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Simplifying expressions Collecting like terms Expanding brackets Factorising Factorising quadratics Difference of two squares Adding and subtracting fractions Simplifying fractionsThis topic is relevant for:

Here we will learn about adding and subtracting algebraic fractions, including algebraic fractions with single and binomial denominators. We will also look at some problems involving quadratics and the difference of two squares.

There are also adding and subtracting algebraic fractions worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.

**Adding and subtracting algebraic fractions **is the skill of adding and subtracting two or more fractions that contain **algebraic terms**.

For example, \cfrac{3}{2x+3}, \, \cfrac{5}{y^{2}} \, , \, or \, \cfrac{2ab}{3a+4b} \, .

In order to add or subtract fractions we must ensure that the fractions have a **common denominator**.

For example, \cfrac{5}{2a^2} + \cfrac{c}{6ab} \, .

Finding the LCM of the algebraic denominators 2a^2 and 6ab using a Venn diagram.

2a^2 = 2\times a \times a

6ab = 2\times 3 \times a \times b

LCM = a \times 2\times a\times 3\times b =6a^{2}b

So the first denominator, 2a^2 , needs multiplying by 3b in order to make 6a^{2}b .

The second denominator, 6ab , needs multiplying by a in order to make 6a^{2}b .

Remember to multiply the numerator and denominator of a fraction by the same thing in order to write an equivalent fraction.

\cfrac{5\times 3b}{2a^2\times 3b} + \cfrac{c\times a}{6ab\times a} = \cfrac{15b}{6a^{2}b} + \cfrac{ac}{6a^{2}b} = \cfrac{15b+ac}{6a^{2}b}

**Step-by-step guide:** Adding and subtracting fractions

**Step-by-step guide:** Algebraic terms

If the fractions already have a **common denominator** then the numerators can be easily added/subtracted and the final answer will also have the same denominator. For algebraic fractions you may need to use additional skills such as simplifying expressions by collecting like terms.

For example,

\cfrac{5}{a}-\cfrac{b}{a} \, =\cfrac{5-b}{a}

\cfrac{2t}{k+1} \, +\cfrac{t}{k+1} \, =\cfrac{2t+t}{k+1} \, =\cfrac{3t}{k+1}

If the fractions have **different denominators,** we then need to find the **lowest common multiple** (LCM) of those denominators. Then we can use **equivalent fractions** to write all the fractions in the question with a **common denominator.**

Finding the LCM of **numerical denominators** will be a familiar skill. Additional skills such as expanding brackets and collecting like terms may be required for the algebraic numerators.

For example,

\cfrac{7a+3}{5}+\cfrac{4a-1}{10} \, .

The LCM of 5 and 10 is 10.

\cfrac{(7a+3)\times 2}{5\times 2} \, + \cfrac{4a-1}{10} \, =\cfrac{14a+6}{10} \, +\cfrac{4a-1}{10} \, =\cfrac{14a+6+4a-1}{10} \, = \cfrac{18a+5}{10}

Finding the **lowest common multiple** of algebraic denominators can involve many different skills including identifying unique algebraic factors and factorising algebraic expressions.

Here are some examples of finding common algebraic denominators.

**Find the LCM of monomial expressions.**In order to work out the LCM of algebraic denominators with**single terms**(known as monomial expressions) you need to identify the**common algebraic factors**and the**unique algebraic factors**of each term.

You may also need to use the skill of finding the LCM of any numerical values involved. You then**multiply together**the

– common algebraic factors– unique algebraic factors

– numerical LCM

to find the overall**LCM for the algebraic terms**.

For example,

**Using Venn diagrams to find the LCM**It can also be helpful to use**Venn diagrams**to find the LCM of algebraic terms. First you need to write the algebraic terms as products of their factors (using prime factors for the numerical part of the term).

Then you put the factors into a Venn diagram, and the LCM will be the result of all the factors in the Venn diagram multiplied together.

For example,

**Multiply the denominators together to a common denominator**Another technique to find a common denominator is to multiply the two denominators together.

This technique does not always find the**lowest**common multiple but does find a common multiple which can be used as the common denominator. This technique can be very useful when there are denominators with more than one term.

For example,

**LCM of binomial and quadratic expressions**Another technique for finding the LCM of algebraic denominators with more than one term is to factorise the expressions first. This can sometimes help to identify the LCM by highlighting common factors.

For example,

**Step-by-step guide:** Factorising

**Step-by-step guide:** Factorising quadratics

**Step-by-step guide:** Difference of two squares

In order to add and subtract algebraic fractions:

**Determine the lowest common multiple of the denominators.****For each fraction, multiply the numerator and the denominator by the same value to obtain the common denominator.****For each fraction, simplify the numerator expression and simplify the denominator expression.****Add/subtract the numerators together and write over the common denominator.**

Get your free adding and subtracting algebraic fractions worksheet of 20+ algebraic fractions questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREEGet your free adding and subtracting algebraic fractions worksheet of 20+ algebraic fractions questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREEWrite as a single fraction in the simplest form,

\cfrac{2}{a} \, +\cfrac{3}{b} \, .

