Solving Equations GCSE Resit Guide

A resit can make familiar maths feel strangely unfamiliar. You may recognise an equation, remember something about changing sides, and still be unsure what to write first. The good news is that solving equations GCSE questions are built from a small set of dependable moves.

Your goal is to isolate the unknown while keeping the equation balanced. Start with one-step equations, progress to two-step and two-sided equations, then add brackets. Show each operation clearly, check your answer by substitution, and the topic becomes far more predictable.

This guide develops that method carefully. You can also use the free GCSE Maths revision hub to find resources for your exam board, including revision lessons, practice questions, past papers and mark schemes.

Your equation-solving checklist

Before working through the examples, remember these principles:

• Treat an equation like a balanced pair of scales.
• Perform the same operation on both sides.
• Undo operations in a sensible order.
• Collect variable terms on one side when the unknown appears twice.
• Expand brackets before trying to isolate the variable.
• Write one algebraic change on each line.
• Substitute your answer into the original equation to check it.

The idea is simple, but the written discipline matters. In a resit paper, correct working can also help an examiner follow your method when a later arithmetic slip occurs.

Before starting, look at the structure of the equation. Does the unknown appear once or on both sides? Are there brackets to expand? Is the unknown multiplied or divided by a number? This brief inspection helps you choose your first operation. You do not need to see the whole solution immediately. You only need to identify a valid next step that makes the equation simpler.

What does solving an equation mean?

An equation says that two expressions have the same value. For example:

$$x+7=19$$

Solving the equation means finding the value of $x$ that makes this statement true. Because $7$ has been added to $x$, subtract $7$ from both sides:

$$x+7-7=19-7$$ $$x=12$$

The language of moving a number to the other side is common, but it can hide what is really happening. The number has not magically moved. You have applied the same operation to both sides, preserving the equality.

For a slower introduction with videos and practice, begin with solving one-step equations.

One-step equations

One-step equations require one inverse operation. Addition is undone by subtraction, multiplication by division, and so on.

Addition and subtraction

Solve:

$$x-8=15$$

Add $8$ to both sides:

$$x-8+8=15+8$$ $$x=23$$

A useful check is:

$$23-8=15$$

The left-hand side equals the right-hand side, so the solution is correct.

Multiplication and division

Solve:

$$5x=35$$

Here, $5x$ means $5\times x$. Divide both sides by $5$:

$$\frac{5x}{5}=\frac{35}{5}$$ $$x=7$$

For an equation such as

$$\frac{x}{4}=6,$$

multiply both sides by $4$:

$$x=24$$

These are central examples of linear equations foundation students should be comfortable solving. Accuracy here gives you the base needed for every harder equation that follows.

Two-step linear equations

A two-step equation contains two operations acting on the unknown. Undo them in reverse order.

Solve:

$$3x+5=20$$

The expression means multiply $x$ by $3$, then add $5$. Undo the addition first:

$$3x=15$$

Now divide both sides by $3$:

$$x=5$$

You can check the answer in the original equation:

$$3\times 5+5=20$$ $$20=20$$

When there is subtraction after the variable term, the structure is the same. For example:

$$4x-7=21$$

Add $7$ before dividing by $4$:

$$4x=28$$ $$x=7$$

MathsGenie's solving equations revision lesson provides further targeted practice once these steps feel secure.

Equations with the unknown on both sides

Students often think a two-sided equation requires a new method. It does not. You still keep the equation balanced, but first collect all variable terms on one side.

Solve:

$$7x-4=3x+16$$

Subtract $3x$ from both sides:

$$4x-4=16$$

Add $4$ to both sides:

$$4x=20$$

Divide by $4$:

$$x=5$$

You could instead subtract $7x$ from both sides, but this would produce a negative variable term. That method can still work, although choosing the smaller variable term to subtract often keeps the arithmetic easier.

The important habit is to write the intermediate line. Jumping directly from the original equation to the answer makes sign errors difficult to spot.

Solving equations with brackets

Brackets are less intimidating when you treat expansion as a preparation step. Multiply every term inside the bracket by the factor outside it.

Solve:

$$3(x+4)=2x+17$$

Expand the bracket:

$$3x+12=2x+17$$

Subtract $2x$ from both sides:

$$x+12=17$$

Subtract $12$:

$$x=5$$

Negative multipliers need particular care. Solve:

$$20-2(x+3)=3x-1$$

The multiplier is $-2$, so both terms inside the bracket change:

$$20-2x-6=3x-1$$

Simplify the left-hand side:

$$14-2x=3x-1$$

Add $2x$ to both sides:

$$14=5x-1$$

Add $1$, then divide by $5$:

$$15=5x$$ $$x=3$$

If expansion is the step causing difficulty, revisit expanding and factorising practice before returning to bracket equations.

