Question

Verify the property : $x+y=y\times x$ by taking:

(i) $x=-\frac{1}{3},y=\frac{2}{7}$

(ii) $x=\frac{-3}{5},y=\frac{-11}{13}$

(iii) $x=2,y=\frac{7}{-8}$

(iv) $x=0,y=\frac{-15}{8}$

Answer 1

(i) $x=-\frac{1}{3},y=\frac{2}{7}x=-\frac{1}{3},y=\frac{2}{7}L.H.S.=x\times y=-\frac{1}{3}\times \frac{2}{7}=\frac{-1\times 2}{3\times 7}=\frac{-2}{21}R.H.S.=y\times x=\frac{2}{7}\times \frac{-1}{3}=\frac{2\times (-1)}{7\times 3}=\frac{-2}{21}\therefore L.H.S.=R.H.S.$

(ii)$x\times y=y\times xx=\frac{-3}{5},y=\frac{-11}{13}L.H.S.=x\times y-\frac{-3}{5}\times \frac{-11}{13}=\frac{(-3)\times (-11)}{5\times 13}=\frac{33}{65}R.H.S.=y\times x=\frac{-11}{13}\times \frac{-3}{5}=\frac{(-11)\times (-3)}{13\times 5}=\frac{33}{65}\therefore L.H.S.=R.H.S.$

(iii)$x\times y=y\times xx=2,y=\frac{7}{-8}=\frac{7\times (-1)}{-8\times (-1)}=\frac{-7}{8}L.H.S.=x\times y=2\times \frac{-7}{8}=\frac{2\times (-7)}{8}=\frac{-7}{4}R.H.S.=y\times x=\frac{-7}{8}\times 2=\frac{-7}{4}\times 1=\frac{-7}{4}\therefore L.H.S.=R.H.S.$

(iv)$x=0,y=\frac{-15}{8}x\times y=y\times xx=0,y=\frac{-15}{8}L.H.S.=x\times y=0\times \frac{-15}{8}=0R.H.S.=y\times x=\frac{-15}{8}\times 0=0\therefore L.H.S.=R.H.S.$