**Determine the lowest common multiple of the denominators.**

Sometimes the lowest common multiple can be found by simply multiplying the two denominators together. This is the case here.

The lowest common multiple of a and b is a\times{b} which we can write as ab.

2**For each fraction, multiply the numerator and the denominator by the same value to obtain the common denominator.**

Multiply the numerator and the denominator of the first fraction by b.

Multiply the numerator and the denominator of the second fraction by a.

3**For each fraction, simplify the numerator expression and simplify the denominator expression.**

The numerator of the first fraction becomes 2\times{b}=2b.

The numerator of the second fraction becomes 3\times{a}=3a.

The common denominator is ab.

This gives us the equivalent calculation \cfrac{2b}{ab} \, +\cfrac{3a}{ab} \, .

4**Add/subtract the numerators together and write over the common denominator.**

\cfrac{2b}{ab} \, +\cfrac{3a}{ab} \, =\cfrac{2b+3a}{ab}

The numerator is 2b+3a.

This expression cannot be simplified because 2b and 3a are not like terms.

The final answer is \cfrac{2b+3a}{ab} \, .

Write as a single fraction in the simplest form,

\cfrac{6}{e}\,-\cfrac{5}{f} \, .

**Determine the lowest common multiple of the denominators.**

Sometimes the lowest common multiple can be found by simply multiplying the two denominators together. This is the case here.

The lowest common multiple of e and f is e\times{f} which we can write as ef.

Multiply the numerator and the denominator of the first fraction by f.

Multiply the numerator and the denominator of the second fraction by e.

**For each fraction, simplify the numerator expression and simplify the denominator expression. **

The numerator of the first fraction becomes 6\times{f}=6f.

The numerator of the second fraction becomes 5\times{e}=5e.

The common denominator is ef.

This gives us the equivalent calculation \cfrac{6f}{ef} \, -\cfrac{5e}{ef} \, .

**Add/subtract the numerators together and write over the common denominator.**

\cfrac{6f}{ef} \, -\cfrac{5e}{ef} \, =\cfrac{6f-5e}{ef}

The numerator is 6f-5e.

This expression cannot be simplified because 6f and - 5e are not like terms.

Then final answer is \cfrac{6f-5e}{ef} \, .

Write as a single fraction in the simplest form,

\cfrac{3x}{4} \, +\cfrac{x}{6} \, .

**Determine the lowest common multiple of the denominators.**

The lowest common multiple of 4 and 6 is 12.

As 12 \div 4 = 3, we need to multiply the numerator and the denominator of the first fraction by 3.

As 12 \div 6 = 2, we need to multiply the numerator and the denominator of the second fraction by 2.

**For each fraction, simplify the numerator expression and simplify the denominator expression.**

The numerator of the first fraction becomes 3x\times{3}=9x.

The numerator of the second fraction becomes x\times{2}=2x.

The common denominator is 12.

This gives us the equivalent calculation \cfrac{9x}{12} \, +\cfrac{2x}{12}\, .

**Add/subtract the numerators together and write over the common denominator.**

\cfrac{9x}{12} \, +\cfrac{2x}{12} \, =\cfrac{9x+2x}{12}

The numerator is 9x+2x \, . These are like terms so we can simplify this expression.

The final answer is \cfrac{11x}{12} \, .

Write as a single fraction in the simplest form,

\cfrac{5x}{7}\,-\cfrac{2x}{3} \, .

**Determine the lowest common multiple of the denominators.**

The lowest common multiple of 7 and 3 is 21.

Multiply the numerator and the denominator of the first fraction by 3.

Multiply the numerator and the denominator of the second fraction by 7.

**For each fraction, simplify the numerator expression and simplify the denominator expression. **

The numerator of the first fraction becomes 5x\times{3}=15x.

The numerator of the second fraction becomes 2x\times{7}=14x.

The common denominator is 21.

This gives us the equivalent calculation \cfrac{15x}{21} \,-\cfrac{14x}{21}\, .

**Add/subtract the numerators together and write over the common denominator.**

\cfrac{15x}{21} \, -\cfrac{14x}{21} \, =\cfrac{15x-14x}{21}

The numerator is 15x-14x .

These are like terms so we can simplify this expression.

\cfrac{15x-14x}{21} \, =\cfrac{x}{21}

The final answer is \cfrac{x}{21} \, .

Write as a single fraction in the simplest form,

\cfrac{2x+1}{3} \, +\cfrac{x+2}{2} \, .

