Solving linear equations in one unknown, including equations with brackets, equations with the unknown on both sides, and equations involving fractions, and forming equations from worded contexts.

What this dot point is asking

Edexcel expects you to solve linear equations, the kind where the unknown appears only to the power one, no matter how they are dressed up: with brackets, fractions, or the unknown on both sides. You should also be able to form an equation from a worded or geometric situation and then solve it. The guiding principle never changes: do the same operation to both sides to keep the equation balanced.

The balance method

An equation is a balance: whatever you do to one side you must do to the other. The aim is to isolate the unknown.

To solve $3x + 7 = 22$: subtract $7$ from both sides to get $3x = 15$, then divide by $3$ to get $x = 5$. Substituting back, $3(5) + 7 = 22$, confirms the solution.

Equations with brackets and fractions

Brackets and fractions are handled before the balance method proper. Expand brackets first: $4(x + 3) = 28$ becomes $4x + 12 = 28$, then $4x = 16$ and $x = 4$.

For fractions, multiply every term by the denominator to clear it. To solve $\dfrac{x}{5} + 2 = 7$, you could subtract $2$ first ($\dfrac{x}{5} = 5$) then multiply by $5$ ($x = 25$). When two fractions appear, multiply through by the lowest common denominator so all fractions disappear at once.

The unknown on both sides

When the unknown appears on both sides, collect all the unknown terms on the side where the coefficient is larger (to keep it positive), and the numbers on the other.

Forming equations

Many marks come from turning a situation into an equation. Define the unknown clearly, translate the relationship, then solve. For "I think of a number, multiply it by $3$ and add $5$ to get $26$", the equation is $3x + 5 = 26$, giving $x = 7$. Geometric facts (angles on a line sum to $180^\circ$, angles in a triangle to $180^\circ$, opposite sides of a rectangle are equal) are common sources of equations, as in the exam question above.

A classic worded type gives a perimeter or a total. If a rectangle is $x$ cm wide and $(x + 4)$ cm long with perimeter $36$ cm, the perimeter is $2(x) + 2(x + 4) = 36$, which simplifies to $4x + 8 = 36$, so $x = 7$. The width is $7$ cm and the length is $11$ cm. The skill being tested is the translation into algebra; once the equation is formed, the solving is routine. Read the question twice to make sure you have set up the relationship the right way round, because a wrong equation cannot earn the solving marks even if your algebra is perfect. A good habit is to write a short sentence defining your unknown, for example "let $x$ be the width in cm", so the examiner can follow your reasoning and award method marks even if the final arithmetic slips.

Try this

Q1. Solve $\dfrac{2x + 1}{3} = 5$. [2 marks]

• Cue. Multiply both sides by $3$: $2x + 1 = 15$, so $2x = 14$ and $x = 7$.

Q2. Solve $4(2x - 1) = 3(x + 7)$. [3 marks]

• Cue. Expand both sides: $8x - 4 = 3x + 21$. Collect: $5x = 25$, so $x = 5$.