9. Homework solutions (22 Sep) — Fractions

22. Adding and subtracting with brackets

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

i) $\displaystyle \dfrac{6}{7} - \left(\dfrac{3}{4} + \dfrac{5}{6}\right) - \left(-\dfrac{7}{5}\right)$

First the bracket: $\dfrac{3}{4} + \dfrac{5}{6} = \dfrac{9}{12} + \dfrac{10}{12} = \dfrac{19}{12}$. And minus times minus is plus:

\dfrac{6}{7} - \dfrac{19}{12} + \dfrac{7}{5}

$\text{LCM}(7, 12, 5) = 420$:

\dfrac{360}{420} - \dfrac{665}{420} + \dfrac{588}{420} = \dfrac{283}{420}

$283$ is prime, so it is irreducible.

\dfrac{283}{420}

Work out the following operations and simplify the result.

c) $\displaystyle \left(\dfrac{9}{10} : \dfrac{36}{40}\right) \cdot \left(\dfrac{85}{7} : \dfrac{15}{21}\right)$

Each division becomes a multiplication by the reciprocal:

\dfrac{9}{10} \cdot \dfrac{40}{36} \cdot \dfrac{85}{7} \cdot \dfrac{21}{15}

Factorise and cancel:

\dfrac{3^2 \cdot (2^3 \cdot 5) \cdot (5 \cdot 17) \cdot (3 \cdot 7)}{(2 \cdot 5) \cdot (2^2 \cdot 3^2) \cdot 7 \cdot (3 \cdot 5)} = 17

Every factor cancels except the $17$ on top.

$$17$$

g) $\displaystyle \left(\dfrac{72}{39} \cdot \dfrac{26}{12}\right) : (-3)$

Dividing by $-3$ is multiplying by $-\dfrac{1}{3}$. The minus sign goes in front:

-\dfrac{72}{39} \cdot \dfrac{26}{12} \cdot \dfrac{1}{3}

Factorise and cancel:

-\dfrac{(2^3 \cdot 3^2) \cdot (2 \cdot 13)}{(3 \cdot 13) \cdot (2^2 \cdot 3) \cdot 3} = -\dfrac{2^2}{3} = -\dfrac{4}{3}

-\dfrac{4}{3}

Carry out the operations, solving each division as a multiplication by the reciprocal, and simplify the result.

c) $\displaystyle \dfrac{98}{72} : \dfrac{48}{66} : \left(-\dfrac{30}{45}\right)$

The first fraction stays and the other two are flipped. There is a single minus sign, so the result is negative:

-\dfrac{98}{72} \cdot \dfrac{66}{48} \cdot \dfrac{45}{30}

Factorise and cancel:

-\dfrac{(2 \cdot 7^2) \cdot (2 \cdot 3 \cdot 11) \cdot (3^2 \cdot 5)}{(2^3 \cdot 3^2) \cdot (2^4 \cdot 3) \cdot (2 \cdot 3 \cdot 5)} = -\dfrac{7^2 \cdot 11}{2^6 \cdot 3} = -\dfrac{539}{192}

-\dfrac{539}{192}

f) $\displaystyle \dfrac{72}{75} \cdot \dfrac{15}{20} : \left(-\dfrac{4}{3}\right)$

The division becomes a multiplication by the reciprocal, with the minus sign in front:

-\dfrac{72}{75} \cdot \dfrac{15}{20} \cdot \dfrac{3}{4}

Factorise and cancel:

-\dfrac{(2^3 \cdot 3^2) \cdot (3 \cdot 5) \cdot 3}{(3 \cdot 5^2) \cdot (2^2 \cdot 5) \cdot 2^2} = -\dfrac{3^3}{2 \cdot 5^2} = -\dfrac{27}{50}

-\dfrac{27}{50}