Solving linear equations and inequations in one variable, including equations with brackets, fractions and the unknown on both sides, and representing inequation solutions.

What this dot point is asking

The SQA wants you to solve linear equations in one variable, including those with brackets, fractions and the unknown on both sides, and to solve linear inequations, remembering the special rule that the inequality sign reverses when you multiply or divide by a negative number.

Solving linear equations

The golden rule is to keep the equation balanced: whatever you do to one side you do to the other. Work through the layers in order, expanding brackets first, then collecting terms.

When there are brackets, expand them before collecting terms; when there are fractions, multiply every term by the denominator to clear them.

Solving linear inequations

An inequation (or inequality) uses one of $<$, $>$, $\le$ or $\ge$ instead of an equals sign. You solve it with the same balancing steps, but one rule changes.

You can often avoid the reversal by keeping the unknown positive: solving $7 - 3x \ge 1$ as $7 - 1 \ge 3x$, that is $6 \ge 3x$, gives $2 \ge x$, the same as $x \le 2$.

Equations with the unknown on both sides

When the unknown appears on both sides, gather all the $x$ terms on the side where the coefficient is larger, so the $x$ term stays positive, then solve as usual.

Equations with brackets and fractions together

The hardest questions combine brackets and fractions. The reliable order is: multiply out to clear fractions first, then expand any brackets, then collect and solve.

A reliable habit is to check your answer by substituting it back into the original equation. For $x = 7$ here, $\dfrac{2 \times 7 + 1}{3} = \dfrac{15}{3} = 5$, which matches, confirming the solution. Checking catches sign slips and is quick enough to do every time.

Showing the solution of an inequation

An inequation usually has a range of answers rather than a single value, so the solution is written as an inequality such as $x > 3$ or $x \le -2$. You may also be asked to show it on a number line: an open circle marks a strict inequality ($<$ or $>$) where the end value is not included, and a filled circle marks $\le$ or $\ge$ where it is included, with an arrow showing the direction of all the values that satisfy the inequation.

Examples in context

Inequations describe limits and budgets. If a stallholder must take more than $\pounds 200$ to break even and sells items at $\pounds 8$ each after paying $\pounds 40$ for the pitch, profit is $8n - 40$, and breaking even needs $8n - 40 > 200$. Solving gives $8n > 240$, so $n > 30$: more than $30$ items must be sold. Equations and inequations turn worded conditions into solvable statements.

Try this

Q1. Solve $3x - 7 = 8$. [2 marks]

• Cue. $3x = 15$, so $x = 5$.

Q2. Solve $2(x + 4) = x + 11$. [2 marks]

• Cue. $2x + 8 = x + 11$, so $x = 3$.

Q3. Solve the inequation $5 - x > 2$. [2 marks]

• Cue. $-x > -3$, so $x < 3$.