Verify the property: $x \times(y + z) = x \times y + x \times z$ 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} $


To do:

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

Solution:

(i) LHS $=x \times(y + z)$

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

$=\frac{-3}{7} \times(\frac{12 \times6+13\times(-5)}{78})$

$=\frac{-3}{7} \times(\frac{72-65}{78})$

$=\frac{-3}{7} \times \frac{7}{78}$

$=\frac{-3 \times 7}{7 \times 78}$

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

RHS $=x \times y + x \times z$

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

$=\frac{-3 \times 12}{7 \times13} + \frac{-3\times(-5)}{7\times6}$

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

$=\frac{-36\times2+5\times13}{182}$

$=\frac{-72+65}{182}$

$=\frac{-7}{182}$

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

LHS $=$ RHS

Therefore,

$x \times(y + z) = x \times y + x \times z$.        

(ii) LHS $=x \times(y + z)$

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

$=\frac{-12}{5} \times(\frac{-15 \times3+8\times4}{12})$

$=\frac{-12}{5} \times(\frac{-45+32}{12})$

$=\frac{-12}{5} \times \frac{-13}{12}$

$=\frac{-12 \times -13}{5 \times 12}$

$=\frac{13}{5}$

RHS $=x \times y + x \times z$

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

$=\frac{-12 \times -15}{5 \times4} + \frac{-12\times8}{5\times3}$

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

$=\frac{9\times5-32\times1}{5}$

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

$=\frac{13}{5}$

LHS $=$ RHS

Therefore,

$x \times(y + z) = x \times y + x \times z$.         

(iii) LHS $=x \times(y + z)$

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

$=\frac{-8}{3} \times(\frac{5 \times2+(-13)\times1}{12})$

$=\frac{-8}{3} \times(\frac{10-13}{12})$

$=\frac{-8}{3} \times \frac{-3}{12}$

$=\frac{-8 \times -3}{3 \times 12}$

$=\frac{2}{3}$

RHS $=x \times y + x \times z$

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

$=\frac{-8 \times 5}{3\times6} + \frac{-8\times(-13)}{3\times12}$

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

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

$=\frac{6}{9}$

$=\frac{2}{3}$

LHS $=$ RHS

Therefore,

$x \times(y + z) = x \times y + x \times z$. 

(iv) LHS $=x \times(y + z)$

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

$=\frac{-3}{4} \times(\frac{-5 \times3+7\times1}{6})$

$=\frac{-3}{4} \times(\frac{-15+7}{6})$

$=\frac{-3}{4} \times \frac{-8}{6}$

$=\frac{-3 \times -8}{4 \times 6}$

$=\frac{1}{1}$

$=1$

RHS $=x \times y + x \times z$

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

$=\frac{-3 \times -5}{4\times2} + \frac{-3\times7}{4\times6}$

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

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

$=\frac{8}{8}$

$=1$

LHS $=$ RHS

Therefore,

$x \times(y + z) = x \times y + x \times z$. 

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Updated on: 10-Oct-2022

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