Linear Equations

A linear equation is one in which the unknown appears only to the first power — no $x^{2}$ terms, no $x$ in a denominator. Solving one means finding the value of the unknown that makes the two sides equal. This lesson covers one- and two-step equations, equations with brackets, equations with the unknown on both sides, equations involving fractions, and — crucially for the harder marks — forming your own equation from a worded context. The mechanical solving is AO1, while turning a real situation into an equation, and justifying each step, draws heavily on AO2 and AO3.

Key Vocabulary

Term	Meaning
Equation	A statement that two expressions are equal, containing an equals sign
Solve	Find the value(s) of the unknown that make the equation true
Solution (root)	The value of the unknown that satisfies the equation
Inverse operation	The operation that undoes another ( $+$ undoes $-$ ; $\times$ undoes $\div$ )
Balance	Keep the equation true by doing the same thing to both sides
Subject	The unknown you are solving for
Form an equation	Translate a worded situation into an equation

The Balance Method

The golden rule is that an equation stays true as long as you do the same thing to both sides. To isolate the unknown, undo the operations around it using inverse operations, working in reverse order.

flowchart TD
    EQ[Linear equation] --> BR{Brackets?}
    BR -->|Yes| EXP[Expand the brackets]
    BR -->|No| BOTH
    EXP --> BOTH{Unknown on both sides?}
    BOTH -->|Yes| MOVE[Collect the unknown on one side]
    BOTH -->|No| FR
    MOVE --> FR{Fractions?}
    FR -->|Yes| MULT[Multiply through by the denominator]
    FR -->|No| ISO[Isolate the unknown with inverse operations]
    MULT --> ISO
    ISO --> ANS[State the solution and check by substituting]

Worked Example 1

Solve $x + 7 = 12$ .

Subtract $7$ from both sides: $x = 12 - 7 = 5$ .

Answer: $x = 5$

Worked Example 2

Solve $3x = 21$ .

Divide both sides by $3$ : $x = \dfrac{21}{3} = 7$ .

Answer: $x = 7$

It often helps to imagine an old-fashioned pair of weighing scales. The equals sign is the pivot, and the two sides are balanced pans. Whatever you do to one pan you must do to the other, or the scales tip and the equation is no longer true. This single image explains every legitimate move in equation-solving: add the same thing to both sides, subtract the same thing, multiply both sides, or divide both sides.

Two-Step Equations

When two operations surround the unknown, undo them one at a time — addition or subtraction first, then multiplication or division. The order is the reverse of BIDMAS: because the original expression would multiply before it adds, we peel the layers off in the opposite order, undoing the addition before the multiplication.

Worked Example 3

Solve $4x + 5 = 17$ .

Subtract $5$ : $4x = 12$ . Divide by $4$ : $x = 3$ .

Answer: $x = 3$

Worked Example 4

Solve $\dfrac{x}{3} - 2 = 4$ .

Add $2$ : $\dfrac{x}{3} = 6$ . Multiply by $3$ : $x = 18$ .

Answer: $x = 18$

Common error: dividing before dealing with the added or subtracted number. In $4x + 5 = 17$ you must remove the $+5$ first.

Worked Example 4b

Solve $20 - 3x = 5$ .

Here the unknown is being subtracted, so take care with signs. Subtract $20$ from both sides: $-3x = -15$ . Now divide by $-3$ : $x = 5$ .

Answer: $x = 5$

An alternative, and for many students a safer route, is to add $3x$ to both sides first so the $x$ -term becomes positive: $20 = 5 + 3x$ . Then subtract $5$ : $15 = 3x$ , so $x = 5$ . Both methods reach the same answer; choose the one you find less error-prone.

Equations with Brackets

Expand the bracket first, then solve as usual.

Worked Example 5

Solve $5(x - 2) = 25$ .

Expand: $5x - 10 = 25$ . Add $10$ : $5x = 35$ . Divide by $5$ : $x = 7$ .

Answer: $x = 7$

Worked Example 6

Solve $3(2x + 1) = 4(x + 5)$ .

Expand both sides: $6x + 3 = 4x + 20$ . Subtract $4x$ : $2x + 3 = 20$ . Subtract $3$ : $2x = 17$ . Divide by $2$ : $x = \dfrac{17}{2} = 8.5$ .

