How to Solve Linear Equations Step by Step

This page condenses a full Year 11 Mathematical Methods lesson on linear equations into the method and the most useful worked examples. It is for students about to start Mathematical Methods or Specialist Mathematics who want solid foundations before calculus.

What is a linear equation?

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A linear equation is one where the variable is only to the first power. In 4x + 8 = 24, the x is really x¹, so the equation is linear.

Compare that with 4x² + 8 = 24. The power of 2 on the x makes it a quadratic, so it is not a linear equation.

Linear equations are a foundation for the Mathematical Methods and Specialist Mathematics content in Years 11 and 12. If they are shaky, you are at risk of struggling when you reach topics such as calculus.

What are the steps to solve a linear equation?

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The aim is always to isolate x. You do that by running the order of operations (BODMAS) in reverse. First undo anything added or subtracted. Then undo anything multiplied or divided. Each time, use the opposite operation and apply it to both sides so the equation stays balanced.

Every type of linear equation in the lesson comes down to the same steps.

1. Clear any fractions. Write every term over the same denominator, then multiply both sides by it.
2. Deal with brackets. Expand them, or divide both sides by the number out the front if that is easier.
3. Group like terms. Move the x terms to one side and the numbers to the other.
4. Factorise x out if x is in two terms with different coefficients.
5. Divide off the coefficient. Divide both sides by whatever is multiplying x, whether that is a number or a bracket.
6. Check. Substitute your answer back into the original equation.

Example: 4x + 8 = 24

There is + 8 on the left, so subtract 8 from both sides.

4x = 16

The 4 is multiplying x, so divide both sides by 4.

x = 4

To check, substitute x = 4 back in. 4 × 4 + 8 = 24, which matches the right-hand side.

How do you solve an equation with x on both sides?

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This adds one step. Group the x terms on one side and the numbers on the other, then solve as normal. It helps to keep your x terms on the left.

Take 8x + 6 = 6x − 10. The 6x on the right is positive, so subtract 6x from both sides.

2x + 6 = −10

You can write 8x − 6x straight away as 2x, because both terms are x to the first power. Now subtract 6 from both sides.

2x = −16

Divide both sides by 2.

x = −8

How do you solve linear equations with brackets?

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There are two ways to handle a number multiplied by a bracket. You can expand the brackets, or you can divide both sides by the number out the front. Have a think about which will be easier for the equation in front of you.

Worked example 1: 6(4x + 10) = 54

1. Expand the brackets: 24x + 60 = 54.
2. Subtract 60 from both sides: 24x = −6.
3. Divide both sides by 24: x = −6/24.
4. Divide the top and bottom by 6 to simplify: x = −1/4.

Leave the answer as −1/4 instead of −0.25. In Methods and Specialist you want exact values, so most teachers will expect the fraction. A decimal makes more sense when there is a reason for it, such as a measurement.

When dividing is quicker

Divide first when the number out the front goes evenly into the other side. In 3(x − 6) = 27, 3 goes into 27, so divide both sides by 3.

x − 6 = 9

x = 15

That is quicker than expanding. Most of the time, though, expanding is the better option.

How do you solve linear equations with fractions?

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An equation with fractions is still linear, as long as x is only to the first power. The approach is to give every term the same denominator, then multiply both sides by that denominator so the fractions disappear. Multiply every term, including any that aren’t fractions.

Example: 3x/15 − 6 = x/3

Since 3 goes into 15, write every term over 15. The 6 becomes 90/15, because there are six 15s in 90. The x/3 becomes 5x/15 when you multiply its top and bottom by 5.

3x/15 − 90/15 = 5x/15

Multiply both sides by 15.

3x − 90 = 5x

Subtract 3x from both sides.

−90 = 2x

x = −45

Why this doesn’t change the equation

Writing 6 as 90/15 is just another way to write 6. When you multiply a fraction’s top and bottom by the same number, you are multiplying by something like 5/5, which equals 1. The value stays the same, so you only do it to that term.

Multiplying a whole side by 15 does change its value. That is why it has to be done to both sides.

Worked example 2: (2x − 6)/4 − (4x − 8)/6 = 10

The denominators are different, so multiply each fraction by the other denominator over itself. That is 6/6 for the first fraction and 4/4 for the second, which puts both over 24.

