Verify the property: x × (y + z) = x × y + x × z by taking:
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) $x=\frac{-3}{4}, y=\frac{-5}{2}, z=\frac{7}{6}$

 

 

Solution:

We have to verify that $x \times(y+z)=x \times y+x \times z$.

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

$\mathbf{x} \times(\mathbf{y}+\mathbf{z})=\frac{-3}{7} \times\left(\frac{12}{13}+\frac{-5}{6}\right)=\frac{-3}{7} \times \frac{72-65}{78}=\frac{-3}{7} \times \frac{7}{78}=\frac{-1}{26}$

$x \times y+x \times z=\frac{-3}{7} \times \frac{12}{13}+\frac{-3}{7} \times \frac{-5}{6}$

$=\frac{-36}{91}+\frac{5}{14}$

$=\frac{-36 \times 2+5 \times 13}{182}=\frac{-72+65}{182}$

$=\frac{-1}{26}$

$\therefore \frac{-3}{7} \times\left(\frac{12}{13}+\frac{-5}{6}\right)=\frac{-3}{7} \times \frac{12}{13}+\frac{-3}{7} \times \frac{-5}{6}$

Hence verified.

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

$x \times(y+z)=\frac{-12}{5} \times\left(\frac{-15}{4}+\frac{8}{3}\right)=\frac{-12}{5} \times \frac{-45+32}{12}=\frac{-12}{5} \times \frac{-13}{12}=\frac{13}{5}$

$x \times y+x \times z=\frac{-12}{5} \times \frac{-15}{4}+\frac{-12}{5} \times \frac{8}{3}$

$=\frac{9}{1}+\frac{-32}{5}$

$=\frac{45-32}{5}$

$=\frac{13}{5}$

$\therefore \frac{-12}{5} \times\left(\frac{-15}{4}+\frac{8}{3}\right)=\frac{-12}{5} \times \frac{-15}{4}+\frac{-12}{5} \times \frac{8}{3}$

Hence verified.

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

$x \times(y+z)=\frac{-8}{3} \times\left(\frac{5}{6}+\frac{-13}{12}\right)=\frac{-8}{3} \times \frac{10-13}{12}=\frac{-8}{3} \times \frac{-3}{12}=\frac{2}{3}$

$x \times y+x \times z=\frac{-8}{3} \times \frac{5}{6}+\frac{-8}{3} \times \frac{-13}{12}$

$=\frac{-20}{9}+\frac{26}{9}$

$=\frac{-20+26}{9}$

$=\frac{2}{3}$

$\therefore \frac{-8}{3} \times\left(\frac{5}{6}+\frac{-13}{12}\right)=\frac{-8}{3} \times \frac{5}{6}+\frac{-8}{3} \times \frac{-13}{12}$

Hence verified.

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

$x \times(y+z)=\frac{-3}{4} \times\left(\frac{-5}{2}+\frac{7}{6}\right)=\frac{-3}{4} \times \frac{-15+7}{6}=\frac{-3}{4} \times \frac{-8}{6}=1$

$x \times y+x \times z=\frac{-3}{4} \times \frac{-5}{2}+\frac{-3}{4} \times \frac{7}{6}$

$=\frac{15}{8}+\frac{-7}{8}$

$=\frac{15-7}{8}$

$=1$

$\therefore \frac{-3}{4} \times\left(\frac{-5}{2}+\frac{7}{6}\right)=\frac{-3}{4} \times \frac{-5}{2}+\frac{-3}{4} \times \frac{7}{6}$

Hence verified.

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