Question

simplify the following
1)(2/3)+(-4/5)+1+(-2/3)+(-11/5)
2)(5/8)+(-8/9)+0+(-13/3)+(17/24)​

Answer 1

$\large\underline{\sf{Solution-}}$

$\sf \: Simplify \: the \: following :$

$\bf (1).. \: \: \sf \bigg(\dfrac{2}{3} \bigg)+ \bigg( \dfrac{ - 4}{5} \bigg)+1+ \bigg(\dfrac{ - 2}{3}\bigg) +\bigg( \dfrac{ - 11}{5} \bigg)$

$\sf \implies \dfrac{2}{3} + \dfrac{ - 4}{5} +1+ \dfrac{ - 2}{3} + \dfrac{ - 11}{5}$

$\sf \implies \dfrac{2}{3} + \dfrac{ - 4}{5} + \dfrac{1}{1} + \dfrac{ - 2}{3} + \dfrac{ - 11}{5}$

$\sf Finding \: LCM \: of \: 3, \: 5, \: 1, \: 3, \: 5$

$\begin{array}{c|c}{\sf { \underline{3}}}&\underline{\sf { \: \: \: 3 - 5 - 1 - 3 - 5 \: \: \: }} \\ {\sf { \underline{5}}}&\underline{\sf {\; \; 1 - 5 - 1 - 1 - 5\; \; \: }} \\{\sf {}}&{\sf \; \;1 - 1 - 1 - 1 - 1\; \; \; } \end{array}$

$\sf LCM = 3 \times 5 = 15$

$\sf \implies \dfrac{2}{3} \times \dfrac{5}{5} = \dfrac{10}{15}$

$\sf \implies \dfrac{ - 4}{5} \times \dfrac{3}{3} = \dfrac{ - 12}{15}$

$\sf \implies \dfrac{ 1}{1} \times \dfrac{15}{15} = \dfrac{15 }{15}$

$\sf \implies \dfrac{ - 2}{3} \times \dfrac{5 }{5} = \dfrac{ - 10}{15}$

$\sf \implies \dfrac{ - 11}{5} \times \dfrac{3}{3} = \dfrac{ - 33}{15}$

$\bf \underline{Now},$

$\sf \implies \dfrac{10}{15} +\dfrac{ - 12}{15} + \dfrac{15 }{15} + \dfrac{ - 10}{15} + \dfrac{ - 33}{15}$

$\sf Taking \: denominator \: as \: common$

$\sf \implies \dfrac{10 + ( - 12) + 15 + ( - 10) + ( - 33)}{15}$

$\sf \implies \dfrac{10 - 12 + 15 - 10 - 33}{15}$

$\sf \implies \dfrac{ - 2 + 15 - 10 - 33}{15}$

$\sf \implies \dfrac{ 13 - 10 - 33}{15}$

$\sf \implies \dfrac{ 3 - 33}{15}$

$\sf \implies \cancel\dfrac{ - 30}{15}$

$\sf \implies - 2$

$\sf \implies \underline{\boxed{ \sf Answer = -2 }}$

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$\bf (2).. \: \: \sf \bigg(\dfrac{5}{8} \bigg)+ \bigg( \dfrac{ - 8}{9} \bigg)+0+ \bigg(\dfrac{ - 13}{3}\bigg) +\bigg( \dfrac{ 17}{24} \bigg)$

$\sf \implies \dfrac{5}{8} + \dfrac{ - 8}{9} +0+ \dfrac{ - 13}{3}+ \dfrac{ 17}{24}$

$\sf \implies \dfrac{5}{8} + \dfrac{ - 8}{9} + \dfrac{0}{1} + \dfrac{ - 13}{3}+ \dfrac{ 17}{24}$

$\sf Finding \: LCM \: of \: 8, \: 9, \: 1, \: 3, \: 24$

$\begin{array}{c|c}{\sf { \underline{2}}}&\underline{\sf { \: \: \: 8 - 9 - 1 - 3 - 24\: \: \: }} \\ {\sf { \underline{2}}}&\underline{\sf {\; \; 4 - 9 - 1 - 3 - 12\; \; \: }} \\ { \sf { \underline{2}}}&\underline{\sf {\; \; 2 - 9 - 1 - 3- 6\; \; \: }} \\ {\sf { \underline{3}}}&\underline{\sf {\; \; 1 - 9- 1 - 3 - 3\; \; \: }} \\ {\sf { \underline{3}}}&\underline{\sf {\; \; 1 - 3- 1 - 1 - 1\; \; \: }} \\{\sf {}}&{\sf \; \;1 - 1 - 1 - 1 - 1\; \; \; } \end{array}$

$\sf LCM = 2 \times 2 \times 2 \times 3 \times 3 = 72$

$\sf \implies \dfrac{5}{8} \times \dfrac{ 9}{9} = \dfrac{45}{72}$

$\sf \implies \dfrac{ - 8}{9} \times \dfrac{ 8}{8} = \dfrac{ - 64}{72}$

$\sf \implies \dfrac{0}{1} \times \dfrac{72 }{72} = \dfrac{0}{72}$

$\sf \implies \dfrac{ - 13}{3} \times \dfrac{24 }{24} = \dfrac{ - 312}{72}$

$\sf \implies \dfrac{ 17}{24} \times \dfrac{ 3}{3} = \dfrac{51}{72}$

$\bf \underline{Now},$

$\sf \implies \dfrac{45}{72} + \dfrac{ - 64}{72} + \dfrac{0}{72} + \dfrac{ - 312}{72} + \dfrac{51}{72}$

$\sf Taking \: denominator \: as \: common$

$\sf \implies \dfrac{45 + ( - 64) + 0 + ( - 312) + 51}{72}$

$\sf \implies \dfrac{45 - 64+ 0 - 312 + 51}{72}$

$\sf \implies \dfrac{-19+ 0 - 312 + 51}{72}$

$\sf \implies \dfrac{-331 + 51}{72}$

$\sf \implies \cancel \dfrac{-280}{72}$

$\sf \implies \dfrac{-35}{9}$

$\sf \implies \underline{\boxed{ \sf Answer = \dfrac{-35}{9} }}$

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${\bf{Additional \: information : }}$

$\begin{gathered}\boxed{\begin{array} {c|c} \sf{ +,- } & \sf{ - } \\ \dfrac{\qquad\qquad}{} & \dfrac{\qquad}{} \\ \sf{- , +} & \sf{ - } \\ \dfrac{\qquad\qquad}{} & \dfrac{\qquad}{} \\ \sf{+,+} & \sf{ + } \\ \dfrac{\qquad\qquad}{} & \dfrac{\qquad }{} \\ \sf{-,-} & \sf{ + }\end{array}}\end{gathered}$