18.
The original equation is satisfied, so x = 5 is correct.
For equations with several solutions, check each solution when practical.
Some equations simplify to a contradiction.
Solve 2x + 5 = 2x + 9.
Subtract 2x from both sides: 5 = 9.
This is impossible. Therefore the equation has no solution.
Other equations simplify to a statement that is always true.
Solve 3(x + 2) = 3x + 6.
Expand the left side: 3x + 6 = 3x + 6.
The statement is true for every value of x. Therefore there are infinitely many solutions.
An equation can also be understood graphically. For a linear equation such as y = 2x + 1, every point on its graph satisfies the equation.
For two equations, their common solution corresponds to the point where their graphs intersect. Thus algebra and graphs provide two different ways of investigating the same mathematical relationship.
Some algebraic equations contain restrictions. A denominator cannot be zero, because division by zero is undefined.
For 1/(x − 2) = 3, x cannot equal 2 because that would make the denominator zero.
Solving gives x − 2 = 1/3, so x = 7/3, which is allowed.
Always consider restrictions before accepting a solution.
Some transformations can introduce values that satisfy a transformed equation but not the original equation. This can occur particularly when squaring both sides or manipulating equations involving roots or denominators.
Therefore, when an equation has restrictions or non-reversible steps, substitution into the original equation is especially important.
Solve 4(2x − 3) + 5 = 3x + 20.
Expand: 8x − 12 + 5 = 3x + 20.
Combine: 8x − 7 = 3x + 20.
Subtract 3x: 5x − 7 = 20.
Add 7: 5x = 27.
Divide by 5: x = 27/5 = 5.4.
Check: 4(10.8 − 3) + 5 = 4(7.8) + 5 = 36.2; right side = 16.2 + 20 = 36.2.
| Rule | Key idea |
|---|---|
| Add/subtract | Perform the same addition or subtraction on both sides |
| Multiply/divide | Perform the same non-zero multiplication or division on both sides |
| Brackets | Distribute a factor to every term inside |
| Like terms | Combine only terms with matching variable parts |
| Fractions | Use a common denominator or multiply through |
| Quadratics | Factorise, complete the square or use the quadratic formula as appropriate |
| Simultaneous equations | Use elimination, substitution or another valid method |
| Checking | Substitute the proposed solution into the original equation |
| Restrictions | Exclude values that make an expression undefined |
Algebraic equations provide a precise language for unknown quantities and relationships. The most important principle is that an equation represents equality: valid operations preserve that equality.
Start with simple inverse operations, then build toward equations involving brackets, fractions, negative values, powers and several variables. More advanced equations may require factorisation, the quadratic formula, simultaneous-equation methods or careful consideration of restrictions.
Strong equation solving combines accurate algebraic manipulation with logical checking. A correct-looking sequence of symbols is not enough: the final value must satisfy the original conditions. Once this habit is developed, algebraic equations become a powerful method for solving both abstract mathematical problems and practical situations.
To develop fluency, practise equations in stages: begin with one-step equations, progress to multi-step equations and brackets, then work with fractions, powers, simultaneous equations and quadratic equations. For every problem, write enough working to show the operation performed and check important answers by substitution.
20 questions covering equations, inverse operations, brackets, fractions, powers, quadratics, simultaneous equations, formulae and interpretation. Answer every question, then submit to see your score instantly.