Operations with Fractions: Problems with Solutions

Solution:
The lowest common denominator is the least common multiple of 2 and 3, which is 6. We expand both fractions: [tex]\frac{1}{2}+\frac{1}{3}=\frac{3}{3\times 2}+\frac{2}{2 \times 3}=\frac{3}{6}+\frac{2}{6}=\frac{5}{6}[/tex].

Solution:
Both fractions already have the same denominator, so we add the numerators: [tex]\frac{3}{2}+\frac{1}{2}=\frac{3+1}{2}=\frac{4}{2}=2[/tex]

Solution:
Since [tex]21=7\times 3[/tex], the lowest common denominator of these fractions is 21. We expand one of them: [tex]\frac{4}{21}+\frac{1}{7}=\frac{4}{21}+\frac{3}{21}=\frac{4+3}{21}=\frac{7}{21}=\frac{1}{3}[/tex]

Solution:
We should first reduce both fractions. The second one already is, but the first is not. We have [tex]\frac{4}{8}+\frac{5}{2}=\frac{1}{2}+\frac{5}{2}=\frac{6}{2}=3[/tex]

Solution:
The least common multiple of 8 and 4 is 8, since [tex]8=2\times 4[/tex]. Then [tex]\frac{3}{8}-\frac{1}{4}=\frac{3}{8}-\frac{2}{2 \cdot 4}=\frac{3}{8}-\frac{2}{8}=\frac{1}{8}[/tex]

Solution:
[tex]\frac{3}{10}+\frac{2}{5}=\frac{3}{10}+\frac{2 \times 2}{5\times2}=\frac{3}{10}+\frac{4}{10}=\frac{7}{10}[/tex]

Solution:
We must first ensure that both fractions are irreducible. [tex]\frac{8}{7}-\frac{2}{14}=\frac{8}{7}-\frac{1}{7}=\frac{8-1}{7}=\frac{7}{7}=1[/tex]

Solution:
[tex]\frac{1}{2}-\frac{5}{17}=\frac{17}{2 \times 17}-\frac{2\times 5}{2 \times 17}=\frac{17-10}{34}=\frac{7}{34}[/tex]

Solution:
[tex]\frac{5}{11}+\frac{13}{22}=\frac{5\times 2}{11\times 2}+\frac{13}{22}=\frac{10}{22}+\frac{13}{22}=\frac{23}{22}[/tex]

Solution:
The lowest common multiple of 2, 3, 6 is 6. So we have [tex]\frac{2}{2\cdot3}-\frac{3}{3\cdot 2}+\frac{1}{6}=\frac{2-3+1}{6}=\frac{0}{6}=0[/tex]

Solution:
[tex]\frac{5}{12}-\frac{1}{3}=\frac{5}{12}-\frac{4}{3.4}=\frac{5-4}{12}=\frac{1}{12}[/tex]

Solution:
[tex]\frac{5}{3}\cdot \frac{15}{10}=\frac{5\cdot 15}{3\cdot 10}=\frac{5\cdot 3\cdot 5}{3\cdot 2\cdot 5}=\frac{5}{2}[/tex]

Solution:
[tex]\frac{5}{18} \cdot \frac{2}{15}=\frac{5}{2 \cdot 9} \cdot \frac{2}{3 \cdot 5}=\frac{2 \cdot 5}{2 \cdot 3 \cdot 5 \cdot 9}=\frac{1}{27}[/tex]

Solution:
Both fractions should be reduced before performing arithmetic operations. We get [tex]\frac{27}{31}-\frac{2 \cdot 27}{2 \cdot 31}=\frac{27}{31}-\frac{27}{31}=0[/tex]

Solution:
We can substitute division by a certian fraction with multiplication by its reciprocal: [tex]\frac{1}{2}:\frac{1}{2}=\frac{1}{2}.\frac{2}{1}=1[/tex]