Question

Verify the property: x × (y + z) = x × y + x × z by taking:

(i) $x=\frac{-3}{7},y=\frac{12}{13},z=\frac{-5}{6}$

(ii) $x=\frac{-12}{5},y=\frac{-15}{4},z=\frac{8}{3}$

(iii) $x=\frac{-8}{3},y=\frac{5}{6},z=\frac{-13}{12}$

(iv) $x=\frac{-3}{4},y=\frac{-5}{2},z=\frac{7}{6}$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}\mathrm{to}\mathrm{verify}\mathrm{that}x\times (y+z)=x\times y+x\times z. \\ \left(\mathrm{i}\right)x=\frac{-3}{7},y=\frac{12}{13},z=\frac{-5}{6} \\ \mathrm{x}\times (\mathrm{y}+\mathrm{z})=\frac{-3}{7}\times (\frac{12}{13}+\frac{-5}{6})=\frac{-3}{7}\times \frac{72-65}{78}=\frac{-3}{7}\times \frac{7}{78}=\frac{-1}{26} \\ x\times y+x\times z=\frac{-3}{7}\times \frac{12}{13}+\frac{-3}{7}\times \frac{-5}{6} \\ =\frac{-36}{91}+\frac{5}{14} \\ =\frac{-36\times 2+5\times 13}{182}=\frac{-72+65}{182} \\ =\frac{-1}{26} \\ \therefore \frac{-3}{7}\times (\frac{12}{13}+\frac{-5}{6})=\frac{-3}{7}\times \frac{12}{13}+\frac{-3}{7}\times \frac{-5}{6} \\ \mathrm{Hence}\mathrm{verified}. \\ \left(\mathrm{ii}\right)x=\frac{-12}{5},y=\frac{-15}{4},z=\frac{8}{3} \\ x\times (y+z)=\frac{-12}{5}\times (\frac{-15}{4}+\frac{8}{3})=\frac{-12}{5}\times \frac{-45+32}{12}=\frac{-12}{5}\times \frac{-13}{12}=\frac{13}{5} \\ x\times y+x\times z=\frac{-12}{5}\times \frac{-15}{4}+\frac{-12}{5}\times \frac{8}{3} \\ =\frac{9}{1}+\frac{-32}{5} \\ =\frac{45-32}{5} \\ =\frac{13}{5} \\ \therefore \frac{-12}{5}\times (\frac{-15}{4}+\frac{8}{3})=\frac{-12}{5}\times \frac{-15}{4}+\frac{-12}{5}\times \frac{8}{3} \\ \mathrm{Hence}\mathrm{verified}. \\ \left(\mathrm{iii}\right)x=\frac{-8}{3},y=\frac{5}{6},\mathrm{z}=\frac{-13}{12} \\ x\times (y+z)=\frac{-8}{3}\times (\frac{5}{6}+\frac{-13}{12})=\frac{-8}{3}\times \frac{10-13}{12}=\frac{-8}{3}\times \frac{-3}{12}=\frac{2}{3} \\ x\times y+x\times z=\frac{-8}{3}\times \frac{5}{6}+\frac{-8}{3}\times \frac{-13}{12} \\ =\frac{-20}{9}+\frac{26}{9} \\ =\frac{-20+26}{9} \\ =\frac{2}{3} \\ \therefore \frac{-8}{3}\times (\frac{5}{6}+\frac{-13}{12})=\frac{-8}{3}\times \frac{5}{6}+\frac{-8}{3}\times \frac{-13}{12} \\ \mathrm{Hence}\mathrm{verified}. \\ \left(\mathrm{iv}\right)x=\frac{-3}{4},y=\frac{-5}{2},\mathrm{z}=\frac{7}{6} \\ x\times (y+z)=\frac{-3}{4}\times (\frac{-5}{2}+\frac{7}{6})=\frac{-3}{4}\times \frac{-15+7}{6}=\frac{-3}{4}\times \frac{-8}{6}=1 \\ x\times y+x\times z=\frac{-3}{4}\times \frac{-5}{2}+\frac{-3}{4}\times \frac{7}{6} \\ =\frac{15}{8}+\frac{-7}{8} \\ =\frac{15-7}{8} \\ =1 \\ \therefore \frac{-3}{4}\times (\frac{-5}{2}+\frac{7}{6})=\frac{-3}{4}\times \frac{-5}{2}+\frac{-3}{4}\times \frac{7}{6} \\ \mathrm{Hence}\mathrm{verified}. \end{gathered}$$