Solving Multi-Step Linear Equations

Hello. In the previous lesson, you translated written situations and labelled diagrams into algebraic expressions and equations. Now you will take the next step: solve the equations you form.

A multi-step linear equation may contain brackets, several terms, the variable on both sides, or fractions. The underlying principle remains simple: an equation is a balance, so every operation applied to one side must also be applied to the other. By the end of this lesson, you should be able to solve these equations cleanly and show exam-ready working.

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A dependable order for solving linear equations

Before doing any “moving,” simplify the equation. Then make the variable stand alone.

Use this sequence:

1. Remove brackets by distributing.
2. Combine like terms on each side.
3. Collect variable terms on one side.
4. Collect constants on the other side.
5. Divide or multiply so the variable’s coefficient is $1$.
6. Check by substituting your answer into the original equation.

The key word is equivalent. Every line of working must have exactly the same solution as the line before it.

The image below captures the same exam routine. Its coloured working helps distinguish distributing, combining terms, and applying the same operation to both sides.

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Brackets: distribute every term, including signs

A bracket means that the number outside multiplies every term inside it.

$$a(b+c)=ab+ac$$

For example,

$$5(3x-4)=15x-20.$$

A negative multiplier needs particular care:

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

The second term becomes positive because

$$-3\times -5=15.$$

Consider the equation

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

Start by distributing the $5$:

$$x+15x-20=12x+4.$$

Combine the like terms on the left:

$$16x-20=12x+4.$$

There are variables on both sides. Subtract $12x$ from both sides to leave the variables on the left:

$$4x-20=4.$$

Add $20$ to both sides:

$$4x=24.$$

Finally, divide both sides by $4$:

$$x=6.$$

A check should use the original equation:

$$6+5(3\cdot 6-4)=12\cdot 6+4.$$

The left side is $76$, and the right side is also $76$. Therefore,

$$x=6$$

is correct.

A useful habit: do not write a new line until you can state what happened to both sides. For instance, from

$$16x-20=12x+4$$

to

$$4x-20=4,$$

you subtracted $12x$ from both sides.

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Variables on both sides: choose a sensible side

Suppose you have

$$5x+9=2x-6.$$

You may collect the variables on either side, but it is usually easier to keep a positive coefficient. Since $5x$ is larger than $2x$, subtract $2x$ from both sides:

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

Then subtract $9$:

$$3x=-15.$$

Divide by $3$:

$$x=-5.$$

The important point is that you are not “sending $2x$ across and changing its sign.” You are subtracting $2x$ from each side. This language reflects the mathematics and prevents sign errors.

Here is a compact view of the typical decisions:

What you see	What to do
$a(b+c)$	Distribute $a$ to both terms in the bracket.
$3x+7x$	Combine to make $10x$.
$5x$ on one side and $2x$ on the other	Add or subtract a variable term from both sides.
A number added to the variable term	Use the opposite operation on both sides.
$ax=b$	Divide both sides by $a$, provided $a\ne 0$.

OpenStax’s general strategy is a useful checklist when an equation looks crowded.

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Fractional coefficients: clear fractions first

A fractional coefficient is a fraction multiplying a variable, such as

$$\frac{x}{3}$$

or

$$\frac{2}{5}x.$$

You can solve fractional equations one small fraction-operation at a time, but that often creates complicated arithmetic. Usually, the cleaner method is to clear the fractions.

Find the lowest common denominator, or LCD, of every denominator in the equation. Then multiply every term on both sides by that LCD. This creates an equivalent equation with no fractional coefficients.

Consider this equation, which has brackets, fractional coefficients, and variables on both sides:

$$\frac{2}{3}(x-3)+\frac{x}{4}=\frac{x}{2}+5.$$

The denominators are $3$, $4$, and $2$. Their LCD is $12$.

Multiply the whole left side and the whole right side by $12$:

$$12\left[\frac{2}{3}(x-3)+\frac{x}{4}\right]=12\left[\frac{x}{2}+5\right].$$

Now simplify each product:

$$8(x-3)+3x=6x+60.$$

The fractions have disappeared. From here, return to the usual strategy.

First distribute:

$$8x-24+3x=6x+60.$$

Combine like terms:

$$11x-24=6x+60.$$

Subtract $6x$ from both sides:

$$5x-24=60.$$

Add $24$ to both sides:

$$5x=84.$$

Divide by $5$:

$$x=\frac{84}{5}.$$

The answer does not need to be a whole number. In decimal form,

$$x=\mathrm{16.8.}$$

To check exactly, substitute $x=\frac{84}{5}$ into the original equation. The left side becomes

$$\frac{2}{3}\left(\frac{84}{5}-3\right)+\frac{84}{5\cdot 4}=\frac{46}{5}+\frac{21}{5}=\frac{67}{5}.$$

The right side becomes

$$\frac{84}{5\cdot 2}+5=\frac{42}{5}+\frac{25}{5}=\frac{67}{5}.$$

Both sides match, so the solution is valid.

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Two unusual outcomes to recognise

Most exam equations have one solution, but collecting variables may reveal otherwise.

No solution

$$3x+7=3x-2.$$

Subtract $3x$ from both sides:

$$7=-2.$$

This statement is false, regardless of $x$. Therefore, the equation has no solution.

Infinitely many solutions

$$4(x+1)=4x+4.$$

Distribute:

$$4x+4=4x+4.$$

Subtract $4x+4$ from both sides:

$$0=0.$$

This is always true, so every value of $x$ satisfies the equation. In a real-world context, the context may still restrict which values make sense, but algebraically there are infinitely many solutions.

These outcomes are not mistakes if your working is valid. They tell you that the two sides describe either incompatible relationships or exactly the same relationship.

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An exam-quality presentation routine

For any multi-step equation, keep your working readable:

• Write one equivalent equation per line.
• Align equals signs vertically where possible.
• Expand brackets before combining terms.
• Keep negative signs attached to their terms.
• When clearing fractions, write the multiplication by the LCD clearly before simplifying.
• Check your final answer in the original equation, especially after several sign changes or fraction operations.

The central pattern is always the same: simplify first, preserve equality at every line, then isolate the variable.

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You can now solve multi-step linear equations involving brackets, signed terms, variables on both sides, and fractional coefficients. The most common mistakes are missing a term during distribution, changing only one side of the equation, and forgetting to multiply every term by the LCD.

Next, you will build on this balance-based reasoning by rearranging formulae to make a specified variable the subject.