# \Find $(x +y) \div (x - y)$. if,(i) $x=\frac{2}{3}, y=\frac{3}{2}$(ii) $x=\frac{2}{5}, y=\frac{1}{2}$(iii) $x=\frac{5}{4}, y=\frac{-1}{3}$(iv) $x=\frac{2}{7}, y=\frac{4}{3}$(v) $x=\frac{1}{4}, y=\frac{3}{2}$

To do:

We have to find $(x +y) \div (x - y)$.

Solution:

(i) $x+y=\frac{2}{3}+\frac{3}{2}$

$=\frac{2\times2+3\times3}{6}$                  (LCM of 3 and 2 is 6)

$=\frac{4+9}{6}$

$=\frac{13}{6}$

$x-y=\frac{2}{3}-\frac{3}{2}$

$=\frac{2\times2-3\times3}{6}$                  (LCM of 3 and 2 is 6)

$=\frac{4-9}{6}$

$=\frac{-5}{6}$

Therefore,

$(x+y)\div(x-y)=\frac{13}{6}\div\frac{-5}{6}$

$=\frac{13}{6}\times\frac{6}{-5}$

$=\frac{13\times6}{6\times-5}$

$=\frac{-13}{5}$

(ii) $x+y=\frac{2}{5}+\frac{1}{2}$

$=\frac{2\times2+1\times5}{10}$                  (LCM of 5 and 2 is 10)

$=\frac{4+5}{10}$

$=\frac{9}{10}$

$x-y=\frac{2}{5}-\frac{1}{2}$

$=\frac{2\times2-1\times5}{10}$                  (LCM of 5 and 2 is 10)

$=\frac{4-5}{10}$

$=\frac{-1}{10}$

Therefore,

$(x+y)\div(x-y)=\frac{9}{10}\div\frac{-1}{10}$

$=\frac{9}{10}\times\frac{10}{-1}$

$=\frac{9\times10}{10\times-1}$

$=\frac{-9}{1}$

$=-9$

(iii) $x+y=\frac{5}{4}+\frac{-1}{3}$

$=\frac{5\times3+(-1)\times4}{12}$                  (LCM of 4 and 3 is 12)

$=\frac{15-4}{12}$

$=\frac{11}{12}$

$x-y=\frac{5}{4}-\frac{-1}{3}$

$=\frac{5\times3-(-1)\times4}{12}$                  (LCM of 4 and 3 is 12)

$=\frac{15+4}{12}$

$=\frac{19}{12}$

Therefore,

$(x+y)\div(x-y)=\frac{11}{12}\div\frac{19}{12}$

$=\frac{11}{12}\times\frac{12}{19}$

$=\frac{11\times12}{12\times19}$

$=\frac{11}{19}$

(iv) $x+y=\frac{2}{7}+\frac{4}{3}$

$=\frac{2\times3+7\times4}{21}$                  (LCM of 7 and 3 is 21)

$=\frac{6+28}{21}$

$=\frac{34}{21}$

$x-y=\frac{2}{7}-\frac{4}{3}$

$=\frac{2\times3-4\times7}{21}$                  (LCM of 7 and 3 is 21)

$=\frac{6-28}{21}$

$=\frac{-22}{21}$

Therefore,

$(x+y)\div(x-y)=\frac{34}{21}\div\frac{-22}{21}$

$=\frac{34}{21}\times\frac{21}{-22}$

$=\frac{17\times21}{21\times(-11)}$

$=\frac{-17}{11}$

(v) $x+y=\frac{1}{4}+\frac{3}{2}$

$=\frac{1\times1+3\times2}{4}$                  (LCM of 4 and 2 is 4)

$=\frac{1+6}{4}$

$=\frac{7}{4}$

$x-y=\frac{1}{4}-\frac{3}{2}$

$=\frac{1\times1-3\times2}{4}$                  (LCM of 4 and 2 is 4)

$=\frac{1-6}{4}$

$=\frac{-5}{4}$

Therefore,

$(x+y)\div(x-y)=\frac{7}{4}\div\frac{-5}{4}$

$=\frac{7}{4}\times\frac{4}{-5}$

$=\frac{7\times4}{4\times(-5)}$

$=\frac{-7}{5}$

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Updated on: 10-Oct-2022

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