**Determine the lowest common multiple of the denominators.**

The lowest common multiple of 3 and 2 is 6.

Multiply the numerator and the denominator of the first fraction by 2.

Multiply the numerator and the denominator of the second fraction by 3.

**For each fraction, simplify the numerator expression and simplify the denominator expression.**

**Note**: This will use skills such as expanding brackets, multiplying terms and collecting like terms. However, do not simplify the fraction by cancelling or dividing.** **

We have to think carefully at this point because the numerators have more than one term. We must remember to multiply all terms in the numerator by the identified value.

It can help to write each numerator in a single bracket before multiplying, like this,

\cfrac{2(2x+1)}{6} \, +\cfrac{3(x+2)}{6} \, .

By expanding the brackets, the numerator of the first fraction becomes 2(2x+1)=4x+2.

By expanding the brackets, the numerator of the second fraction becomes 3(x+2)=3x+6.

This gives us the updated calculation \cfrac{4x+2}{6} \, +\cfrac{3x+6}{6}.

**Add/subtract the numerators together and write over the common denominator. Simplify the numerator expression by collecting like terms.**

Imagine that each numerator is within a single set of brackets.

When we add the numerators together (because the denominators are now the same), we get (4x+2)+(3x+6).

\cfrac{4x+2}{6} \, +\cfrac{3x+6}{6} \, =\cfrac{(4x+2)+(3x+6)}{6}

Expanding each bracket gives us the expression 4x+2+3x+6 which can be simplified to 7x+8 by collecting like terms.

\cfrac{(4x+2)+(3x+6)}{6} \, =\cfrac{7x+8}{6}

This step is important when subtracting fractions (see example 6).

The final answer is \cfrac{7x+8}{6} \, .

Write as a single fraction in the simplest form,

\cfrac{2x+1}{4} \, -\cfrac{x-1}{5} \, .

**Determine the lowest common multiple of the denominators.**

The lowest common multiple of 4 and 5 is 20.

Multiply the numerator and the denominator of the first fraction by 5.

Multiply the numerator and the denominator of the second fraction by 4.

**For each fraction, simplify the numerator expression and simplify the denominator expression. **

**Note: **This will use skills such as expanding brackets, multiplying terms and collecting like terms. However, do not simplify the fraction by cancelling or dividing.** **

We have to think carefully at this point because the numerators have more than one term. We must remember to multiply all terms in the numerator by the identified value.

Writing each numerator in a single bracket before multiplying makes our calculation look like this.

\cfrac{5(2x+1)}{20} \, -\cfrac{4(x-1)}{20}

By expanding the brackets, the numerator of the first fraction becomes 5(2x+1)=10x+5.

By expanding the brackets, the numerator of the second fraction becomes 4(x-1)=4x-4.

This gives us the equivalent calculation \cfrac{10x+5}{20} \, -\frac{4x-4}{20}\, .

Imagine that each numerator is within a single set of brackets.

When we subtract the numerators, we get (10x+5)-(4x-4).

\cfrac{10x+5}{20} \, -\cfrac{4x-4}{20} \, =\cfrac{(10x+5)-(4x-4)}{20}=\cfrac{10x+5-4x+4}{20}

Expanding each bracket gives us the expression 10x+5-4x+4.

Notice how the last term has changed from - 4 to +4 .

This is because we must subtract each term that is in the second bracket. When you subtract negative four it is equivalent to adding four.

Simplifying this expression by collecting like terms gives us 10x+5-4x+4=6x+9.

\cfrac{10x+5-4x+4}{20} \, =\cfrac{6x+9}{20}

The final answer is \cfrac{6x+9}{20} \, .

We could also write \cfrac{3(2x+3)}{20} \, as our solution if we factorised the numerator.

Write as a single fraction in the simplest form,

\cfrac{5}{x} \, +\cfrac{2}{x+1} \, .

**Determine the lowest common multiple of the denominators.**

Sometimes the lowest common multiple can be found by simply multiplying the two denominators together. This is the case here.

The lowest common multiple of x and x+1 is x(x+1) which we can write as x\times{(x+1)} or x^{2}+x.

**Top tip**: Leave the denominator in its factorised form as this may help you simplify your solution at the end of the question.

Multiply the numerator and the denominator of the first fraction by (x+1).

Multiply the numerator and the denominator of the second fraction by x.

**For each fraction, simplify the numerator expression and simplify the denominator expression. **

**Note: **This will use skills such as expanding brackets, multiplying terms and collecting like terms. However, do not simplify the fraction by cancelling or dividing.

The numerator of the first fraction becomes 5\times{(x+1)}=5(x+1)=5x+5.

The numerator of the second fraction becomes 2\times{x}=2x.

The common denominator is x(x+1).

