Solving Linear Equations (Edexcel IGCSE Maths A (Modular))

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Mark Curtis

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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

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

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

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

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

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

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

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

The solution to the equation is x = 4
Adding 1 was undone by subtracting 1
Multiplying by 2 was undone by dividing by 2
- Addition and subtraction are said to be inverse (opposite) operations
- Multiplication and division are also inverse operations

Does the order of steps matter?

As long as each step is applied correctly, the order in which inverse operations are applied does not matter
- Applying the operations in one order may be easier than another
Consider $4 x + 8 = 12$
- It is easier to first subtract 8 from both sides

$4 x = 4$

Then divide both sides by 4

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

If you want to first divide by 4, a common mistake is to write $x + 8 = 3$
- This is incorrect as 8 has not been divided by 4
- You must divide every term 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

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

How do I solve linear equations with negative numbers?

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

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

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

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

Then divide both sides by -3 and simplify

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

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

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

Some people prefer to write $2 - 3 x = 10$ as $- 3 x + 2 = 10$ , which is equivalent
- You then subtract 2 and divide by -3 as before

Exam Tip

Substitute your answer back into the original equation to check you got it right!

Worked Example

Solve the equation

$9 - 7 x = 5$

Subtract 9 from both sides of the equation

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

Divide both sides by -7
Remember that a negative divided by a negative will result in a positive number

$x = \frac{- 4}{- 7}$

$x = \frac{4}{7}$

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

$2 x - 6 = 10$

Then solve as shown previously
- Add 6 then divide by 2

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

Expanding brackets first will always work, but you can also divide first
- Dividing both sides of $2 (x - 3) = 10$ by 2 gives $(x - 3) = 5$
  - which gives $x = 8$
- 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, $\frac{x}{5} + 4 = \frac{9}{2}$
- The lowest common denominator of 5 and 2 is 10
- Multiply all terms on both sides by 10

$\begin{array}{rcl} (10 \times \frac{x}{5}) + (10 \times 4) & = & 10 \times \frac{9}{2} \\ \frac{10 x}{5} + 40 & = & \frac{90}{2} \end{array}$

Simplify the fractions

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

Now solve as before, by subtracting 40, then dividing by 2

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

Unless the question specifies otherwise, you can leave the answer like this
- A decimal or mixed number would also be accepted

What if the unknown is on the denominator?

For example $\frac{4}{x - 2} = 3$
- 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

$4 = 3 (x - 2) 4 = 3 x - 6$

Now solve as before, by adding 6 to both sides, then dividing by 3

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

Worked Example

(a) Solve the equation

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

Expand the bracket

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

Simplify

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

It can be helpful to consider 16 - 20x as -20x + 16
Subtract 16 from both sides

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

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

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

Collect the x terms (or whichever variable is involved) together on one side
- To do this, remove all the x terms from one side
  - It is easiest to remove the smallest x term to avoid negatives
For example, $4 x - 7 = 11 + x$
Remove the x term on the right-hand side, by subtracting 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
This now has the same form as previously seen
- Solve by adding 7 then dividing by 3

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

Worked Example

Solve the equation

$4 - 5 x = 6 x - 29$

Remove the x terms from either side
We will remove them from the left as -5x is smaller than 6x

Add 5x to both sides

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

Get 11x on its own by adding 29 to both sides

$33 = 11 x$

Divide both sides by 11 to find x

$3 = x$

$x = 3$

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Solving Linear Equations (Edexcel IGCSE Maths A (Modular))

Revision Note

Solving Linear Equations

What are linear equations?

How do I solve linear equations?

Does the order of steps matter?

How do I solve linear equations with negative numbers?

Exam Tip

Worked Example

How do I solve linear equations with brackets?

How do I solve linear equations with fractions?

What if the unknown is on the denominator?

Worked Example

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

Worked Example

You've read 0 of your 10 free revision notes

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Author: Mark Curtis