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Linear Equations (CIE IGCSE Maths: Core)

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Mark

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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
- 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
- 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

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

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

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

- - 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 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

$x = 6$

Equations with Brackets & Fractions

How do I solve linear equations with brackets?

If a linear equation involves brackets, expand the brackets first
For example, solve $2 (x - 3) = 10$
- Expand the brackets

$2 x - 6 = 10$

- This now has the same form as before
  - Solve by adding 6 then dividing by 2

$\begin{array}{rcl} 2 x & = & 16 \\ x & = & \frac{16}{2} \\ x & = & 8 \end{array}$

Expanding brackets first will always work, but you can also divide first
- Dividing both sides of by 2 gives
  - which solves to $x = 8$
- This method works but can lead to harder fractions

How do I solve linear equations with fractions?

If a linear equation contains fractions, multiply both sides by the lowest common denominator
For example, solve $\frac{x}{5} + 4 = \frac{9}{2}$
- The lowest common denominator of 5 and 2 is 10
- Multiply all terms on both sides by 10
  - Don't forget to do it to the 4

$\begin{array}{rcl} 10 \times \frac{x}{5} + 10 \times 4 & = & 10 \times \frac{9}{2} \end{array}$

- Simplify the fractions
  - Imagine the 10 is in the numerator and cancel

$\begin{array}{rcl} 2 x + 40 & = & 45 \end{array}$

- This now has the same form as before
  - Solve by subtracting 40 then dividing by 2

$\begin{array}{rcl} 2 x & = & 5 \\ x & = & \frac{5}{2} \end{array}$

- Unless the question asks, you can leave the answer like this
  - A decimal or mixed number would also be accepted
Sometimes equations have the unknown on the denominator, for example
- Multiply both sides of the equation by the denominator

$\frac{4}{x - 2} \times (x - 2) = 3 (x - 2)$

- Simplify the fractions, and expand any brackets

$4 = 3 (x - 2)$

$4 = 3 x - 6$

- This now looks like a standard linear equation
  - Solve by adding 6 to both sides, then dividing by 3

$10 = 3 x \frac{10}{3} = x$

Exam Tip

In the exam, always substitute your answer back into the equation to check you got it right!

Worked example

(a)

Solve the equation

$5 (3 - 4 x) + 1 = 26$

First expand the brackets on the left-hand side

$\begin{array}{rcl} 15 - 20 x + 1 & = & 26 \end{array}$

Simplify the left-hand side (by adding 15 and 1)

$\begin{array}{rcl} 16 - 20 x & = & 26 \end{array}$

Remember 16 - 20x is the same as -20x + 16
Subtract 16 from both sides

$\begin{array}{rcl} - 20 x & = & 26 - 16 \\ - 20 x & = & 10 \end{array}$

Divide both sides by -20 and simplify

$\begin{array}{rcl} x & = & \frac{10}{- 20} \\ x & = & - \frac{1}{2} \end{array}$

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

-0.5 is also accepted

(b)

Solve the equation

$\frac{5 x}{4} = \frac{1}{2}$

The lowest common denominator of 4 and 2 is 4
Multiply both sides by 4

$\begin{array}{rcl} 4 \times \frac{5 x}{4} & = & 4 \times \frac{1}{2} \end{array}$

Simplify (cancel) the fractions

$\begin{array}{rcl} 5 x & = & 2 \end{array}$

To solve this equation, divide both sides by 5

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

0.4 is also accepted

Unknown on Both Sides

How do I solve linear equations with x terms on both sides?

If a linear equation contains the unknown variable on both sides, collect the x terms together on one side
- To do this, choose a side you wish to remove the x term from
  - this should be the side with the lowest value x term
- Apply the opposite of that term to both sides
For example, solve $4 x - 7 = 11 + x$
- Let's remove the x term on the right-hand side
- x has been added, so subtract x from both sides

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

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

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

- - There are no longer any x terms on the right
  - The 4x has become 3x on the left
- This now has the same form as equations with one variable
  - Solve it by adding 7 then dividing by 3

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

Exam Tip

In the exam, substitute your answer back into the original 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 remove this term first
Add 5x to both sides and simplify (collect like terms)

$\begin{array}{rcl} 4 - 5 x + 5 x & = & 6 x - 29 + 5 x \\ 4 & = & 11 x - 29 \end{array}$

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 can reflect the equation to present your final answer as x = ...

$x = 3$

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Linear Equations (CIE IGCSE Maths: Core)

Revision Note

What are linear equations?

How do I solve linear equations?

Does the order of operations matter?

How do I solve linear equations with negative numbers?

How do I solve linear equations with brackets?

How do I solve linear equations with fractions?

How do I solve linear equations with x terms on both sides?

You've read 0 of your 0 free revision notes

Get unlimited access

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

Author: Mark