a)

$\displaystyle\frac{\var{a_coprime}}{\var{b_coprime}}+\frac{\var{c_coprime}}{\var{d_coprime}}$

To add or subtract fractions, we need to have a common denominator on both fractions.

To get a common denominator, we need to find the lowest common multiple of the two denominators.

The lowest common multiple of $\var{b_coprime}$ and $\var{d_coprime}$ is $\var{lcm}.$

This will be the new denominator, and we need to multiply each fraction individually to ensure we get this denominator. 

For $\displaystyle\frac{\var{a_coprime}}{\var{b_coprime}}$, we need to multiply the fraction by $\displaystyle\frac{\var{lcm_b}}{\var{lcm_b}}$ to give $\displaystyle\frac{\var{alcm_b}}{\var{lcm}}.$

For $\displaystyle\frac{\var{c_coprime}}{\var{d_coprime}}$, we need to multiply the fraction by $\displaystyle\frac{\var{lcm_d}}{\var{lcm_d}}$ to give $\displaystyle\frac{\var{clcm_d}}{\var{lcm}}.$

Now that we have each fraction in terms of a common denominator, we can now add the fractions together. 

$\displaystyle\frac{\var{alcm_b}}{\var{lcm}}+\frac{\var{clcm_d}}{\var{lcm}}=\frac{(\var{alcm_b}+\var{clcm_d})}{\var{lcm}}=\frac{\var{alcmclcm}}{\var{lcm}}.$

From this, we can try to simplify the result down by finding the greatest common divisor of the numerator and denominator and dividing the whole fraction by this amount. 

The greatest common divisor of $\var{alcmclcm}$ and $\var{lcm}$ is $\var{gcd}.$

Therefore, the expression cannot be simplified further, and $\displaystyle\frac{\var{num}}{\var{denom}}$ is the final answer.

b)

$\displaystyle\frac{\var{f_coprime}}{\var{g_coprime}}-\frac{\var{h_coprime}}{\var{j_coprime}}+2.$

The two fractions can be individually multiplied to achieve a common denominator of the lowest common multiple, $\var{lcm2}.$

$\displaystyle\frac{\var{f_coprime}}{\var{g_coprime}}$ becomes $\displaystyle\frac{\var{flcm2_g}}{\var{lcm2}}$ and $\displaystyle\frac{\var{h_coprime}}{\var{j_coprime}}$ becomes $\displaystyle\frac{\var{hlcm2_j}}{\var{lcm2}}.$

We can now subtract the second fraction from the first.

$\displaystyle\frac{\var{flcm2_g}}{\var{lcm2}}-\frac{\var{hlcm2_j}}{\var{lcm2}}=\frac{\var{flcmhlcm}}{\var{lcm2}}.$

From this, the question asks us to add $2$. We need to change the mixed number, $2$, into an improper fraction. 

$\displaystyle2=2\bigg(\frac{\var{lcm2}}{\var{lcm2}}\bigg)=\frac{\var{twolcm2}}{\var{lcm2}}.$

We can now continue with the question.

$\displaystyle\frac{\var{flcmhlcm}}{\var{lcm2}}+\frac{\var{twolcm2}}{\var{lcm2}}=\frac{\var{num2unsim}}{\var{lcm2}}.$

We can look to simplify by dividing by the greatest common divisor of $\var{num2unsim}$ and $\var{lcm2}$ which is $\var{gcd2}.$

Therefore, no further simplification is possible, and $\displaystyle\simplify{{num2unsim}/{lcm2}}$ is the final answer.

c)

$\displaystyle\var{k}+\frac{\var{l_coprime}}{\var{m_coprime}}-\frac{\var{n_coprime}}{\var{o_coprime}}.$

We need to convert the decimal into a fraction and to do this, we need to multiply it by $10$ for every decimal place.

