Solving Linear Equations (Cambridge (CIE) IGCSE International Maths)

Revision Note

Author

Mark Curtis

Expertise

Maths

Solving Linear Equations

What are linear equations?

A linear equation is one that can be written in the form $a x + b = c$
- $a, b,$ and $c$ are numbers and $x$ is the variable
  - 2x + 3 = 5
  - 3x + 4 = 1
  - x - 5 = -3
The greatest power of x is 1
- There are no terms like x²

How do I solve linear equations?

You need to use operations like adding, subtracting, multiplying and dividing to get x on its own
Any operation you do to one side of the equation must also be done to the other side
For example, to solve $2 x + 1 = 9$ look at the +1 on the left
- Undo this by subtracting 1 from both sides and simplifying

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

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

$\begin{array}{rcl} 2 x + 1 - 1 & = & 9 - 1 \\ 2 x & = & 8 \end{array}$

This equation is now easier to solve
2x is 2 × x so undo this by dividing both sides by 2 and simplifying

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

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

$\begin{array}{rcl} \frac{2 x}{2} & = & \frac{8}{2} \\ x & = & 4 \end{array}$

The solution to the equation is x = 4
Note that +1 was undone by subtracting 1
- addition and subtraction are said to be inverse (opposite) operations
Similarly, multiplying by 2 was undone by dividing by 2
- multiplication and division are also inverse operations

Does the order of operations matter?

Take $4 x + 8 = 12$
- It is easier to first subtract 8 from both sides

$\begin{array}{rcl} 4 x + 8 - 8 & = & 12 - 8 \\ 4 x & = & 4 \end{array}$

then divide both sides by 4

$\begin{array}{rcl} \frac{4 x}{4} & = & \frac{4}{4} \\ x & = & 1 \end{array}$

If you want to first divide by 4, a common mistake is to write $x + 8 = 3$
- You cannot divide just the 4x and the 12 by 4
  - You must also divide the 8 by 4

$\begin{array}{rcl} \frac{4 x}{4} + \frac{8}{4} & = & \frac{12}{4} \\ x + 2 & = & 3 \end{array}$

Then subtract 2 from both sides

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

How do I solve linear equations with negative numbers?

For example, solve the equation $2 - 3 x = 10$
- Subtract 2 from both sides and simplify

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

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

$\begin{array}{rcl} 2 - x - 2 & = & 10 - 2 \\ - 3 x & = & 8 \end{array}$

Be careful not to lose the negative sign on the 3
Then divide both sides by -3 and simplify

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

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

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

An alternative way to think about $2 - 3 x = 10$ is by reordering the left-hand side
- $2 - 3 x$ is the same as $- 3 x + 2$
- Reordering like this does not change the right-hand side
  - So the equation is $- 3 x + 2 = 10$
- Some people prefer to see it like this
  - You then subtract 2 and divide by -3 as before

Exam Tip

In the exam, substitute your answer back into the original 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 30 ÷ 5
Keep the x on the left-hand side

$x = 6$

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