Sum and subtraction of fractions

1. $\dfrac{1}{6}+\dfrac{2}{6}+\dfrac{4}{6}=\dfrac{1+2+4}{6}=\dfrac{7}{6}$
2. $4+\dfrac{2}{1}-\dfrac{-3}{1}=\dfrac{4}{1}+\dfrac{2-(-3)}{1}=\dfrac{4+2+3}{1}=\dfrac{9}{1}=9$
3. We begin by simplifying the first fraction: $\dfrac{3}{6}=\dfrac{3:3}{6:3}=\dfrac{1}{2}.$ Then, $\left.\begin{array}{l}2=2 \\ 15=3\cdot5 \end{array} \right\} \Rightarrow m.c.m(2,15)=2\cdot3\cdot5=30$

Now we find the value $m$ for the two fractions: $m_1=\dfrac{30}{2}=15$ and $m_2=\dfrac{30}{15}=2.$ Then: $$\dfrac{1}{2}-\dfrac{2}{15}=\dfrac{1\cdot15}{2\cdot15}-\dfrac{2\cdot2}{15\cdot2}=\dfrac{15}{30}-\dfrac{4}{30}=\dfrac{15-4}{30}=\dfrac{11}{30}.$$

4. The fractions are already simplified, so we calculate the less common multiple: $$\dfrac{19}{24}+\dfrac{1}{18}=\dfrac{19\cdot3}{24\cdot3}+\dfrac{1\cdot4}{18\cdot4}=\dfrac{57}{72}+\dfrac{4}{72}=\dfrac{57+4}{72}=\dfrac{61}{72}.$$

5. We simplify the two fractions: $$\dfrac{-5}{5}=\dfrac{-1}{1}=-1$ and $\dfrac{2}{6}=\dfrac{1}{3}.$$ Then, $$-1+\dfrac{1}{5}+\dfrac{1}{3}-\dfrac{4}{15}=\dfrac{-1\cdot15}{15}+\dfrac{3}{5\cdot3}+\dfrac{5}{3\cdot5}-\dfrac{4}{15}=$$ $$\dfrac{-15+3+5-4}{15}=\dfrac{-11}{15}$$

1. To find an equivalent fraction to $\dfrac{1}{-2}$ with denominator $-8$, we start calculating the value $m$: $m=\dfrac{-8}{-2}=4$. The value $m$ exists and it's integer, so this means that the fraction exists: $\dfrac{1}{-2}=\dfrac{1\cdot4}{-2\cdot4}=\dfrac{4}{-8}.$
2. In this case, we do the same process but when we try to calculate the number $m$ we get: $m=\dfrac{-8}{3}=2,\widehat{6}.$ The value $m$ is not an integer, so there isn't an equivalent fraction to $\dfrac{-1}{3}$ which has $-8$ as denominator.