1. Common denominator: 6(2x − 6)/24 − 4(4x − 8)/24 = 10.
2. Multiply both sides by 24: 6(2x − 6) − 4(4x − 8) = 240.
3. Expand: 12x − 36 − 16x + 32 = 240.
4. Collect like terms: −4x − 4 = 240.
5. Add 4 to both sides: −4x = 244.
6. Divide both sides by −4: x = −61.

Keep 4(4x − 8) in brackets as a whole. The minus sign in front applies to all of it, which is why −4 × −8 gives +32.

Clear the fractions before you expand. Expanding the brackets while they are still over a denominator makes the working messier.

What is a literal equation and how do you solve it?

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A literal equation uses letters in place of numbers. Solving it for x is also called making x the subject. The process is exactly the same as before. The answer is an expression in the other letters instead of a number.

Worked example 3: ax + b = cx + d

1. Subtract cx from both sides: ax − cx + b = d.
2. Subtract b from both sides: ax − cx = d − b.
3. Factorise x out of the left side: x(a − c) = d − b.
4. Divide both sides by (a − c): x = (d − b)/(a − c).

Step 3 is the key skill. To factorise x out, ask what x has to be multiplied by to get each term back. x × a gives ax and x × c gives cx, so the bracket is (a − c).

ax and cx have different coefficients, so you can’t subtract them the way you would with 2x − x. Factorising is how you deal with that.

What shortcuts can you use to solve linear equations faster?

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The practice questions show two techniques that save time. Use them if you want to, or stick with the standard steps.

Flip the signs

Flipping every sign is the same as multiplying both sides by −1. In 5 − x = −4, subtract 5 from both sides.

−x = −9

Flip the signs.

x = 9

You can also flip the signs at the start.

7 − x = 3

x − 7 = −3

x = 4

Multiply by the reciprocal

In −7x/4 = −5, you could multiply both sides by −4 and then divide by 7. Multiplying both sides by the reciprocal, −4/7, does both in one step.

x = −5 × (−4/7)

x = 20/7

Getting rid of the negative on the x early is worth it, because you will have to clear it in the end anyway. The same move solves px/q = r in one step. Multiply both sides by q/p to get x = qr/p.

How do you solve harder literal equations?

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Harder literal equations use the same moves. Most of the work is choosing a sensible order.

Divide before you expand

In p(x + q) = r, expanding would only mean factorising again later. Divide both sides by p instead.

x + q = r/p

x = r/p − q

Group the x terms, then factorise

In px + q = rx, subtract q and subtract rx from both sides.

px − rx = −q

x(p − r) = −q

x = −q/(p − r)

Factorise x out of two fractions

In x/p + x/q = 1, x is multiplied by 1/p in the first term and by 1/q in the second.

x(1/p + 1/q) = 1

x = 1/(1/p + 1/q)

The bracket adds to one fraction, (p + q)/(pq). Dividing 1 by a fraction flips it.

x = pq/(p + q)

Factorise out 1/x

In p/x + q/x = 1, both terms are something multiplied by 1/x.

(1/x)(p + q) = 1

1/x = 1/(p + q)

Each side is now a single fraction with 1 on top, so flip both sides.

x = p + q

Worked example 4: (q − rx)/p + (p − rx)/q + 3 = 0

1. Multiply the first fraction by q/q and the second by p/p so both are over pq: q(q − rx)/(pq) + p(p − rx)/(pq) + 3 = 0.
2. Move the 3 to the right: q(q − rx)/(pq) + p(p − rx)/(pq) = −3.
3. Multiply both sides by pq: q(q − rx) + p(p − rx) = −3pq.
4. Expand: q² − qrx + p² − prx = −3pq.
5. Keep only the x terms on the left: −qrx − prx = −3pq − q² − p².
6. Factorise x out: x(−qr − pr) = −3pq − q² − p².
7. Divide both sides by (−qr − pr): x = (−3pq − q² − p²)/(−qr − pr).
8. A negative over a negative is positive, and r factorises out of the bottom: x = (3pq + q² + p²)/(r(p + q)).

Related resources

• 33 minVideoHow to Factorise Monic Quadratics Using the Cross Method
• 23 minVideoHow to Factorise by Grouping Step by Step
• 55 minVideoHow to Factorise Using the Highest Common Factor (HCF)