Solving Linear Equations (Cambridge O Level Maths)

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Amber

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24 October 2023

Solving Linear Equations

What are linear equations?

A linear equation is an equation that will produce a straight line when plotted on a graph
The greatest power of $x$ in a linear equation is 1
- This means there are no terms of $x^{2}$ or a higher order
A linear equation is normally in form $a x + b = c$
- where $a, b,$ and $c$ are constants and $x$ is a variable

How do I solve a linear equation?

To solve a linear equation you need to isolate the variable, usually $x$ , by carrying out inverse operations to both sides of the equation
- Inverse operations are just the opposite operations to what has already happened to the variable
The order in which the inverse operations are carried out is important
- Most of the time, this will be BIDMAS in reverse
- However it depends on the order in which the operations were applied to the variable to form the equation

How do I solve a linear equation of the form ax + b = c?

The operations that have been applied to $x$ here are:
- STEP 1
  Multiply by $a$

- STEP 2
  Add $b$
To solve this, you must carry out the inverse operations in reverse order
- STEP 1
  Subtract $b$
- STEP 2
  Divide by $a$
For example, to solve the equation $2 x + 1 = 9$
- STEP 1
  Subtract 1

$\begin{array}{rcl} 2 x + 1 & = & 9 \end{array}$

$\begin{array}{rcl} (- 1) \end{array} \begin{array}{rcl} (- 1) \end{array}$

$\begin{array}{rcl} 2 x & = & 8 \end{array}$

- STEP 2
  Divide by 2

$\begin{array}{rcl} 2 x & = & 8 \end{array}$

$\begin{array}{rcl} (\div 2) \end{array} \begin{array}{rcl} (\div 2) \end{array}$

$\begin{array}{rcl} x & = & 4 \end{array}$

Be extra careful if any of the terms have negatives
For example, to solve the equation $2 - 3 x = 10$
- STEP 1
  Subtract 2

$\begin{array}{rcl} 2 - 3 x & = & 10 \end{array}$

$\begin{array}{rcl} (- 2) \end{array} \begin{array}{rcl} (- 2) \end{array}$

$\begin{array}{rcl} - 3 x & = & 8 \end{array}$

- - Be careful not to drop the negative sign
- STEP 2
  Divide by -3

$\begin{array}{rcl} - 3 x & = & 8 \end{array}$

$\begin{array}{rcl} (\div - 3) \end{array} \begin{array}{rcl} (\div - 3) \end{array}$

$\begin{array}{rcl} x & = & - \frac{8}{3} \end{array}$

Examiner Tip

If you have time in the exam, you should substitute your answer back into the equation to check you got it right

Worked example

Solve the equation

$5 x - 8 = 22$

Add 8 to both sides of the equation

$\begin{array}{rcl} 5 x & = & 22 + 8 \end{array}$

Work out 22 + 8

$5 x = 30$

Divide both sides by 5

$x = \frac{30}{5}$

Work out $\frac{30}{5}$ (30 ÷ 5)
Keep the x on the left-hand side

$x = 6$

Equations with Brackets & Fractions

How do I solve a linear equation that contains brackets?

If a linear equation contains brackets on one, or both, sides start by expanding the brackets
For example, to solve the equation $2 (x - 1) = 10 - (3 + x)$
- STEP 1
  Expand the brackets on both sides

$2 x - 2 = 10 - 3 - x$

- STEP 2
  Collect the $x$ terms to one side by subtracting the term with the smaller coefficient of $x$

$\begin{array}{rcl} 2 x - 2 & = & 7 - x \end{array}$

$\begin{array}{rcl} - (- x) - (- x) \end{array}$

$\begin{array}{rcl} 3 x - 2 & = & 7 \end{array}$

- - Be extra careful if any of the terms have negatives
- STEP 3
  Solve the linear equation using the method above

$\begin{array}{rcl} 3 x - 2 & = & 7 \\ 3 x & = & 9 \\ x & = & 3 \end{array}$

How do I solve a linear equation that contains fractions?

