Solve linear equations with one unknown, including those with brackets, fractions and the unknown on both sides, and form equations from worded contexts.

What this dot point is asking

Solving linear equations is the central algebra skill at GCSE. WJEC asks you to solve equations with one unknown, including those with brackets, fractions and the unknown on both sides, and to form equations from worded contexts and then solve them. The technique, doing the same inverse operation to both sides until the unknown stands alone, underpins simultaneous equations, rearranging formulae and much of geometry, so it is worth making completely automatic. It appears on both components.

The balance method

An equation says two expressions are equal. Whatever you do to one side you must do to the other, keeping the balance. To isolate the unknown, undo the operations surrounding it in reverse BIDMAS order.

For $3x + 5 = 20$: subtract $5$ (undo the addition) to get $3x = 15$, then divide by $3$ (undo the multiplication) to get $x = 5$.

Equations with brackets

Expand the brackets first, then solve as usual.

For $4(x - 2) = 12$, expand to $4x - 8 = 12$, add $8$ to get $4x = 20$, then divide to get $x = 5$. Alternatively, since the right side divides exactly, you could divide both sides by $4$ first; either order works, but expanding is the more general method.

Equations with fractions

A fraction is cleared by multiplying both sides by its denominator.

So $\dfrac{x}{2} + \dfrac{x}{3} = 5$ multiplies through by $6$ to give $3x + 2x = 30$, so $5x = 30$ and $x = 6$.

The unknown on both sides

When the unknown appears on both sides, the first job is to gather it onto one side.

Move the smaller unknown term to avoid a negative coefficient, which keeps the arithmetic cleaner.

Forming equations from words

WJEC often gives a worded situation and expects you to build the equation. Choose a letter for the unknown, translate each statement into algebra, and form an equation, often from a total or an equality. "I think of a number, multiply it by $3$ and add $4$ to get $19$" becomes $3n + 4 = 19$, solving to $n = 5$. Perimeter, angle-sum and money problems are all set this way, and the algebra then solves exactly as above.

A common geometry version gives angles on a straight line or in a triangle in terms of $x$. If three angles on a straight line are $2x$, $x + 10$ and $50$, they sum to $180$, so $2x + (x + 10) + 50 = 180$, giving $3x + 60 = 180$, so $x = 40$. The skill is recognising the relationship that creates the equation (angles on a line sum to $180$, a perimeter is the sum of the sides), then solving the linear equation that results.

Equations with the unknown in the denominator

Occasionally the unknown sits on the bottom of a fraction, such as $\dfrac{12}{x} = 4$. Multiply both sides by $x$ to lift it out of the denominator: $12 = 4x$, so $x = 3$. The same first move, multiplying through by whatever is on the denominator, clears the fraction and turns the problem back into a standard linear equation. Always multiply the whole of both sides, and check the answer by substituting it back into the original fraction.

Why this matters

Linear equations are the workhorse of GCSE algebra and reappear constantly: simultaneous equations are two linear equations solved together, rearranging a formula is solving for a chosen letter, and many geometry and ratio problems reduce to forming and solving a linear equation. Because WJEC rewards clear, balanced steps with method marks, laying out each line, and checking by substitution, secures marks even under exam pressure.