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Express each of the following as a rational number of the form $\frac{p}{q}$
(i) $ -\frac{8}{3}+\frac{-1}{4}+\frac{-11}{6}+\frac{3}{8}-3 $
(ii) $ \frac{6}{7}+1+\frac{-7}{9}+\frac{19}{21}+\frac{-12}{7} $
(iii) $ \frac{15}{2}+\frac{9}{8}+\frac{-11}{3}+6+\frac{-7}{6} $
(iv) $ \frac{-7}{4}+0+\frac{-9}{5}+\frac{19}{10}+\frac{11}{14} $
(v) $ \frac{-7}{4}+\frac{5}{3}+\frac{-1}{2}+\frac{-5}{6}+2 $
To do:
We have to express the given expressions as a rational number of the form $\frac{p}{q}$.
Solution:
(i) \( -\frac{8}{3}+\frac{-1}{4}+\frac{-11}{6}+\frac{3}{8}-3 \)
LCM of $3, 4, 6$ and $8=24$
$-\frac{8}{3}+\frac{-1}{4}+\frac{-11}{6}+\frac{3}{8}-3 = \frac{(-8)(8)+(-1)(6)+(-11)(4)+3(3)-3(24)}{24}$
$= \frac{-64-6-44+9-72}{24}$
$=\frac{-177}{24}$
$=\frac{-59}{8}$
Therefore,
$\frac{8}{3}+\frac{-1}{4}+\frac{-11}{6}+\frac{3}{8}-3 =\frac{-59}{8}$.
(ii) \( \frac{6}{7}+1+\frac{-7}{9}+\frac{19}{21}+\frac{-12}{7} \)
LCM of $7, 9, 21$ and $7=63$
$\frac{6}{7}+1+\frac{-7}{9}+\frac{19}{21}+\frac{-12}{7} = \frac{(6)(9)+(1)(63)+(-7)(7)+19(3)+(-12)(9)}{63}$
$= \frac{54+63-49+57-108}{63}$
$=\frac{17}{63}$
Therefore,
$\frac{6}{7}+1+\frac{-7}{9}+\frac{19}{21}+\frac{-12}{7} =\frac{17}{63}$.
(iii) \( \frac{15}{2}+\frac{9}{8}+\frac{-11}{3}+6+\frac{-7}{6} \)
LCM of $2, 8, 3$ and $6=24$
$\frac{15}{2}+\frac{9}{8}+\frac{-11}{3}+6+\frac{-7}{6}= \frac{(15)(12)+(9)(3)+(-11)(8)+6(24)+(-7)(4)}{24}$
$= \frac{180+27-88+144-28}{24}$
$=\frac{235}{24}$
Therefore,
$\frac{15}{2}+\frac{9}{8}+\frac{-11}{3}+6+\frac{-7}{6} =\frac{235}{24}$.
(iv) \( \frac{-7}{4}+0+\frac{-9}{5}+\frac{19}{10}+\frac{11}{14} \)
LCM of $4, 5, 10$ and $14=140$
$\frac{-7}{4}+0+\frac{-9}{5}+\frac{19}{10}+\frac{11}{14}= \frac{(-7)(35)+(0)(140)+(-9)(28)+19(14)+(11)(10)}{140}$
$= \frac{-245+0-252+266+110}{140}$
$=\frac{-121}{140}$
Therefore,
$\frac{-7}{4}+0+\frac{-9}{5}+\frac{19}{10}+\frac{11}{14}=\frac{-121}{140}$.
(v) \( \frac{-7}{4}+\frac{5}{3}+\frac{-1}{2}+\frac{-5}{6}+2 \)
LCM of $4, 3, 2$ and $6=12$
$\frac{-7}{4}+\frac{5}{3}+\frac{-1}{2}+\frac{-5}{6}+2= \frac{(-7)(3)+(5)(4)+(-1)(6)+(-5)(2)+(2)(12)}{12}$
$= \frac{-21+20-6-10+24}{12}$
$=\frac{7}{12}$
Therefore,
$\frac{-7}{4}+\frac{5}{3}+\frac{-1}{2}+\frac{-5}{6}+2=\frac{7}{12}$.
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