Question

$\text{Verify that}(x+y)\times z=x\times z+y\times z,if\left(i\right)x=\frac{4}{5},y=\frac{-2}{3}\text{~}andz=-4\left(ii\right)x=2,y=\frac{4}{5}\text{~}andz=\frac{3}{-10}$

Answer 1

$\left(i\right)x=\frac{4}{5},y=\frac{-2}{3}\text{~}andz=-4\text{Using}(x+y)\times z=x\times z+y\times z\Rightarrow (\frac{4}{5}+\frac{-2}{3})\times -4=\frac{4}{5}\times -4+\frac{-2}{3}\times -4\Rightarrow (\frac{4\times 3}{5\times 3}-\frac{2\times 5}{3\times 5})\times -4=\frac{-16}{5}+\frac{8}{3}\Rightarrow \frac{12-10}{15}\times -4=\frac{-48+40}{15}=\frac{-8}{15}=\frac{-8}{15}\left(ii\right)x=2,y=\frac{4}{5}\text{~}andz=\frac{3}{-10}\text{Using,}(x+y)\times z=x\times z+y\times z\Rightarrow (\frac{2}{1}+\frac{4}{5})\times \frac{3}{-10}=2\times \frac{3}{-10}+\frac{4}{5}\times \frac{3}{-10}\Rightarrow (\frac{2\times 5}{1\times 5}+\frac{4\times 1}{5\times 1})\times \frac{3}{-10}=\frac{3}{-5}+\frac{6}{-25}\Rightarrow \left(\frac{10+4}{5}\right)\times \frac{3}{-10}=\frac{-3\times 5}{5\times 5}+\frac{-6\times 1}{5\times 5}\Rightarrow \frac{14}{5}\times \frac{3}{-10}=\frac{-15-6}{25}\Rightarrow \frac{-21}{25}=\frac{-21}{25}$