$\displaystyle\frac{\var{k}}{1}\times\frac{100}{100}=\frac{\var{100k}}{100}.$

We should look to simplify by dividing by the greatest common divisor which is $\var{gcd_k100}.$

Therefore, it is not possible to simplify any further, and the fraction stays as

\[\simplify{{{100k}}/{100}}\text{.}\]

The original expression is now $\displaystyle\frac{\var{k_simp}}{\var{simp}}+\frac{\var{l_coprime}}{\var{m_coprime}}-\frac{\var{n_coprime}}{\var{o_coprime}}.$

We can multiply each fraction individually to achieve the common denominator $\var{gcd3}$.

\[\frac{\var{k_simp}}{\var{simp}}\text{ becomes }\frac{\var{k_simp*term1}}{\var{gcd3}}\text{, }\frac{\var{l_coprime}}{\var{m_coprime}}\text{ becomes }\frac{\var{l_coprime*term2}}{\var{gcd3}}\text{ and }\frac{\var{n_coprime}}{\var{o_coprime}}\text{ becomes }\frac{\var{n_coprime*term3}}{\var{gcd3}}\text{.}\]

We can now complete the addition. 

\[\frac{\var{k_simp*term1}}{\var{gcd3}}+\frac{\var{l_coprime*term2}}{\var{gcd3}}-\frac{\var{n_coprime*term3}}{\var{gcd3}}=\frac{\var{(k_simp*term1)+(l_coprime*term2)-(n_coprime*term3)}}{\var{gcd3}}\text{.}\]

We should look to simplify this fraction by dividing by the highest common divisor, $\var{gcd_numgcd3}.$

Therefore, it is not possible to simplify the fraction any further and the final answer is

\[\simplify{{num1}/{gcd3}}\text{.}\]

a)

To multiply $\displaystyle\frac{\var{a_coprime}}{\var{c_coprime}}\times\frac{\var{b_coprime}}{\var{d_coprime}}$, address the numerators and denominators separately.

Multiply the numerators across both fractions.

$\var{a_coprime}\times\var{b_coprime}=\var{ab}$,

and then multiply the denominators across both fractions.

$\var{c_coprime}\times\var{d_coprime}=\var{cd}$.

The values of the multiplied numerators and denominators will be the numerator and denominator of the new fraction: $\displaystyle\frac{\var{ab}}{\var{cd}}$.

This answer may need simplifying down, and to do this, find the greatest common divisor in both the numerator and denominator and divide by this number.

The greatest common divisor of $\var{ab}$ and $\var{cd}$ is $\var{gcd}$.

By using $\var{gcd}$ to cancel down the fraction, the final answer is $\displaystyle\simplify{{ab}/{cd}}$.

b)

To multiply $\displaystyle\simplify{{k_coprime}/{j_coprime}}\times\var{f}\frac{\var{g_coprime}}{\var{h_coprime}}$, we first need to change the mixed number term into an improper fraction. 

To do this, we need to multiply $(\var{f}\times\var{h_coprime}=\var{fh})$ and add it to what was already on the numerator of the fraction, $\var{g_coprime}$.

$\displaystyle\frac{(\var{fh}+\var{g_coprime})}{\var{h_coprime}}= \displaystyle\frac{\var{numif}}{\var{h_coprime}}$.

Next, we multiply the numerators and denominators across both fractions separately, as done in part a). 

$\var{k_coprime}\times\var{numif} = \var{num}$,

$\var{j_coprime}\times\var{h_coprime}=\var{denom}$.

This gives the unsimplified version of the new fraction $\displaystyle\frac{\var{num}}{\var{denom}}$.

To simplify, find the greatest common divisor in both the numerator and denominator and divide by this number. 

The greatest common divisor of $\var{num}$ and $\var{denom}$ is $\var{gcdb}$.

By using $\var{gcdb}$ to cancel down the fraction, the final answer is $\displaystyle\simplify{{num}/{denom}}$.

c)

To square a fraction means to multiply the fraction by itself. To do this, multiply the numerators and denominators across individually.

$\displaystyle\bigg(\frac{\var{l_coprime}}{\var{m_coprime}}\bigg)^2=\frac{\var{l_coprime}}{\var{m_coprime}}\times\frac{\var{l_coprime}}{\var{m_coprime}}=\frac{\var{l_coprime^2}}{\var{m_coprime^2}}.$

From this, we should look if it is possible to simplify by finding the highest common divisor of $\var{l_coprime^2}$ and $\var{m_coprime^2}.$

The greatest common divisor is $\var{gcd_lcmc}$.