If a linear equation contains a fraction on one or both sides, remove the fractions by multiplying both sides by everything on the denominator
- Remember to put brackets around the expression first
For example, to solve the equation $\frac{2}{x + 3} = \frac{3}{4 - 2 x}$ you will need to multiply both sides of the equation by both $(x + 3)$ and $(4 - 2 x)$
- STEP 1
  Multiply both sides by $(x + 3)$

$\begin{array}{rcl} \frac{2}{x + 3} (x + 3) & = & \frac{3}{4 - 2 x} (x + 3) \\ 2 & = & \frac{3 (x + 3)}{4 - 2 x} \end{array}$

- STEP 2
  Multiply both sides by $(4 - 2 x)$

$\begin{array}{rcl} 2 (4 - 2 x) & = & \frac{3 (x + 3)}{4 - 2 x} (4 - 2 x) \\ 2 (4 - 2 x) & = & 3 (x + 3) \end{array}$

- STEP 3
  Expand the brackets on both sides

$8 - 4 x = 3 x + 9$

- STEP 4
  Collect the $x$ terms to one side by subtracting the term with the smaller coefficient of $x$

$\begin{array}{rcl} 8 - 4 x & = & 3 x + 9 \end{array}$

$\begin{array}{rcl} - (- 4 x) - (- 4 x) \end{array}$

$\begin{array}{rcl} 8 & = & 7 x + 9 \end{array}$

- STEP 5
  Solve the equation
  - You can swap the sides if it makes solving the equation easier

$\begin{array}{rcl} 7 x + 9 & = & 8 \\ 7 x & = & - 1 \\ x & = & - \frac{1}{7} \end{array}$

Examiner Tip

If you have time in the exam, you should substitute your answer back into the equation to check you got it right

Worked example

Solve the equation.

$\frac{3 x - 2}{4 - x} = - 1$

Get rid of the fraction by multiplying both sides by the denominator

(4 - x)

$\begin{array}{rcl} 3 x - 2 & = & - (4 - x) \end{array}$

Expand the brackets.

$\begin{array}{rcl} 3 x - 2 & = & - 4 + x \end{array}$

Bring the x terms to one side of the equation by subtracting x from both sides.

$\begin{array}{rcl} 2 x - 2 & = & - 4 \end{array}$

Get the x term by itself by adding 2 to both sides.

$2 x = - 2$

Solve by dividing both sides by 2.

$x = - 1$

Solve the equation.

$2 + \frac{x + 1}{3} = 4$

Isolate the fraction by subtracting 2 from both sides.

$\begin{array}{rcl} \frac{x + 1}{3} & = & 2 \end{array}$

Get rid of the fraction by multiplying both sides by the denominator (3).

$\begin{array}{rcl} x + 1 & = & 6 \end{array}$

Get the x term by itself by subtracting 1 from both sides.

$x = 5$

Unknown on Both Sides

How do I solve a linear equation with the unknown variable, x, on both sides?

If a linear equation contains the unknown variable, usually $x$ on both sides start by collecting these terms together on one side of the equation
- Moving the $x$ term with the smallest coefficient (number in front of $x$ ) is easiest
For example, to solve the equation $4 x - 7 = 11 + x$
- STEP 1
  Move the $x$ term with the smallest coefficient of $x$
  - The coefficients are 4 and 1 so move the $x$ -term on the right hand side

$\begin{array}{rcl} 4 x - 7 & = & 11 + x \end{array}$

$\begin{array}{rcl} (- x) (- x) \end{array}$

$\begin{array}{rcl} 3 x - 7 & = & 11 \end{array}$

- STEP 2
  Solve the linear equation using the method above

$\begin{array}{rcl} 3 x - 7 & = & 11 \\ 3 x & = & 18 \\ x & = & 6 \end{array}$

Examiner Tip

If you have time in the exam, you should substitute your answer back into the equation to check you got it right

Worked example

Solve the equation

$4 - 5 x = 6 x - 29$

5x is smaller than 6x so it's easier to move this term first
Add 5x to both sides

$4 = 6 x - 29 + 5 x$

Simplify the right-hand side by adding the 6x and the 5x

$4 = 11 x - 29$

This could also have been written as 4 = -29 + 11x
Leave the x's on the right-hand side (avoid 'reflecting' the equation)
Get 11x on its own (by adding 29 to both sides)

$4 + 29 = 11 x$

Work out 4 + 29

$33 = 11 x$

Get x on its own (by dividing both sides by 11)

$\frac{33}{11} = x$

Work out 33 ÷ 11

$3 = x$

The answer currently has x on the right-hand side
At this point, you may reflect the equation to give your answer as x = ...

$x = 3$

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Solving Linear Equations (Cambridge O Level Maths)

Revision Note

What are linear equations?

How do I solve a linear equation?

How do I solve a linear equation of the form ax + b = c?

How do I solve a linear equation that contains brackets?

How do I solve a linear equation that contains fractions?

How do I solve a linear equation with the unknown variable, x, on both sides?

You've read 0 of your 5 free revision notes this week

Sign up now. It’s free!

Join the 100,000+ Students that ❤️ Save My Exams

Author: Amber