For $x=\frac{3}{4} $and $y=\frac{-9}{8}$, insert a rational number between:

$(x+y)^{-1} and x^{-1}+y^{-1} $

Given: 
 $x=\frac{3}{4} $and $y=\frac{-9}{8}$

To do ;  Insert a rational number between:

(i) $(x+y)^{-1}  and  x^{-1}+y^{-1} $

(ii) $ (x-y)^{-1}  and x^{-1}-y^{-1} $.

Solution:

(i) $(x+y)^{-1}  and  x^{-1}+y^{-1} $

 $(\frac{3}{4}+\frac{-9}{8})^{-1} $ and  $(\frac{3}{4})^{-1}+(\frac{-9}{8})^{-1} $

LCM OF 8 AND 4 IS 8

=$(\frac{6-9}{8})^{-1}$ and $\frac{4}{3} -\frac{8}{9}$

=$\frac{-8}{3} and \frac{12-8}{9}$ [LCM OF 3 and 9 is 9]

=$\frac{-8}{3} and \frac{4}{9}$

=$\frac{-24}{9} and \frac{4}{9}$

One rational  number between $\frac{-24}{9} and \frac{4}{9}$ = $\frac{1}{9}$