This gives us the equivalent calculation

\cfrac{5(x+1)}{x(x+1)} \, +\cfrac{2x}{x(x+1)} \, =\cfrac{5x+5}{x(x+1)} \, +\cfrac{2x}{x(x+1)} \, .

5x+5+2x=7x+5 by collecting like terms.

\cfrac{5x+5}{x(x+1)} \, + \cfrac{2x}{x(x+1)} \, =\cfrac{7x+5}{x(x+1)}

Then final answer is \cfrac{7x+5}{x(x+1)} \, .

**Note**: By expanding the denominator, \cfrac{7x+5}{x^{2}+x} \, is also a valid solution.

Write as a single fraction in the simplest form,

\cfrac{3}{x+2} \, -\cfrac{6}{3x} \, .

**Determine the lowest common multiple of the denominators.**

The lowest common multiple of x+2 and 3x is (x+2) \times 3x which we can write as 3x(x+2) or 3x^{2}+6x.

Remember it is often helpful to leave the denominator in its factorised form.

Multiply the numerator and the denominator of the first fraction by 3x.

Multiply the numerator and the denominator of the second fraction by (x+2).

**For each fraction, simplify the numerator expression and simplify the denominator expression. **

**Note: **This will use skills such as expanding brackets, multiplying terms and collecting like terms. However, do not simplify the fraction by cancelling or dividing.

The numerator of the first fraction becomes 3\times{3x}=9x.

The numerator of the second fraction becomes

6\times{(x+2)}=6(x+2)=6x+12.

This gives us the equivalent calculation

\cfrac{9x}{3x(x+2)} \, -\cfrac{6x(x+2)}{3x(x+2)} \, =\cfrac{9x}{3x(x+2)} \, -\cfrac{6x+12}{3x(x+2)} \, .

9x-(6x+12)=9x-6x-12

Then, by collecting like terms, the numerator is equal to 3x-12.

\cfrac{9x}{3x(x+2)} \, -\cfrac{6x+12}{3x(x+2)} \, =\cfrac{9x-(6x+12)}{3x(x+2)} \, =\cfrac{9x-6x-12}{3x(x+2)} \, =\cfrac{3x-12}{3x(x+2)}

The numerator can factorise to be 3(x-4).

This gives us the fraction \cfrac{3x-12}{3x(x+2)} \, =\cfrac{3(x-4)}{3x(x+2)} \, .

The numerator and the denominator share a common factor of 3 and so we can cancel this factor.

\cfrac{3(x-4)}{3x(x+2)} \, =\cfrac{x-4}{x(x+2)}

This gives the final answer \cfrac{x-4}{x(x+2)} \, .

By expanding brackets, another form of the solution is \cfrac{x-4}{x^{2}+2x} \, .

Write as a single fraction in the simplest form,

\cfrac{5}{x+3}+\frac{1}{2x+6} \, .

**Determine the lowest common multiple of the denominators.**

When working with binomial denominators it is helpful to first see if they factorise. Here 2x+6 factorises to 2(x+3).

We can now see that the denominators share a common factor of (x+3) and the lowest common multiple is 2(x+3).

To write the fractions with a common denominator, we only need to multiply the numerator and the denominator of the first fraction by 2.

**For each fraction, simplify the numerator expression and simplify the denominator expression.**

**Note: **This will use skills such as expanding brackets, multiplying terms and collecting like terms. However, do not simplify the fraction by cancelling or dividing. ** **

The numerator of the first fraction becomes 5\times{2}=10.

The common denominator is 2(x+3).

This gives us the equivalent calculation \cfrac{10}{2(x+3)} \, +\cfrac{1}{2(x+3)} \, .

10+1=11

\cfrac{10}{2(x+3)} \, +\cfrac{1}{2(x+3)} \, =\cfrac{10+1}{2(x+3)} \, =\cfrac{11}{2(x+3)}

The final answer is \cfrac{11}{2(x+3)} \, .

Write as a single fraction in the simplest form,

\cfrac{5}{x+2} \, +\cfrac{3}{x-2} \, .

**Determine the lowest common multiple of the denominators.**

(x+2)\times(x-2)=(x+2)(x-2)

Remember you can leave the denominator in its factorised form.

Multiply the numerator and the denominator of the first fraction by (x-2).

Multiply the numerator and the denominator of the second fraction by (x+2).

**For each fraction, simplify the numerator expression and simplify the denominator expression. **

**Note: **This will use skills such as expanding brackets, multiplying terms and collecting like terms. However, do not simplify the fraction by cancelling or dividing.

The numerator of the first fraction becomes 5\times{(x-2)}=5(x-2)=5x-10.

The numerator of the second fraction becomes 3\times{(x+2)}=3(x+2)=3x+6 \, .

The common denominator is (x+2)(x-2).

