Problems from Sum and subtraction of fractions

Solve the following operations:

  1. $$\dfrac{1}{6}+\dfrac{2}{6}+\dfrac{4}{6}$$
  2. $$4+\dfrac{2}{1}-\dfrac{-3}{1}$$
  3. $$\dfrac{3}{6}-\dfrac{2}{15}$$
  4. $$\dfrac{19}{24}+\dfrac{1}{18}$$
  5. $$\dfrac{-5}{5}+\dfrac{1}{5}+\dfrac{2}{6}-\dfrac{4}{15}$$
See development and solution

Development:

  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, $$$-1+\dfrac{1}{5}+\dfrac{1}{3}-\dfrac{4}{15}=\dfrac{-1\cdot15}{15}+\dfrac{3}{5\cdot3}+\dfrac{5}{3\cdot5}-\dfrac{4}{15}=$$$ $$$\dfrac{-15+3+5-4}{15}=\dfrac{-11}{15}$$$

Solution:

  1. $$\dfrac{7}{6}$$
  2. $$9$$
  3. $$\dfrac{11}{30}$$
  4. $$\dfrac{61}{72}$$
  5. $$-\dfrac{11}{15}$$
Hide solution and development

Is there any equivalent fraction to the following ones with denominator $$-8$$?

  1. $$\dfrac{1}{-2}$$
  2. $$\dfrac{-1}{3}$$
See development and solution

Development:

  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.

Solution:

  1. Yes, the fraction $$-\dfrac{4}{8}.$$
  2. No.
Hide solution and development
View theory