Solving linear equations with the unknown on one or both sides, equations with brackets and fractions, and changing the subject of a formula.

What this dot point is asking

AQA wants you to solve linear equations (where the unknown appears only to the first power) by balancing, including equations with brackets, fractions and the unknown on both sides, and to rearrange a formula to make a different variable the subject. This is the most heavily used algebra skill in the whole specification: almost every multi-step problem ends with an equation to solve.

The balancing principle

An equation is a balance: whatever you do to one side you must do to the other. The aim is to isolate the unknown using inverse operations. To solve $3x + 5 = 20$, subtract $5$ from both sides to get $3x = 15$, then divide both sides by $3$ to get $x = 5$. You can always check by substituting back: $3(5) + 5 = 20$, which is correct.

Equations with brackets and the unknown on both sides

When brackets appear, expand them first. When the unknown is on both sides, collect all the unknown terms on one side and all the numbers on the other.

Equations with fractions

Clear fractions by multiplying every term by the common denominator. To solve $\dfrac{x}{3} + \dfrac{x}{4} = 7$, multiply through by $12$ (the LCM of $3$ and $4$): $4x + 3x = 84$, so $7x = 84$ and $x = 12$. When the unknown is in the denominator, multiply both sides by that denominator: $\dfrac{12}{x} = 4$ gives $12 = 4x$, so $x = 3$.

Changing the subject of a formula

Rearranging a formula uses the same balancing moves, but you isolate a chosen letter. Apply inverse operations in the reverse order of BIDMAS. To make $u$ the subject of $v = u + at$, subtract $at$ to get $u = v - at$. To make $a$ the subject of $v = u + at$, subtract $u$ then divide by $t$: $a = \dfrac{v - u}{t}$.

When the new subject appears in more than one place, factorise it out. To make $x$ the subject of $y = \dfrac{x + 2}{x - 1}$, multiply by $(x - 1)$ to get $y(x - 1) = x + 2$, expand to $yx - y = x + 2$, collect $x$ terms: $yx - x = y + 2$, factorise: $x(y - 1) = y + 2$, divide: $x = \dfrac{y + 2}{y - 1}$.

Forming equations from words

Many marks come from turning a worded situation into an equation before solving it. The trick is to name the unknown, then translate each statement into algebra. "I think of a number, multiply it by $3$, add $4$, and the result is $25$" becomes $3x + 4 = 25$, solved to give $x = 7$. A consecutive-number problem such as "three consecutive integers sum to $72$" becomes $x + (x + 1) + (x + 2) = 72$, so $3x + 3 = 72$ and $x = 23$, giving $23, 24, 25$. Defining the variable clearly at the start is what secures the method marks.

Checking and the role of inverse operations

Every linear equation can be checked by substituting the solution back into the original, which costs seconds and catches arithmetic slips. The deeper idea is that solving and rearranging both undo a chain of operations in reverse order. If $y = 3x + 4$ builds $y$ by "multiply by $3$, then add $4$", solving for $x$ reverses it as "subtract $4$, then divide by $3$". Recognising that inverse operations are applied last-first is exactly the BIDMAS-in-reverse principle that also governs changing the subject of a formula.