Solving Linear Equations with Brackets and Unknowns on Both Sides

Good to see you again. In the previous lesson, you expanded single brackets by multiplying every term inside and then collected like terms. Now we use those same skills to solve full equations—especially ones where $x$ appears on both sides of the equals sign.

Allow about 40 minutes. By the end, you should be able to expand brackets, simplify each side, get all the $x$-terms onto one side without breaking the equation’s balance, and check your answer in the original equation.

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An equation is a balance

An equals sign means the two sides have exactly the same value. Think of it as a balanced scale: if you do something to the left side, you must do the same thing to the right side.

For the equation in the image,

$$3x+4=x+12$$

we want $x$ on just one side. There is one $x$ on the right, so subtract $x$ from both sides:

$$3x+4-x=x+12-x$$ $$2x+4=12$$

Now remove the $+4$ by subtracting $4$ from both sides:

$$2x+4-4=12-4$$ $$2x=8$$

Finally, divide both sides by $2$:

$$x=4$$

The crucial idea is not “moving terms across and changing signs.” That shortcut can be useful later, but it causes mistakes if you do not understand it. What is really happening is that you are applying the same operation to both sides of a balanced equation.

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The full routine: brackets first, then solve

When an equation has brackets and unknowns on both sides, use this order:

1. Expand every bracket.
2. Simplify each side by collecting like terms.
3. Remove an $x$-term from one side by adding or subtracting the same $x$-term on both sides.
4. Remove the remaining number term.
5. Divide by the coefficient of $x$.
6. Check your answer in the original equation.

The first two steps come directly from the previous lesson. Only after the brackets are gone should you start solving.

Watch Maths Genie’s Solving equations with an unknown on both sides. The selected example combines the two skills: expanding brackets accurately, then eliminating an unknown term from one side.

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Worked example: brackets on both sides

Solve:

$$3(2x+1)=4(x-2)+17$$

1. Expand both brackets

On the left, multiply $3$ by both terms. On the right, multiply $4$ by both terms:

$$6x+3=4x-8+17$$

2. Simplify each side

The right side has two number terms:

$$6x+3=4x+9$$

At this point, the brackets have gone and each side is as simple as possible.

3. Put all $x$-terms on one side

There is $6x$ on the left and $4x$ on the right. Subtract $4x$ from both sides:

$$6x+3-4x=4x+9-4x$$ $$2x+3=9$$

Choosing to subtract $4x$, the smaller positive $x$-term, leaves a positive $2x$. This is usually the neatest choice.

4. Remove the number next to $x$

Subtract $3$ from both sides:

$$2x+3-3=9-3$$ $$2x=6$$

5. Divide

$$x=3$$

6. Check in the original equation

Use $x=3$:

$$3(2x+1)=4(x-2)+17$$ $$3(2\times 3+1)=4(3-2)+17$$ $$3(7)=4(1)+17$$ $$21=21$$

Both sides match, so:

$$\boxed{x=3}$$

Checking is especially valuable when brackets and negative signs are involved. It can catch an expansion or sign error that otherwise looks convincing.

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Why “remove the smaller $x$-term” usually works

Suppose that, after expanding and simplifying, you have:

$$6x-3=x+17$$

Subtracting $x$ from both sides gives:

$$5x-3=17$$

This is usually easier than subtracting $6x$, which would give:

$$-3=-5x+17$$

That second route is still mathematically correct, but it brings in negative coefficients unnecessarily.

So, when both $x$-terms are positive, a good exam habit is:

Remove the $x$-term with the smaller coefficient.

It is a convenience, not a law. When one $x$-term is negative, you often need to add instead.

For example:

$$4(x+3)=18-2x$$

Expand first:

$$4x+12=18-2x$$

The right side has $-2x$. To cancel it, add $2x$ to both sides:

$$4x+12+2x=18-2x+2x$$ $$6x+12=18$$

Then subtract $12$:

$$6x=6$$

Finally divide by $6$:

$$x=1$$

Notice that adding $2x$ is not random: it creates the zero pair

$$-2x+2x=0$$

on the right-hand side.

The BBC Bitesize article gives a visual explanation of this balance idea and reinforces the strategy of eliminating an $x$-term from both sides.

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A complete exam-style example

Solve:

$$5(x-2)+3=2(x+4)+12$$

Write one tidy line per algebra step. Do not try to expand, collect terms, and solve all at once.

Expand

$$5x-10+3=2x+8+12$$

Collect like terms on each side

$$5x-7=2x+20$$

Remove the smaller $x$-term

Subtract $2x$ from both sides:

$$5x-7-2x=2x+20-2x$$ $$3x-7=20$$

Remove the number term

Add $7$ to both sides:

$$3x-7+7=20+7$$ $$3x=27$$

Divide by $3$

$$x=9$$

Check

Substitute $x=9$ into the original equation:

$$5(9-2)+3=2(9+4)+12$$ $$5(7)+3=2(13)+12$$ $$35+3=26+12$$ $$38=38$$

So the answer is:

$$\boxed{x=9}$$

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Errors that lose easy marks

Expanding only one term in a bracket

Incorrect:

$$3(2x-5)=6x-5$$

Correct:

$$3(2x-5)=6x-15$$

The $3$ multiplies both $2x$ and $-5$.

Forgetting to simplify after expanding

For example:

$$2(x+3)+4x=18$$

must become:

$$2x+6+4x=18$$

then:

$$6x+6=18$$

Do not treat $2x+6+4x$ as if it were already simplified.

Doing an operation to only one side

Incorrect:

$$5x-7=2x+20$$ $$3x-7=20$$

This looks like $2x$ disappeared from the left, but the operation has not been shown on both sides. Write it clearly:

$$5x-7-2x=2x+20-2x$$

Then simplify to get:

$$3x-7=20$$

Checking the wrong equation

Your answer should be checked in the equation you were originally given, including its brackets. If you only check a simplified line, you might miss an early mistake in expansion.

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A quick self-check before you finish

Before boxing an answer, scan your working:

• Have all brackets been expanded correctly?
• Have you collected like terms on each side?
• Did every added or subtracted term appear on both sides?
• Is $x$ alone in your final line?
• Does substituting your answer into the original equation make both sides equal?

If your final answer is a fraction or negative number, do not assume it is wrong. Equations can perfectly validly have answers such as:

$$x=-4$$

or

$$x=-\frac{5}{2}$$

The check tells you whether the value works.

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Key takeaways

• Expand brackets before trying to isolate the unknown.
• Simplify each side separately by collecting like terms.
• Keep the equation balanced: whatever you add, subtract, multiply, or divide by on one side must also be done on the other.
• Usually, eliminating the smaller positive $x$-term avoids unnecessary negative coefficients.
• If an $x$-term is negative, add its positive opposite to both sides.
• Substitute your answer back into the original equation to check it.

Next, you will practise another powerful algebra skill: rearranging formulae to make a different variable the subject. The balance principle from this lesson will still be central—except the goal will be to leave a letter, rather than a number, on its own.