Algebra
UK Secondary (7-11)
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Addition and subtraction of rational expressions II
Lesson

In How Low Can You Go, we looked at how to find the lowest common denominator (LCD) between algebraic terms. This is the first step we take to add and subtract algebraic fractions.

In this chapter, we'll look at the rest of the process. It's just like adding and subtracting fractions.

Let's go through it using an example: $\frac{2x}{5\left(x+4\right)}-\frac{1}{x+4}$2x5(x+4)1x+4

1. Find the LCD between the denominators.

The LCD between the two fractions is $5\left(x+4\right)$5(x+4)

2. Multiply each term by the necessary multiple to reach the LCD. Remember, whatever you do to the denominator, you must also do to the numerator to keep the fractions equivalent.

We don't need to change the first term as it already has the LCD as the denominator. The second term we need to multiply by $5$5:

$\frac{1}{x+4}\times\frac{5}{5}=\frac{5}{5\left(x+4\right)}$1x+4×55=55(x+4)

3. Now we can solve the problem by adding/ subtracting the numerators. Remember to simplify your answer if you can.

$\frac{2x}{5\left(x+4\right)}-\frac{5}{5\left(x+4\right)}=\frac{2x-5}{5\left(x-4\right)}$2x5(x+4)55(x+4)=2x55(x4)

 

Remember!

We can only add and subtract fractions with like denominators. However, whatever we do to the denominator, we need to do to the numerator to keep our fractions equivalent.

 

Worked Examples

Question 1

Simplify $\frac{5t}{9}-\frac{7t}{2}$5t97t2.

Question 2

Simplify $\frac{4}{3p}-\frac{2}{p}$43p2p.

Question 3

Simplify the following expression: $\frac{4}{3\left(x+9\right)}+\frac{1}{x\left(x+9\right)}$43(x+9)+1x(x+9)

question 4

Simplify $\frac{5y}{3y-3}-\frac{4y}{9y+7}$5y3y34y9y+7. Give your final answer in factorised form.

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