Operations with Fractions: Difficult Problems with Solutions

Solution:
We expand both fractions to their lowest common denominator: [tex]\frac{3}{2}+\frac{5}{7}=\frac{3 \cdot 7}{2 \cdot 7}+\frac{5 \cdot 2}{7 \cdot 2}=\frac{21}{14}+\frac{10}{14}=\frac{31}{14}[/tex].

Solution:
[tex]\frac{8}{15}-\frac{1}{3}=\frac{8}{15}-\frac{5}{3 \cdot 5}=\frac{8-5}{15}=\frac{3}{15}=\frac{1}{5}[/tex]

Solution:
Since both 17 and 19 are prime numbers, their lowest common multiple is also their product. Then [tex]\frac{8}{17}-\frac{8}{19}=\frac{8 \cdot 19}{17 \cdot 19}-\frac{8 \cdot 17}{17 \cdot 19}=\frac{8 \cdot 19-8 \cdot 17}{323}=\frac{8(19-17)}{323}=\frac{8 \cdot 2}{323}=\frac{16}{323}[/tex]

Solution:
We must first assure that both fractions are reduced and then expand to the lowest common denominator. [tex]\frac{18}{15}-\frac{1}{2}=\frac{3 \cdot 6}{3 \cdot 5}-\frac{1}{2}=\frac{6}{5}-\frac{1}{2}=\frac{6 \cdot 2}{5 \cdot 2}-\frac{5}{5 \cdot 2}=\frac{12-5}{10}=\frac{7}{10}[/tex].

Solution:
[tex]\frac{3}{7} \cdot 21-\frac{15}{2}=\frac{3 \cdot 3 \cdot 7}{7}-\frac{15}{2}=9-\frac{15}{2}=\frac{2 \cdot 9}{2}-\frac{15}{2}=\frac{18-15}{2}=\frac{3}{2} = 1\frac{1}{2}[/tex]

Solution:
[tex]\frac{15}{3} \cdot \frac{9}{10}-\frac{2}{3} \cdot \frac{9}{8}=\frac{5 \cdot 3 \cdot 9}{3 \cdot 2 \cdot 5}-\frac{ \cdot 9}{3 \cdot 8}=\frac{9}{2}-\frac{3}{4}=\frac{18}{4}-\frac{3}{4}=\frac{15}{4}=3\frac{3}{4}[/tex]

Solution:
[tex]\frac{3}{7} \cdot \frac{15}{9}-\frac{2}{3}=\frac{3 \cdot 15}{7 \cdot 9}-\frac{2}{3}=\frac{3 \cdot 3 \cdot 5}{7 \cdot 9}-\frac{2}{3}=\frac{5}{7}-\frac{2}{3}=\frac{15}{21}-\frac{14}{21}=\frac{1}{21}[/tex]

Solution:
[tex]\frac{5^2-1}{18}\cdot \frac{10}{4}=\frac{24\cdot 10}{18\cdot 4}=\frac{6\cdot 4\cdot 10}{6\cdot 3\cdot 4}=\frac{10}{3}[/tex]

Solution:
[tex]\frac{3}{8} \cdot \frac{24}{5} \cdot \frac{10}{27}=\frac{3 \cdot 24 \cdot 10}{8 \cdot 5 \cdot 27}=\frac{3 \cdot 3 \cdot 8 \cdot 2 \cdot 5}{8 \cdot 5 \cdot 3 \cdot 3 \cdot 3}=\frac{2}{3}[/tex]

Solution:
[tex]\frac{3\cdot 5\cdot 7}{5\cdot 7\cdot 9}-\frac{2\cdot 4\cdot 6}{4\cdot 6\cdot 8}=\frac{3}{9}-\frac{2}{8}=\frac{1}{3}-\frac{1}{4}=\frac{4}{12}-\frac{3}{12}=\frac{1}{12}[/tex]

Solution:
[tex]\frac{2^5}{15}-\frac{2^4}{15}=2^4(\frac{2}{15}-\frac{1}{15})=\frac{2^4}{15}=\frac{16}{15}[/tex]

Solution:
[tex]\frac{4}{9}.\frac{27}{8}+\frac{9}{8}.\frac{16}{3}=\frac{3}{2}+6=\frac{3}{2}+\frac{12}{2}=\frac{15}{2}[/tex]

Solution:
[tex]\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}=\frac{2}{4}-\frac{1}{4}+\frac{2}{16}-\frac{1}{16}=\frac{1}{4}+\frac{1}{16}=\frac{4}{16}+\frac{1}{16}=\frac{5}{16}[/tex]

Solution:
Girls are: 2/3 from 21 or 21 * 2/3 = 42/7 = 14.
21 - 14 = 7
Boys are 7.

Solution:
[tex]\frac{2}{7}.\frac{14}{2}+1=\frac{2.14}{7.2}+1=\frac{14}{7}+1=2+1=3[/tex]