This gives us the equivalent calculation

\cfrac{5(x+2)}{(x+2)(x-2)} \, +\cfrac{3(x+2)}{(x+2)(x-2)} \, =\cfrac{5x-10}{(x+2)(x-2)} \, +\cfrac{3x+6}{(x+2)(x-2)} \, .

**Add/subtract the numerators together and write over the common denominator.**

(5x-10)+(3x+6)=5x-10+3x+6 and so by collecting like terms, the numerator is equal to 8x-4 which we can factorise to be 4(2x-1).

\cfrac{5x-10}{(x+2)(x-2)} \, +\cfrac{3x+6}{(x+2)(x-2)} \, =\cfrac{5x-10+3x+6}{(x+2)(x-2)} \, =\cfrac{8x-4}{(x+2)(x-2)}

The numerator can factorise to be 4(2x-1)

\cfrac{8x-4}{(x+2)(x-2)} \, =\cfrac{4(2x-1)}{(x+2)(x-2)} \, .

The final answer is \cfrac{4(2x-1)}{(x+2)(x-2)} \, .

Write as a single fraction in the simplest form,

\cfrac{x}{x-2} \, -\cfrac{2x}{x+3} \, .

**Determine the lowest common multiple of the denominators.**

(x-2)\times(x+3)=(x-2)(x+3)

Multiply the numerator and the denominator of the first fraction by (x+3).

Multiply the numerator and the denominator of the second fraction by (x-2).

**For each fraction, simplify the numerator expression and simplify the denominator expression.**

**Note: **This will use skills such as expanding brackets, multiplying terms and collecting like terms. However, do not simplify the fraction by cancelling or dividing.

The numerator of the first fraction becomes x\times{(x+3)}=x^{2}+3x.

The numerator of the second fraction becomes 2x\times{(x-2)}=2x^{2}-4x.

The common denominator is (x-2)(x+3).

\cfrac{x(x+3)}{(x-2)(x+3)} \, -\cfrac{2x(x-2)}{(x-2)(x+3)} \, =\cfrac{x^{2}+3x}{(x-2)(x+3)} \, -\cfrac{2x^{2}-4x}{(x-2)(x+3)}

This gives us the equivalent calculation \cfrac{x^{2}+3x}{(x-2)(x+3)} \, -\cfrac{2x^{2}-4x}{(x-2)(x+3)} \, .

x^{2}+3x-(2x^{2}-4x)=x^{2}+3x-2x^{2}+4x then by collecting like terms, the numerator is equal to -x^{2}+7x.

\cfrac{x^{2}+3x}{(x-2)(x+3)} \, -\cfrac{2x^{2}-4x}{(x-2)(x+3)} \, =\cfrac{x^2+3x-(2x^2-4x)}{(x-2)(x+3)} \, =\cfrac{-x^2+7x}{(x-2)(x+3)}

The numerator can factorise to be x(7-x).

\cfrac{-x^2+7x}{(x-2)(x+3)} \, =\cfrac{x(7-x)}{(x-2)(x+3)}

The final answer is \cfrac{x(7-x)}{(x-2)(x+3)} \, .

Write as a single fraction in the simplest form,

\cfrac{x+5}{x^{2}-1} \, +\cfrac{3x-2}{x+1} \, .

**Determine the lowest common multiple of the denominators.**

The denominator of the first fraction is in the difference of two squares form. Factoring this expression gives us a pair of double brackets.

x^{2}-1=(x+1)(x-1)

Now we can see that the denominators share the common binomial factor of (x+1).

Therefore the lowest common multiple of (x^{2}-1) and (x+1) is (x+1)(x-1).

To write the fractions with a common denominator we only need to multiply the numerator and the denominator of the second fraction by (x-1).

**For each fraction, simplify the numerator expression and simplify the denominator expression. **

**Note: **This will use skills such as expanding brackets, multiplying terms and collecting like terms. However, do not simplify the fraction by cancelling or dividing.

The numerator of the second fraction becomes (3x-2)(x-1)=3x^{2}-5x+2.

The common denominator is (x+1)(x-1).

This gives us the equivalent calculation

\cfrac{x+5}{x^2-1} \, +\cfrac{(3x-2)(x-1)}{(x+1)(x-1)} \, =\cfrac{x+5}{(x+1)(x-1)} \, +\cfrac{3x^{2}-5x+2}{(x+1)(x-1)} \, .

**Add/Subtract the numerators together and write over the common denominator. Simplify the numerator expression by collecting like terms.**

x+5+(3x^{2}-5x+2)=3x^{2}-4x+7 . This gives us the fraction

\cfrac{x+5+(3x^2-5x+2)}{(x+1)(x-1)} \, =\cfrac{3x^{2}-4x+7}{(x+1)(x-1)} \, .

There is a chance that this fraction will simplify if we can factorise the quadratic numerator into a pair of double brackets and show that the numerator and denominator share a common binomial factor.

