Question

check the validity of the result, (x + y)^-1 ≠ x^-1 + y^-1 for x = 1/3, y = -2/7​

Answer 1

$\star\sf\underline\blue{Explanation:-}$

$\sf {(x + y)^{-1} = \bigg(\dfrac{1}{3} + \dfrac{-2}{7} \bigg)^{-1}}$

$\longrightarrow \sf{\bigg[\dfrac{7 + (-6)}{21} \bigg]^{-1}}$

$\longrightarrow \sf{\bigg[\dfrac{7 - 6}{21} \bigg]^{-1}}$

$\longrightarrow \sf{\bigg[\dfrac{1}{21} \bigg]^{-1}}$

$\longrightarrow \boxed{\sf{\purple{21}}}$

$\sf\underline{\green{\:\:\: Now,\:\:\:}}$

$\sf {x^{-1} + y^{-1} = \bigg(\dfrac{1}{3}\bigg)^{-1} + \bigg(\dfrac{-2}{7}\bigg)^{-1}}$

$\longrightarrow \sf {3 + \dfrac{7}{-2}}$

$\longrightarrow \sf {\dfrac{3}{1} + \dfrac{-7}{2}}$

$\longrightarrow \sf {\dfrac{6 - 7}{2}}$

$\longrightarrow \boxed{\sf{\purple{\dfrac{-1}{2}}}}$

$\sf\underline{\pink{\:\:\: Hence,\:\:\:}}$

$\:\:\:\:\dag\bf{\underline{\underline \orange{(x + y)^{-1} \: \neq \: x^{-1} \:+\: y^{-1}}}}$