Equations with Brackets & Fractions (Cambridge (CIE) IGCSE International Maths)

Revision Note

Author

Mark Curtis

Expertise

Maths

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 $2 (x - 3) = 10$ by 2 gives $(x - 3) = 5$
  - 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 $\frac{4}{x - 2} = 3$
- 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

$4 = 3 (x - 2)$

$4 = 3 x - 6$

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

You've read 0 of your 10 free revision notes

Unlock more, it's free!

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

the (exam) results speak for themselves:

Next topic

Did this page help you?