3x^{2}-4x+7 does not factorise.

The final answer is therefore \cfrac{3x^{2}-4x+7}{(x+1)(x-1)} \, .

**Not recognising double negatives when subtracting algebraic fractions**

Some students fail to recognise the double negative when subtracting fractions in a question such as this.

\cfrac{2x}{7} \, -\cfrac{x-1}{7}

For example a student may incorrectly write the following working out.

In these steps the student has failed to subtract both of the terms of the numerator of the second fraction. To avoid this mistake it is advisable to write the numerators in brackets, as this helps to remind you that all the terms in the bracket need to be subtracted. The correct solution looks like this.

\cfrac{2x}{7}\,-\cfrac{x-1}{7}= \cfrac{(2x)-(x-1)}{7} \, = \cfrac{2x-x+1}{7} \, =\cfrac{x+1}{7}

**Adding or subtracting a value to the numerator and the denominator of a fraction in an attempt to write an equivalent fraction**

To write an equivalent fraction we must multiply or divide both the numerator and denominator by the same value. Sometimes students incorrectly assume that they can do the same with addition or subtraction. For example, a student may incorrectly write the following working out.

The fraction \cfrac{3}{x} \, is not equivalent to \cfrac{3+1}{x+1} \, =\cfrac{4}{x+1} \, .

To avoid this mistake you must remember that you can only multiply or divide the numerator and denominator by the same value to write an equivalent fraction.

The lowest common multiple of the denominators in this question can be found by multiplying the denominators together (x)\times(x+1) = x(x+1). The correct solution looks like this.

1. Write as a single fraction in the simplest form,

\cfrac{1}{c} \, +\cfrac{4}{d} \, .

\cfrac{5}{c+d}

\cfrac{d+4}{c+d}

\cfrac{5}{cd}

\cfrac{4c+d}{cd}

\begin{aligned} &\cfrac{1}{c} \, +\cfrac{4}{d}\\\\ &=\cfrac{1\times{d}}{c\times{d}} \, +\cfrac{4\times{c}}{d\times{c}}\\\\ &=\cfrac{d}{cd}\, +\cfrac{4c}{cd}\\\\ &=\cfrac{d+4c}{cd}\\\\ &=\cfrac{4c+d}{cd} \end{aligned}

2. Write as a single fraction in the simplest form,

\cfrac{9}{g} \, -\cfrac{7}{h} \, .

\cfrac{9g-7h}{gh}

\cfrac{2}{g-h}

\cfrac{9h-7g}{gh}

\cfrac{2}{gh}

\begin{aligned} &\cfrac{9}{g} \, -\frac{7}{h}\\\\ &=\cfrac{9\times{h}}{g\times{h}} \, -\cfrac{7\times{g}}{h\times{g}}\\\\ &=\cfrac{9h}{gh} \, -\cfrac{7g}{gh}\\\\ &=\cfrac{9h-7g}{gh} \end{aligned}

3. Write as a single fraction in the simplest form,

\cfrac{2x}{3} \, +\cfrac{5x}{6} \, .

\cfrac{3x}{2}

\cfrac{7x}{9}

\cfrac{x}{2}

\cfrac{7x^{2}}{9}

\begin{aligned} &\cfrac{2x}{3} \, +\cfrac{5x}{6} \\\\ &=\cfrac{2x\times{2}}{3\times{2}} \, +\cfrac{5x}{6}\\\\ &=\cfrac{4x}{6} \, +\cfrac{5x}{6}\\\\ &=\cfrac{9x}{6} \end{aligned}

4. Write as a single fraction in the simplest form,

\cfrac{4x}{7} \, -\cfrac{2x}{5} \, .

\cfrac{6x}{35}

x

\cfrac{18x}{35}

3x

\begin{aligned} &\cfrac{4x}{7} \, -\cfrac{2x}{5}\\\\ &=\cfrac{4x\times{5}}{7\times{5}} \, -\cfrac{2x\times{7}}{5\times{7}}\\\\ &=\cfrac{20x}{35} \, -\cfrac{14x}{35}\\\\ &=\cfrac{20x-14x}{35}\\\\ &=\cfrac{6x}{35} \end{aligned}

5. Write as a single fraction in the simplest form,

\cfrac{3x+5}{4} \, +\cfrac{x+1}{2} \, .

