Solving Linear Equations (Edexcel GCSE Maths: Foundation)

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Mark

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21 February 2024

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

Examiner 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

Examiner 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

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

Examiner 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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Solving Linear Equations (Edexcel GCSE Maths: Foundation)

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 10 free revision notes

Unlock more, it's free!

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Author: Mark