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2n ESO · Fractions Exercises 16 Sep 2026

7. Homework solutions (16 Sep)

Solutions to exercises 21 b, c and g and 22 a and b: adding and subtracting fractions with whole numbers and brackets.

21. Operations with fractions

Carry out these operations and give the result as an irreducible fraction.

b) $\displaystyle -\dfrac{45}{20} - \dfrac{90}{50} - \dfrac{39}{9}$

Simplify each fraction:

$$-\dfrac{45}{20} - \dfrac{90}{50} - \dfrac{39}{9} = -\dfrac{9}{4} - \dfrac{9}{5} - \dfrac{13}{3}$$

$\text{LCM}(4, 5, 3) = 60$. Scale everything up to denominator $60$:

$$-\dfrac{135}{60} - \dfrac{108}{60} - \dfrac{260}{60} = -\dfrac{503}{60}$$

$503$ is prime, so the fraction is already irreducible.

$$-\dfrac{503}{60}$$
c) $\displaystyle \dfrac{2}{18} + \dfrac{6}{12} - \dfrac{4}{15}$

Simplify each fraction:

$$\dfrac{2}{18} + \dfrac{6}{12} - \dfrac{4}{15} = \dfrac{1}{9} + \dfrac{1}{2} - \dfrac{4}{15}$$

$\text{LCM}(9, 2, 15) = 90$:

$$\dfrac{10}{90} + \dfrac{45}{90} - \dfrac{24}{90} = \dfrac{31}{90}$$
$$\dfrac{31}{90}$$
g) $\displaystyle 5 - \dfrac{4}{9} - 3 + \dfrac{12}{30}$

Write the whole numbers over $1$ and simplify $\dfrac{12}{30} = \dfrac{2}{5}$:

$$\dfrac{5}{1} - \dfrac{4}{9} - \dfrac{3}{1} + \dfrac{2}{5}$$

$\text{LCM}(9, 5) = 45$:

$$\dfrac{225}{45} - \dfrac{20}{45} - \dfrac{135}{45} + \dfrac{18}{45} = \dfrac{88}{45}$$
$$\dfrac{88}{45}$$

22. Adding and subtracting with brackets

Carry out the following additions and subtractions of fractions and simplify the result.

a) $\displaystyle \dfrac{26}{75} - \dfrac{76}{30} - \left(\dfrac{7}{6}\right) - 2$

Drop the bracket (nothing changes: a minus sign in front and a positive number inside) and simplify $\dfrac{76}{30} = \dfrac{38}{15}$:

$$\dfrac{26}{75} - \dfrac{38}{15} - \dfrac{7}{6} - \dfrac{2}{1}$$

$\text{LCM}(75, 15, 6) = 150$:

$$\dfrac{52}{150} - \dfrac{380}{150} - \dfrac{175}{150} - \dfrac{300}{150} = -\dfrac{803}{150}$$

$803 = 11 \cdot 73$ and $150 = 2 \cdot 3 \cdot 5^2$ share no factor, so it is irreducible.

$$-\dfrac{803}{150}$$
b) $\displaystyle \left(\dfrac{9}{10} - \dfrac{13}{32}\right) - \left(\dfrac{3}{5} - \dfrac{1}{6}\right)$

Remove the brackets. The minus sign in front of the second one changes the sign of everything inside:

$$\dfrac{9}{10} - \dfrac{13}{32} - \dfrac{3}{5} + \dfrac{1}{6}$$

$\text{LCM}(10, 32, 5, 6) = 480$:

$$\dfrac{432}{480} - \dfrac{195}{480} - \dfrac{288}{480} + \dfrac{80}{480}$$

You can group the positive and negative terms first:

$$\dfrac{512}{480} - \dfrac{483}{480} = \dfrac{29}{480}$$
$$\dfrac{29}{480}$$