Answer: $x = 8.5$

Worked Example 6b

Solve $2(x + 4) + 3(x - 1) = 25$ .

Expand both brackets: $2x + 8 + 3x - 3 = 25$ . Collect like terms on the left: $5x + 5 = 25$ . Subtract $5$ : $5x = 20$ . Divide by $5$ : $x = 4$ .

Answer: $x = 4$

Notice the order of operations here: expand every bracket, then tidy the side into a single $x$ -term and a single constant, and only then begin undoing operations. Rushing to "solve" before the left-hand side is fully simplified is a common cause of lost marks.

Equations with the Unknown on Both Sides

Collect all the unknown terms on one side and all the numbers on the other. It is usually tidier to move the smaller $x$ -term so the coefficient stays positive. The principle is unchanged — do the same to both sides — but now you apply it twice: once to gather the $x$ -terms together, and once to gather the constants.

Worked Example 7

Solve $7x - 4 = 3x + 12$ .

Subtract $3x$ : $4x - 4 = 12$ . Add $4$ : $4x = 16$ . Divide by $4$ : $x = 4$ .

Answer: $x = 4$

Worked Example 8

Solve $2x + 9 = 8x - 3$ .

To keep the $x$ -coefficient positive, subtract $2x$ : $9 = 6x - 3$ . Add $3$ : $12 = 6x$ . Divide by $6$ : $x = 2$ .

Answer: $x = 2$

Common error: subtracting the larger $x$ -term and creating a negative coefficient, then losing a sign at the final division.

Forming Equations from Context

Some of the most valuable marks on an OCR paper come from turning a worded situation into an equation and then solving it. The method is always the same: choose a letter for the quantity you want to find, write down every relationship the question gives you in terms of that letter, build the equation, solve it, and finally answer the actual question (with units). Defining your variable in a sentence — "let the width be $w$ cm" — is itself worth marks and keeps your working clear.

Worked Example 8b

A number is multiplied by $5$ and then $7$ is added; the result is $42$ . Work out the number.

Let the number be $n$ . The description gives $5n + 7 = 42$ . Subtract $7$ : $5n = 35$ . Divide by $5$ : $n = 7$ .

Answer: the number is $7$ .

Equations with Fractions

Multiply every term by the denominator (or the lowest common denominator) to clear the fractions, then solve. Clearing fractions early turns an awkward-looking equation into an ordinary one; the only thing to remember is that every term, on both sides, must be multiplied — including any whole numbers.

Worked Example 9

Solve $\dfrac{2x - 1}{3} = 5$ .

Multiply both sides by $3$ : $2x - 1 = 15$ . Add $1$ : $2x = 16$ . Divide by $2$ : $x = 8$ .

Answer: $x = 8$

Worked Example 10

Solve $\dfrac{x}{2} + \dfrac{x}{5} = 7$ .

The lowest common denominator is $10$ . Multiply every term by $10$ : $5x + 2x = 70$ , so $7x = 70$ and $x = 10$ .

Answer: $x = 10$

Common error: multiplying only the fraction terms by the denominator and forgetting to multiply the number on the right-hand side.

Extended Worked Examples

Worked Example 11 — Fractions on both sides

Solve $\dfrac{3x + 1}{4} = \dfrac{x + 5}{2}$ .

Multiply every term by $4$ (the LCD): $3x + 1 = 2(x + 5)$ . Expand: $3x + 1 = 2x + 10$ . Subtract $2x$ : $x + 1 = 10$ , so $x = 9$ .

Answer: $x = 9$

Worked Example 12 — Forming an equation from a perimeter

The perimeter of a rectangle is $46\ \text{cm}$ . Its length is $5\ \text{cm}$ more than its width. Work out the width.

Linear Equations

Linear Equations

Key Vocabulary

The Balance Method

Worked Example 1

Worked Example 2

Two-Step Equations

Worked Example 3

Worked Example 4

Worked Example 4b

Equations with Brackets

Worked Example 5

Worked Example 6

Worked Example 6b

Equations with the Unknown on Both Sides

Worked Example 7

Worked Example 8

Forming Equations from Context

Worked Example 8b

Equations with Fractions

Worked Example 9

Worked Example 10

Extended Worked Examples

Worked Example 11 — Fractions on both sides

Worked Example 12 — Forming an equation from a perimeter

More in Mathematics