A worked example
A big fraction with everything mixed in: first resolve what is inside brackets and powers, then multiplications and divisions, and finally sums and differences. The numerator and the denominator are worked separately.
$$\dfrac{\dfrac{3}{4}-\dfrac{1}{2}\cdot\left(\dfrac{2}{5}:\dfrac{8}{7}-1\right)}{\dfrac{4}{3}+\dfrac{1}{2}:\dfrac{3}{4}+\left(\dfrac{1}{2}\right)^{2}}$$
- Work out the divisions and the power: $\dfrac{2}{5}:\dfrac{8}{7}=\dfrac{14}{40}$, $\ \dfrac{1}{2}:\dfrac{3}{4}=\dfrac{4}{6}$ and $\left(\dfrac{1}{2}\right)^2=\dfrac{1}{4}$.$$\dfrac{\dfrac{3}{4}-\dfrac{1}{2}\cdot\left(\dfrac{14}{40}-1\right)}{\dfrac{4}{3}+\dfrac{4}{6}+\dfrac{1}{4}}$$
- Simplify: $\dfrac{14}{40}=\dfrac{7}{20}$ and $\dfrac{4}{6}=\dfrac{2}{3}$.$$\dfrac{\dfrac{3}{4}-\dfrac{1}{2}\cdot\left(\dfrac{7}{20}-1\right)}{\dfrac{4}{3}+\dfrac{2}{3}+\dfrac{1}{4}}$$
- Inside the bracket, common denominator: $\dfrac{7}{20}-\dfrac{20}{20}=-\dfrac{13}{20}$. In the denominator, $\dfrac{4}{3}+\dfrac{2}{3}+\dfrac{1}{4}=\dfrac{16+8+3}{12}=\dfrac{27}{12}$.$$\dfrac{\dfrac{3}{4}-\dfrac{1}{2}\cdot\left(-\dfrac{13}{20}\right)}{\dfrac{27}{12}}$$
- Multiply and simplify the denominator: $\dfrac{1}{2}\cdot\left(-\dfrac{13}{20}\right)=-\dfrac{13}{40}$ and $\dfrac{27}{12}=\dfrac{9}{4}$.$$\dfrac{\dfrac{3}{4}+\dfrac{13}{40}}{\dfrac{9}{4}}$$
- In the numerator: $\dfrac{3}{4}+\dfrac{13}{40}=\dfrac{30+13}{40}=\dfrac{43}{40}$.$$\dfrac{\dfrac{43}{40}}{\dfrac{9}{4}}=\dfrac{43}{40}:\dfrac{9}{4}$$
- Dividing is multiplying by the inverse, and we simplify $40$ and $4$ (by $4$):$$\dfrac{43}{40}\cdot\dfrac{4}{9}=\dfrac{43}{10\cdot 9}=\dfrac{43}{90}$$
Result
$\dfrac{43}{90}$
Exercise 31 (done on the board)
31Do the following operations and simplify the result
Each part has its solution folded: do it in your notebook and then check it.
a) $\left(\dfrac{3}{5}-\dfrac{1}{2}\cdot\dfrac{4}{3}\right)^{2}+\dfrac{3}{5}:5-\left(\dfrac{3}{2}\cdot\dfrac{7}{5}+\dfrac{1}{3}\right)$
Products, divisions and powers, from the inside out:
$$\left(\dfrac{3}{5}-\dfrac{2}{3}\right)^{2}+\dfrac{3}{25}-\left(\dfrac{21}{10}+\dfrac{1}{3}\right)$$
Common denominator inside the brackets:
$$\left(\dfrac{9}{15}-\dfrac{10}{15}\right)^{2}+\dfrac{3}{25}-\left(\dfrac{63}{30}+\dfrac{10}{30}\right)=\left(-\dfrac{1}{15}\right)^{2}+\dfrac{3}{25}-\dfrac{73}{30}=\dfrac{1}{225}+\dfrac{3}{25}-\dfrac{73}{30}$$
Common denominator $450$:
$$\dfrac{2}{450}+\dfrac{54}{450}-\dfrac{1095}{450}=-\dfrac{1039}{450}$$
b) $\dfrac{5}{9}-\left(-\dfrac{3}{4}+\dfrac{1}{2}\cdot 3\right)-\dfrac{1}{3}:\left(\dfrac{1}{4}\right)^{2}$
Power, product and division:
$$\dfrac{5}{9}-\left(-\dfrac{3}{4}+\dfrac{3}{2}\right)-\dfrac{1}{3}:\dfrac{1}{16}=\dfrac{5}{9}-\left(-\dfrac{3}{4}+\dfrac{6}{4}\right)-\dfrac{16}{3}=\dfrac{5}{9}-\dfrac{3}{4}-\dfrac{16}{3}$$
Common denominator $36$:
$$\dfrac{20}{36}-\dfrac{27}{36}-\dfrac{192}{36}=-\dfrac{199}{36}$$