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Simplify each of the following and write as a rational number of the form:
(i) $ \frac{3}{4}+\frac{5}{6}+\frac{-7}{8} $
(ii) $ \frac{2}{3}+\frac{-5}{6}+\frac{-7}{9} $
(iii) $ \frac{-11}{2}+\frac{7}{6}+\frac{-5}{8} $
(iv) $ \frac{-4}{5}+\frac{-7}{10}+\frac{-8}{15} $
(v) $ \frac{-9}{10}+\frac{22}{15}+\frac{13}{-20} $
(vi) $ \frac{5}{3}+\frac{3}{-2}+\frac{-7}{3}+3 $
To do:
We have to simplify and write the given expressions as rational numbers.
Solution:
(i) $\frac{3}{4}+\frac{5}{6}+\frac{-7}{8}$
LCM of $4, 6$ and $8=24$
$\frac{3}{4}+\frac{5}{6}+\frac{-7}{8}=\frac{3(6)+5(4)+(-7)(3)}{24}$
$=\frac{18+20-21}{24}$
$=\frac{38-21}{24}$
$=\frac{17}{24}$
Therefore,
$\frac{3}{4}+\frac{5}{6}+\frac{-7}{8}=\frac{17}{24}$.
(ii) \( \frac{2}{3}+\frac{-5}{6}+\frac{-7}{9} \)
LCM of $3, 6$ and $9=18$
$\frac{2}{3}+\frac{-5}{6}+\frac{-7}{9} = \frac{2(6)+(-5)(3)+(-7)(2)}{18}$
$=\frac{12-15-14}{18}$
$=\frac{12-29}{18}$
$=\frac{-17}{18}$
Therefore,
$\frac{2}{3}+\frac{-5}{6}+\frac{-7}{9}=\frac{-17}{18}$.
(iii) \( \frac{-11}{2}+\frac{7}{6}+\frac{-5}{8} \)
LCM of $2, 6$ and $8=24$
$\frac{-11}{2}+\frac{7}{6}+\frac{-5}{8} = \frac{(-11)(12)+(7)(4)+(-5)(3)}{24}$
$= \frac{-132+28-15}{24}$
$=\frac{-147+28}{24}$
$=\frac{-119}{24}$
Therefore,
$\frac{-11}{2}+\frac{7}{6}+\frac{-5}{8}=\frac{-119}{24}$.
(iv) \( \frac{-4}{5}+\frac{-7}{10}+\frac{-8}{15} \)
LCM of $5, 10$ and $15=30$
$\frac{-4}{5}+\frac{-7}{10}+\frac{-8}{15} = \frac{(-4)(6)+(-7)(3)+(-8)(2)}{30}$
$= \frac{-24-21-16}{30}$
$=\frac{-61}{30}$
Therefore,
$\frac{-4}{5}+\frac{-7}{10}+\frac{-8}{15}=\frac{-61}{30}$.
(v) \( \frac{-9}{10}+\frac{22}{15}+\frac{13}{-20} \)
LCM of $10, 15$ and $20=60$
$\frac{-9}{10}+\frac{22}{15}+\frac{13}{-20} = \frac{(-9)(6)+(22)(4)+(-13)(3)}{60}$
$= \frac{-54+88-39}{60}$
$=\frac{-5}{60}$
$=\frac{-1}{12}$
Therefore,
$\frac{-9}{10}+\frac{22}{15}+\frac{13}{-20} =\frac{-1}{12}$.
(vi) \( \frac{5}{3}+\frac{3}{-2}+\frac{-7}{3}+3 \)
LCM of $3, 2, 3$ and $1=6$
$\frac{5}{3}+\frac{3}{-2}+\frac{-7}{3}+3 = \frac{(5)(2)+(-3)(3)+(-7)(2)+3(6)}{6}$
$= \frac{10-9-14+18}{6}$
$=\frac{5}{6}$
Therefore,
$\frac{5}{3}+\frac{3}{-2}+\frac{-7}{3}+3 =\frac{5}{6}$.
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