3x

\cfrac{5x+7}{4}

\cfrac{2x+3}{3}

\cfrac{5x+6}{4}

\begin{aligned} &\cfrac{3x+5}{4} \, +\cfrac{x+1}{2}\\\\ &=\cfrac{3x+5}{4} \, +\cfrac{(x+1)\times{2}}{2\times{2}}\\\\ &=\cfrac{3x+5}{4}\, +\cfrac{2x+2}{4}\\\\ &=\cfrac{3x+5+2x+2}{4}\\\\ &=\cfrac{5x+7}{4} \end{aligned}

6. Write as a single fraction in the simplest form,

\cfrac{3x+2}{2} \, -\cfrac{x-1}{4} \, .

x+1

5x+3

\cfrac{5(x+1)}{4}

\cfrac{5x+3}{4}

\begin{aligned} &\cfrac{3x+2}{2} \, -\cfrac{x-1}{4}\\\\ &=\cfrac{(3x+2)\times{2}}{2\times{2}} \, -\cfrac{x-1}{4}\\\\ &=\cfrac{6x+4}{4} \, -\cfrac{x-1}{4}\\\\ &=\cfrac{6x+4-(x-1)}{4}\\\\ &=\cfrac{6x+4-x+1}{4}\\\\ &=\cfrac{5x+5}{4}\\\\ &=\cfrac{5(x+1)}{4} \end{aligned}

7. Write as a single fraction in the simplest form,

\cfrac{1}{2x} \, +\cfrac{4}{x+2} \, .

\cfrac{5}{3x+2}

\cfrac{11x}{2x(x+2)}

\cfrac{5}{2x(x+2)}

\cfrac{9x+2}{2x(x+2)}

\begin{aligned} &\cfrac{1}{2x} \, +\cfrac{4}{x+2}\\\\ &=\cfrac{1\times{(x+2)}}{2x\times{(x+2)}} \, +\cfrac{4\times{2x}}{(x+2)\times{2x}}\\\\ &=\cfrac{x+2}{2x(x+2)} \, +\cfrac{8x}{2x(x+2)}\\\\ &=\cfrac{x+2+8x}{2x(x+2)}\\\\ &=\cfrac{9x+2}{2x(x+2)} \end{aligned}

8. Write as a single fraction in the simplest form,

\cfrac{3}{x+4} \, -\cfrac{5}{2x} \, .

\cfrac{x+20}{2x(x+4)}

\cfrac{x-20}{2x(x+4)}

\cfrac{2}{x-4}

\cfrac{x+4}{2x(x+4)}

\begin{aligned} &\cfrac{3}{x+4} \, -\cfrac{5}{2x}\\\\ &=\cfrac{3\times{2x}}{(x+4)\times{2x}} \, -\cfrac{5\times{(x+4)}}{2x\times{(x+4)}}\\\\ &=\cfrac{6x}{2x(x+4)} \, -\cfrac{5x+20}{2x(x+4)}\\\\ &=\cfrac{6x-(5x+20)}{2x(x+4)}\\\\ &=\cfrac{6x-5x-20)}{2x(x+4)}\\\\ &=\cfrac{x-20}{2x(x+4)} \end{aligned}

9. Write as a single fraction in the simplest form,

\cfrac{4}{3(x-1)} \, -\cfrac{1}{x-1} \, .

\cfrac{1}{3(x-1)}

\cfrac{1}{x-1}

\cfrac{3}{2(x-1)}

\cfrac{1}{3x-1}

\begin{aligned} &\cfrac{4}{3(x-1)} \, -\cfrac{1}{x-1}\\\\ &=\cfrac{4}{3(x-1)} \, -\cfrac{1\times{3}}{(x-1)\times{3}}\\\\ &=\cfrac{4}{3(x-1)} \, -\cfrac{3}{3(x-1)}\\\\ &=\cfrac{4-3}{3(x-1)}\\\\ &=\cfrac{1}{3(x-1)} \end{aligned}

10. Write as a single fraction in the simplest form,

\cfrac{10}{x-3} \, -\cfrac{4}{x-1} \, .

\cfrac{2(3x+1)}{(x-3)(x-1)}

-3

\cfrac{2(3x+1)}{(x-3)(x-1)}

\cfrac{8x}{(x-3)(x-1)}

\begin{aligned} &\cfrac{10}{x-3} \, -\cfrac{4}{x-1}\\\\ &=\cfrac{10\times{(x-1)}}{(x-3)\times{(x-1)}} \, -\cfrac{4\times{(x-3)}}{(x-1)\times{(x-3)}}\\\\ &=\cfrac{10x-10}{(x-3)(x-1)} \, -\cfrac{4x-12}{(x-3)(x-1)}\\\\ &=\cfrac{10x-10-(4x-12)}{(x-3)(x-1)}\\\\ &=\cfrac{10x-10-4x+12}{(x-3)(x-1)}\\\\ &=\cfrac{6x+2}{(x-3)(x-1)}\\\\ &=\cfrac{2(3x+1)}{(x-3)(x-1)} \end{aligned}

11. Write as a single fraction in the simplest form,

\cfrac{3x}{x+5} \, +\cfrac{x}{x+1} \, .

