Fraction Addition and Subtraction: Difficult Problems with Solutions

Solution:
[tex]\frac{1}{15}+\frac{4}{5}=\frac{1}{15}+\frac{3\cdot4}{3\cdot5}=\frac{1}{15}+\frac{12}{15}=\frac{1+12}{15}=\frac{13}{15}[/tex]

Solution:
[tex]\frac{1}{2}+\frac{1}{4}+\frac{1}{8}=\frac{4}{4\cdot2}+\frac{2}{2\cdot4}+\frac{1}{8}=\frac{4}{8}+\frac{2}{8}+\frac{1}{8}=\frac{4+2+1}{8}=\frac{7}{8}[/tex]

Solution:
[tex]\frac{3}{9}+\frac{9}{27}=\frac{3\cdot1}{3\cdot3}+\frac{9\cdot1}{9\cdot3}=\frac{1}{3}+\frac{1}{3}=\frac{1+1}{3}=\frac{2}{3}[/tex]

Solution:
[tex]\frac{2}{3}+\frac{5}{12}=\frac{4\cdot2}{4\cdot3}+\frac{5}{12}=\frac{8}{12}+\frac{5}{12}=\frac{13}{12}[/tex]

Solution:
[tex]\frac{1}{2}+\frac{5}{14}=\frac{7}{7\cdot2}+\frac{5}{14}=\frac{7}{14}+\frac{5}{14}=\frac{12}{14}=\frac{2\cdot6}{2\cdot7}=\frac{6}{7}[/tex]

Solution:
[tex]\frac{5}{21}+\frac{1}{3}=\frac{5}{21}+\frac{7}{3\cdot7}=\frac{5}{21}+\frac{7}{21}=\frac{5+7}{21}=\frac{12}{21}=\frac{3\cdot4}{3\cdot7}=\frac{4}{7}[/tex]

Solution:
[tex]\frac{2}{7}+\frac{13}{28}=\frac{4\cdot2}{4\cdot7}+\frac{13}{28}=\frac{8}{28}+\frac{13}{28}=\frac{8+13}{28}=\frac{21}{28}=\frac{3\cdot7}{4\cdot7}=\frac{3}{4}[/tex]

Solution:
[tex]\frac{15}{25}+\frac{8}{15}=\frac{3\cdot5}{5\cdot5}+\frac{8}{15}=\frac{3}{5}+\frac{8}{15}=\frac{3\cdot3}{3\cdot5}+\frac{8}{15}=\frac{9}{15}+\frac{8}{15}=\frac{9+8}{15}=\frac{17}{15}=1\frac{2}{15}[/tex]

Solution:
[tex]\frac{5}{21}-\frac{1}{7}=\frac{5}{21}-\frac{3}{3\cdot7}=\frac{5}{21}-\frac{3}{21}=\frac{5-3}{21}=\frac{2}{21}[/tex]

Solution:
[tex]\frac{2}{14}+\frac{5}{21}=\frac{2\cdot1}{2\cdot7}+\frac{5}{21}=\frac{1}{7}+\frac{5}{21}=\frac{3}{3\cdot7}+\frac{5}{21}=\frac{3}{21}+\frac{5}{21}=\frac{8}{21}[/tex]

Solution:
[tex]\frac{5}{13}+\frac{3}{26}=\frac{2\cdot5}{2\cdot13}+\frac{3}{26}=\frac{10}{26}+\frac{3}{26}=\frac{13}{26}=\frac{13}{2\cdot13}=\frac{1}{2}[/tex]

Solution:
[tex]\frac{52}{77}+\frac{2}{11}=\frac{52}{77}+\frac{7\cdot2}{7\cdot11}=\frac{52}{77}+\frac{14}{77}=\frac{52+14}{77}=\frac{66}{77}=\frac{6\cdot11}{7\cdot11}=\frac{6}{7}[/tex]