Forming equations from a problem

An exam question may not provide the equation. It may describe lengths, prices, ages or angles and ask you to create one.

Suppose a rectangle has length $x+2$, width $x$ and perimeter $28$. The perimeter is the total of all four sides:

$$2(x+2)+2x=28$$

Expand and simplify:

$$2x+4+2x=28$$ $$4x+4=28$$ $$4x=24$$ $$x=6$$

The algebra is not harder than an ordinary linear equation. The main challenge is translating the words correctly. Write what each expression represents before forming the equation. Then ask whether your final answer needs to be interpreted, perhaps as a length, price or number of objects.

Use forming and solving equations practice to develop this translation skill.

How rearranging formulae connects to equations

Rearranging formulae uses the same balancing principles. The difference is that you isolate a chosen letter rather than calculate a numerical value.

For example, make $v$ the subject of:

$$d=vt$$

Divide both sides by $t$:

$$v=\frac{d}{t}$$

For a two-step example, make $x$ the subject of:

$$y=3x+4$$

Subtract $4$:

$$y-4=3x$$

Divide by $3$:

$$x=\frac{y-4}{3}$$

Thinking of the target letter as the unknown makes rearrangement much less mysterious. Higher-tier students can extend this idea through rearranging harder formulae.

Common mistakes in GCSE equation questions

Changing a sign without showing an operation

In

$$x+6=14,$$

writing that $6$ becomes $-6$ may produce the right answer, but it encourages mistakes in longer equations. Think: subtract $6$ from both sides.

Dividing only one term

From

$$4x+8=20,$$

you should normally subtract $8$ first. If you divide the entire equation by $4$, every term must be divided:

$$x+2=5$$

Dividing only $4x$ changes the equation and is incorrect.

Expanding only the first bracket term

The correct expansion is:

$$3(x+5)=3x+15$$

It is not $3x+5$. The outside factor multiplies everything inside the bracket.

Losing a negative sign

Be especially careful with expressions such as:

$$-2(x-4)=-2x+8$$

The second term becomes positive because $-2\times -4=8$.

Stopping before isolating the unknown

A line such as $2x=10$ is progress, not the final answer. Complete the division and write $x=5$.

Exam-style practice

Try these without looking at the answers. Show every line of working and check each solution by substitution.

A. Solve $x+9=23$.

B. Solve $6x=42$.

C. Solve $\frac{x}{5}=9$.

D. Solve $4x+3=27$.

E. Solve $9-2x=-5$.

F. Solve $7x+2=4x+20$.

G. Solve $5(x-2)=2x+11$.

H. Solve $18-3(2x-1)=2x+5$.

I. Make $F$ the subject of $C=\frac{5}{9}(F-32)$.

J. A rectangle has width $x$, length $x+2$ and perimeter $28$. Find $x$.

Answers

A. $x=14$

B. $x=7$

C. $x=45$

D. $x=6$

E. $x=7$

F. $x=6$

G. $x=7$

H. $x=2$

I. $F=\frac{9C}{5}+32$

J. $x=6$

After marking your answers, do not simply count the correct ones. Identify the first line where each incorrect solution went wrong. That line tells you what to practise next.

If substitution gives different results on the two sides, return to your working and check each transition. Look first for an operation applied to only one side, a missed bracket term or a negative sign copied incorrectly. Correct that line, then work forward again. Checking becomes much more useful when it helps you locate an error rather than merely noticing that an answer is wrong.

Build a reliable resit routine

Equation solving improves through short, repeated practice rather than one exhausting session. Begin with a revision lesson, complete a small group of questions, mark them immediately, and return to errors after a day or two. Then apply the skill inside a full paper, where recognising the topic is part of the challenge.

Keep a brief record of recurring mistakes alongside your revision plan. Describe the action you need to remember, such as multiplying every bracket term or dividing both sides. At your next session, attempt a similar question without looking at the correction. Once you can explain your method and repeat it independently, mix that question type with others so you practise choosing the method yourself.

The Edexcel GCSE Maths revision page organises topics by grade and includes practice resources. Whichever exam board you take, MathsGenie can help you combine revision lessons, mini tests, predicted papers, past papers, mark schemes and video solutions into a focused resit plan.

Start with the easiest equation you cannot yet solve confidently. Master it, then add one layer of difficulty. Confidence in algebra rarely arrives all at once. It grows each time your working remains balanced and your checked answer works.