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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 youβre 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.
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:
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} \, .
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.
2For 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.
3For 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} \, .
4Add/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.
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 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.Β
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.
For each fraction, multiply the numerator and the denominator by the same value to obtain the common denominator.
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.
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 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.Β
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.
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 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.
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 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.Β
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}\, .
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 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.
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 (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.Β
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)} \, .
Add/subtract the numerators together and write over the common denominator. Simplify the numerator expression by collecting like terms.
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.
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+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.
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 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.Β
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)} \, .
Add/subtract the numerators together and write over the common denominator. Simplify the numerator expression by collecting like terms.
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).
For each fraction, multiply the numerator and the denominator by the same value to obtain the common denominator.
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)} \, .
Add/subtract the numerators together and write over the common denominator. Simplify the numerator expression by collecting like terms.
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.
Sometimes the lowest common multiple can be found by simply multiplying the two denominators together. This is the case here.
(x+2)\times(x-2)=(x+2)(x-2)
Remember you can leave the denominator in its factorised form.
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 (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.Β
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.
Sometimes the lowest common multiple can be found by simply multiplying the two denominators together. This is the case here.
(x-2)\times(x+3)=(x-2)(x+3)
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 (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)} \, .
Add/subtract the numerators together and write over the common denominator. Simplify the numerator expression by collecting like terms.
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).
For each fraction, multiply the numerator and the denominator by the same value to obtain the common denominator.
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.Β
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)} \, .
1. Write as a single fraction in the simplest form,
\cfrac{1}{c} \, +\cfrac{4}{d} \, .
\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} \, .
\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} \, .
\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} \, .
\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} \, .
\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} \, .
\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} \, .
\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} \, .
\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} \, .
\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} \, .
\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} \, .
\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} \, .
\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.Β Write as a single fraction, \cfrac{1}{x} \, +\cfrac{1}{y} \, -\cfrac{1}{z} \, .
(2 marks)
\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.Β Lucia is answering the question,
(a)Β Describe the mistake that Lucia has made in her working out.
(b)Β Determine the correct solution to \cfrac{2}{x+1} \, +\cfrac{1}{x} \, .
(4 marks)
(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.Β Write 2-x+\cfrac{x^{2}-4}{x+4} \, as a single fraction of the form \cfrac{a(b-x)}{x+c} \, Β where a, b, and c are integers.
(4 marks)
\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)
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