\cfrac{4x(x+2)}{(x+5)(x+1)}

\cfrac{4x}{(x+5)(x+1)}

\cfrac{2x(x+2)}{x+3}

\cfrac{3x^{2}}{2(x+3)}

\begin{aligned} &\cfrac{3x}{x+5} \, +\cfrac{x}{x+1}\\\\ &=\cfrac{3x\times{(x+1)}}{(x+5)\times(x+1)} \, +\cfrac{x\times{(x+5)}}{(x+1)\times{(x+5)}}\\\\ &=\cfrac{3x^{2}+3x}{(x+5)(x+1)} \, +\cfrac{x^{2}+5x}{(x+5)(x+1)}\\\\ &=\cfrac{3x^{2}+3x+x^{2}+5x}{(x+5)(x+1)}\\\\ &=\cfrac{4x^{2}+8x}{(x+5)(x+1)}\\\\ &=\cfrac{4x(x+2)}{(x+5)(x+1)} \end{aligned}

12. Write as a single fraction in the simplest form,

\cfrac{x-1}{x+2} \, -\cfrac{x-1}{x^{2}-4} \, .

\cfrac{x^{2}-1}{x^{2}-4}

\cfrac{(x-5)(x-1)}{(x+2)(x-2)}

\cfrac{(x-3)(x-1)}{(x+2)(x-2)}

x-2

\begin{aligned} &\cfrac{x-1}{x+2} \, -\cfrac{x-1}{x^{2}-4}\\\\ &=\cfrac{(x-1)\times{(x-2)}}{(x+2)\times{(x-2)}} \, -\cfrac{x-1}{(x+2)(x-2)}\\\\ &=\cfrac{(x-1)(x-2)}{(x+2)(x-2)} \, -\cfrac{x-1}{(x+2)(x-2)}\\\\ &=\cfrac{x^{2}-3x+2}{(x+2)(x-2)} \, -\cfrac{x-1}{(x+2)(x-2)}\\\\ &=\cfrac{x^{2}-3x+2-(x-1)}{(x+2)(x-2)}\\\\ &=\cfrac{x^{2}-3x+2-x+1}{(x+2)(x-2)}\\\\ &=\cfrac{x^{2}-4x+3}{(x+2)(x-2)}\\\\ &=\cfrac{(x-3)(x-1)}{(x+2)(x-2)} \end{aligned}

1. Write as a single fraction, \cfrac{1}{x} \, +\cfrac{1}{y} \, -\cfrac{1}{z} \, .

**(2 marks)**

Show answer

\cfrac{yz}{xyz} \, +\cfrac{xz}{xyz} \, -\cfrac{xy}{xyz}

**(1)**

\cfrac{yz+xz-xy}{xyz}

**(1)**

**Alternative method 1**

\cfrac{y+x}{xy} \, -\cfrac{1}{z}

**(1)**

\cfrac{z(y+x)-xy}{xyz} \; or \, \cfrac{yz+xz-xy}{xyz}

**(1)**

**Alternative method 2**

\cfrac{1}{x} \, +\cfrac{z-y}{yz}

**(1)**

\cfrac{yz+x(z-y)}{xyz} \; or \, \cfrac{yz+xz-xy}{xyz}

**(1)**

**Alternative method 3**

\cfrac{1}{y} \, +\cfrac{z-x}{xz}

**(1)**

\cfrac{xz+y(z-x)}{xyz} \; or \cfrac{xz+yz-xy}{xyz}

**(1)**

2. Lucia is answering the question,

(a) Describe the mistake that Lucia has made in her working out.

(b) Determine the correct solution to \cfrac{2}{x+1} \, +\cfrac{1}{x} \, .

**(4 marks)**

Show answer

(a) She has added 1 to the numerator and denominator of the second fraction in an attempt to make a common denominator of x+1.

However \cfrac{2}{x+1} is not equivalent to \cfrac{1}{x} \, .

**(1)**

(b)

\cfrac{2\times{x}}{x(x+1)} \, +\cfrac{1\times{(x+1)}}{x(x+1)}

**(1)**

\cfrac{2x+x+1}{x(x+1)}

**(1)**

\cfrac{3x+1}{x(x+1)}

**(1)**

3. Write 2-x+\cfrac{x^{2}-4}{x+4} \, as a single fraction of the form \cfrac{a(b-x)}{x+c} \, where a, b, and c are integers.

**(4 marks)**

Show answer

\cfrac{(2-x)(x+4)}{x+4} \, +\cfrac{x^{2}-4}{x+4}

**(1)**

\cfrac{2x+8-x^{2}-4x}{x+4} \, +\cfrac{x^{2}-4}{x+4}

**(1)**

\cfrac{4-2x}{x+4}

**(1)**

\cfrac{2(2-x)}{x+4} or a=2, b=2, c=4

**(1)**

You have now learned how to:

- Add and subtract algebraic fractions

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