Therefore, it is not possible to simplify this further, and the final answer is

$\displaystyle\frac{\var{l_coprime2}}{\var{m_coprime2}}$.

d)

Helen was on holiday for $28$ days and spent $\displaystyle\frac{\var{aa}}{7}$ of her time in Spain. 

$\displaystyle\frac{\var{aa}}{7}\times\frac{28}{1}=\frac{\var{bb}}{7}=\var{cc}$ days in Spain. 

Whilst in Spain, she spends $\displaystyle\frac{\var{dd}}{4}$ of her time in Barcelona.

$\displaystyle\frac{\var{dd}}{4}\times\frac{\var{cc}}{1}=\frac{\var{ddcc}}{4}=\var{ee}$ days in Barcelona. 

a)

When faced with dividing fractions, it much easier to switch one of the fractions around and multiply them together instead of divide them.

\[ \left( \frac{\var{f_coprime}}{\var{g_coprime}}\div\frac{\var{h_coprime}}{\var{j_coprime}} \right) \equiv \left( \frac{\var{f_coprime}}{\var{g_coprime}}\times\frac{\var{j_coprime}}{\var{h_coprime}} \right) = \frac{\var{fj}}{\var{gh}} \]

Then, simplify by finding the highest common divisor in the numerator and denominator which in this case is $\var{gcd1}$. 

This gives a final answer of $\displaystyle\simplify{{fj}/{gh}}$.

b)

\[ \frac{\var{f1_coprime}}{\var{g1_coprime}}\div\frac{\var{h1_coprime}}{\var{j1_coprime}} \equiv \left( \frac{\var{f1_coprime}}{\var{g1_coprime}}\times\frac{\var{j1_coprime}}{\var{h1_coprime}} \right)=\frac{\var{f1j1}}{\var{g1h1}} \]

Then, simplify by finding the highest common divisor in the numerator and denominator which in this case is $\var{gcd2}$.

This gives a final answer of $\displaystyle\simplify{{f1j1}/{g1h1}}$.

c)

\[ {\var{f3}\frac{\var{g3_coprime}}{\var{h3_coprime}}}\div{\var{f4}\frac{\var{g4_coprime}}{\var{h4_coprime}}} \]

The first thing to do is to change the mixed numbers into improper fractions.

An improper fraction is a fraction where the numerator is greater than the denominator. To change a mixed fraction to an improper fraction, multiply the integer part of the mixed number by the denominator, and add it to the existing numerator to make the new numerator of the improper fraction. The denominator will stay the same. 

\[ {\var{f3}\frac{\var{g3_coprime}}{\var{h3_coprime}}}\equiv\frac{(\var{f3}\times\var{h3_coprime})+\var{g3_coprime}}{\var{h3_coprime}}=\frac{\var{f3h3}+\var{g3_coprime}}{\var{h3_coprime}}=\frac{\var{f3h3+g3_coprime}}{\var{h3_coprime}} \]

\[ {\var{f4}\frac{\var{g4_coprime}}{\var{h4_coprime}}}\equiv\frac{(\var{f4}\times\var{h4_coprime})+\var{g4_coprime}}{\var{h4_coprime}}=\frac{\var{f4h4}+\var{g4_coprime}}{\var{h4_coprime}}=\frac{\var{f4h4+g4_coprime}}{\var{h4_coprime}} \]

We now have our mixed numbers as improper fractions.

\[ \frac{\var{f3h3+g3_coprime}}{\var{h3_coprime}}\div\frac{\var{f4h4+g4_coprime}}{\var{h4_coprime}} \]

Now, use the same method as in parts a) and b) to divide by switching around one fraction and changing the division symbol to multiplication.

\[ \frac{\var{f3h3+g3_coprime}}{\var{h3_coprime}}\div\frac{\var{f4h4+g4_coprime}}{\var{h4_coprime}}\equiv\frac{\var{f3h3+g3_coprime}}{\var{h3_coprime}}\times\frac{\var{h4_coprime}}{\var{f4h4+g4_coprime}}=\frac{\var{num}}{\var{denom}} \]

Finally, the last thing to do is to simplify your answer down by finding the highest common divisor in the numerator and denominator, which in this case is $\var{gcd3}$.

