\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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