Solution:
[tex]\frac{18}{65}+\frac{7}{13}=\frac{18}{65}+\frac{7\cdot5}{13\cdot5}=\frac{18}{65}+\frac{35}{65}=\frac{18+35}{65}=\frac{53}{65}[/tex]

Solution:
[tex]\frac{5}{74}-\frac{2}{37}=\frac{5}{74}-\frac{2\cdot2}{37\cdot2}=\frac{5}{74}-\frac{4}{74}=\frac{5-4}{74}=\frac{1}{74}[/tex]

Solution:
[tex]\frac{33}{28}-\frac{13}{14}=\frac{33}{28}-\frac{2\cdot13}{2\cdot14}=\frac{33}{28}-\frac{26}{28}=\frac{33-26}{28}=\frac{7}{28}=\frac{1}{4}[/tex]

Solution:
The lowest common denominator is [tex]LCM(2,3,6)=6[/tex].
[tex]\frac{1}{2}+\frac{1}{3}+\frac{7}{6}=\frac{3}{3\cdot2}+\frac{2}{2\cdot3}+\frac{7}{6}=\frac{3+2+7}{6}=\frac{12}{6}=2[/tex]

Solution:
[tex]\frac{5}{16}+\frac{7}{14}=\frac{5}{16}+\frac{7}{7\cdot2}=\frac{5}{16}+\frac{1}{2}=\frac{5}{16}+\frac{8}{16}=\frac{13}{16}[/tex]

Solution:
[tex]\frac{28}{35}+\frac{63}{15}=\frac{4\cdot 7}{5\cdot7}+\frac{21\cdot3}{5\cdot3}=\frac{4}{5}+\frac{21}{5}=\frac{21+4}{5}=\frac{25}{5}=5[/tex]

Solution:
[tex]\frac{18}{24}+\frac{23}{53}=\frac{3\cdot6}{4\cdot6}+\frac{23}{53}=\frac{3}{4}+\frac{23}{53}=\frac{3\cdot53}{4\cdot53}+\frac{23\cdot4}{53\cdot4}=\frac{159}{212}+\frac{92}{212}=\frac{159+92}{212}=\frac{251}{212}[/tex]

Solution:
[tex]\frac{19}{51}+\frac{11}{17}=\frac{19}{51}+\frac{3\cdot11}{3\cdot17}=\frac{19}{51}+\frac{33}{51}=\frac{19+33}{51}=\frac{52}{51}[/tex]

Solution:
[tex]\frac{349}{23}-\frac{234}{23}=\frac{349-234}{23}=\frac{115}{23}=\frac{5\cdot23}{23}=5[/tex]

Solution:
[tex]\frac{137}{134}-\frac{15}{67}=\frac{137}{134}-\frac{2.15}{2.67}=\frac{137}{134}-\frac{30}{134}=\frac{107}{134}[/tex]

Solution:
[tex]\frac{1}{11}-\frac{2}{11}+\frac{3}{11}-\frac{4}{11}+\frac{5}{11}=\frac{1-2+3-4+5}{11}=\frac{(5-4)+(3-2)+1}{11}=\frac{1+1+1}{11}=\frac{3}{11}[/tex]

Solution:
[tex]\frac{5}{18}-\frac{2}{9}+\frac{7}{36}=\frac{2\cdot5}{2\cdot18}-\frac{2\cdot4}{9\cdot4}+\frac{7}{36}=\frac{10}{36}-\frac{8}{36}+\frac{7}{36}=\frac{10-8+7}{36}=\frac{9}{36}=\frac{1}{4}[/tex]

Solution:
[tex]\frac{234}{165}-\frac{124}{165}=\frac{234-124}{165}=\frac{110}{165}=\frac{11\cdot10}{11\cdot15}=\frac{10}{15}=\frac{2\cdot5}{3\cdot5}=\frac{2}{3}[/tex]

Solution:
Friendly + unfriendly ghosts are 1/4 + 3/5 = 5/20 + 12/20 = 17/20
So neutral ghosts are 1 - 17/20 = 20/20 - 17/20 = 3/20