By doing this, you will get a final answer of

\[ \simplify{{num}/{denom}} \]

d)

\[ \frac{\var{a}}{(\simplify[all,!collectNumbers]{{b}-{c}/{d}})} \]

Consider the denominator first, as following the rules of BODMAS, you should address brackets first.

You need to get a common denominator for both terms on the denominator, like this:

\[ \var{b}\times\frac{\var{d}}{\var{d}} = \frac{\var{bd}}{\var{d}} \]

This now allows you to complete the addition or subtraction as both terms have a common denominator. 

\[ {\simplify[all,!collectNumbers]{{bd}/{d}-{c}/{d}}} = \frac{\var{bd_c}}{\var{d}} \]

This means that the expression is now:

\[ \frac{\var{a}}{\frac{\var{bd_c}}{\var{d}}} \]

Dealing with this requires a bit of manipulation. You have to change the divisor of the denominator to be a mulitplier of the numerator. The denominator, ${\var{bd_c}}$ was being divided by ${\var{d}}$ but by flipping it around, the numerator, ${\var{a}}$ will be mulitplied by ${\var{d}}$. The value of the expression remains the same.

\[ \frac{\var{a}}{\frac{\var{bd_c}}{\var{d}}}\equiv \frac{(\var{a})\times(\var{d})}{\var{bd_c}}= \frac{\var{ad}}{\var{bd_c}} \]

From this, you can try to cancel the expression down by finding the highest common factor of the numerator and denominator, to give a final answer of

\[ \simplify{{ad}/{bd_c}} \]

To find the odd fraction out, reduce each fraction to lowest terms.

\begin{align}
\frac{\var{mone}}{\var{none}} &= \frac{\var{m_coprime}}{\var{n_coprime}}\times\frac{\var{one}}{\var{one}} \\[0.5em]
\frac{\var{mtwo}}{\var{ntwo}} &= \frac{\var{m_coprime}}{\var{n_coprime}}\times\frac{\var{two}}{\var{two}} \\[0.5em]
\frac{\var{mthree}}{\var{nthree}} &= \frac{\var{m_coprime}}{\var{n_coprime}}\times\frac{\var{three}}{\var{three}} \\[0.5em]
\frac{\var{mfour}}{\var{nfour}} &= \frac{\var{m_coprime}}{\var{n_coprime}}\times\frac{\var{four}}{\var{four}} \\[0.5em]
\frac{\var{mfive}}{\var{nfive}} &= \frac{\var{mfive/gcd(mfive,nfive)}}{\var{nfive/gcd(mfive,nfive)}} \times \frac{\var{gcd(mfive,nfive)}}{\var{gcd(mfive,nfive)}}
\end{align}

All but one of these fractions are equivalent to $\displaystyle\frac{\var{m_coprime}}{\var{n_coprime}}$.

The odd fraction out is $\displaystyle\frac{\var{mfive}}{\var{nfive}}$.

To find the odd fraction out, reduce each fraction to lowest terms.

\begin{align}
\frac{\var{osix}}{\var{psix}}=\frac{\var{o_coprime}}{\var{p_coprime}}\times\frac{\var{six}}{\var{six}} \\[0.5em]
\frac{\var{oseven}}{\var{pseven}}=\frac{\var{o_coprime}}{\var{p_coprime}}\times\frac{\var{seven}}{\var{seven}} \\[0.5em]
\frac{\var{oeight}}{\var{peight}}=\displaystyle\frac{\var{o_coprime}}{\var{p_coprime}}\times\frac{\var{eight}}{\var{eight}} \\[0.5em]
\frac{\var{onine}}{\var{pnine}}=\frac{\var{o_coprime}}{\var{p_coprime}}\times\frac{\var{nine}}{\var{nine}} \\[0.5em]
\frac{\var{oten}}{\var{pten}} = \frac{\var{oten/gcd(oten,pten)}}{\var{pten/gcd(oten,pten)}} \times \frac{\var{gcd(oten,pten)}}{\var{gcd(oten,pten)}}
\end{align}

The odd fraction out is $\displaystyle\frac{\var{oten}}{\var{pten}}$.

a)

To convert a decimal into a fraction, firstly place the decimal over $1$ as a fraction, and then multiply both the numerator and denominator by $10$ for however many decimal places the decimal has. For example, if the decimal was $0.1$, you would multiply the fraction by $10$ as there is one decimal place. If the decimal was $0.01$, you would multiply it by $100$, as there are two decimal places.  

i)

$\var{a}$

\[
\frac{\var{a}}{1}\times\frac{10}{10}=\frac{\var{10a}}{10}\text{.}
\]

ii)

$\var{b}$

\[
\frac{\var{b}}{1}\times\frac{100}{100}=\frac{\var{100b}}{100}=\simplify{{100b}/{100}}\text{.}
\]

iii)

$\var{d}$

\[
\frac{\var{d}}{1}\times\frac{10}{10}=\frac{\var{10d}}{10}=\simplify{{10d}/{10}}\text{.}
\]

iv)

$0.\dot{\var{c}}$

To convert a recurring decimal to a fraction, the first step is to set up a simple equation where

\[
x=0.\dot{\var{c}}\text{.}
\]

By multiplying both sides by $10$, we can gain another simple equation where

\[
10x=\var{c}.\dot{\var{c}}\text{.}
\]

By subtracting one equation from the other, we can find the fraction equivalent of the recurring decimal. 

\[
\begin{align}
&&\var{c}.\dot{\var{c}}&={10}x\\
-&&{0.\dot{\var{c}}}&=x\\
&&\overline{\qquad} & \overline{\qquad}\\
&&{\var{c}}&=9x\\
\\
&&\frac{\var{c}}{9}&=x
\end{align}
\]

$\displaystyle\frac{\var{c}}{9}$ simplifies to $\simplify{{{c}}/{9}}$ by dividing by $3$ and therefore, $0.\dot{\var{c}}=\simplify{{{c}}/{9}}$ in its fractional form.}

b)

$\displaystyle\var{f}$

\[
\var{f}\times\frac{\var{f1000}}{\var{f1000}}=\frac{\var{f2}}{\var{f1000}}\text{.}
\]

From this, we can look to see if we can cancel down the fraction by finding the highest common divisor between the numerator and denominator. This is $\var{mygcd}$.

Simplifying by this amount gives the final answer

\[\frac{\var{f3}}{\var{f4}}.\]

c)

$\var{h}.\dot{\var{j}}\dot{\var{k}}.$

To convert a recurring decimal to a fraction, the first step is to set up a simple equation where,

$x=\var{h}.\dot{\var{j}}\dot{\var{k}}.$

By multiplying both sides by $100$ to isolate the recurring section on the left hand side of the decimal point, we can gain another simple equation

$100x=\var{h}\var{j}\var{k}.\dot{\var{j}}\dot{\var{k}}.$

Now that we have two equations in terms of $x$, we can subtract one from the other and solve to get a value of $x$.

\[
\begin{align}
&&\var{h}\var{j}\var{k}.\dot{\var{j}}\dot{\var{k}}&=100x\\
-&&\var{h}.\dot{\var{j}}\dot{\var{k}}&=x\\
&&\overline{\qquad} & \overline{\qquad} 
\\
&&{{\var{h}}\var{j}\var{k-h}}&=99x\\
\\
&&\frac{\var{numerator}}{\var{g}}&=x\text{.}\\
\end{align}
\]

From this, we should look to see if it is possible to simplify by finding the greatest common divisor of the numerator and the denominator. The greatest common divisor is $\var{gcd1 }.$

Therefore, it is not possible to simplify and so

\[
\begin{align}
\var{h}.\dot{\var{j}}\dot{\var{k}}=\simplify{{{numerator}}/{g}}\text{ in its fractional form.}\\
\end{align}
\]