Verify that a  (b+c) ≠(ab) + (ac)for each of the following

1.	Verify that a  (b+c) ≠(ab) + (ac)for each of the following values of a,band c a = 4 , b= -3 ,c = 5
Answer» $\tt{ a \div (b + c) \not = (a \div b) + (a \div c)}$ $\tt{now\:put\:the\:values\:for\:a,b\:and\:c}$ $\tt{4 \div ( - 3 + 5) \not = (4 \div - (3)) + (4 \div 5)}$ $\tt{ calculate \: the \: sum}$ $\tt{4 \div \red2 \not = (4 \div - (3)) + (4 \div 5)}$ $\small{ \tt divide \: a \: positive \: and \: a \: negative \: equals \: a \: negative} \ratio( + ) \div ( - ) = ( - )$ $\tt{4 \div 2 \not = ( \red - 4 \div 3+ (4 \div 5)}$ $\tt write \: {\orange {\underline{divison \: as \: a \: fraction.}}}$ $⬇⬇⬇⬇⬇⬇$ $\tt4 \div 2 \not = ( \red{ \frac{ - 4}{3}} ) + (4 \div 5)$ $\tt divide \: no.$ $\tt ⬇⬇⬇⬇⬇$ $\tt4 \div 2 \not = ( \frac{ - 4}{3} ) + \red{0.8}$ $\tt \red{ 4 \div 2 }\not = ( \frac{ - 4}{3} ) + {0.8}$ $\tt divide \: no.$ $\tt ⬇⬇⬇⬇⬇$ $\tt \red{2 }\not = ( \frac{ - 4}{3} ) + {0.8}$ $4 \div 2 \not = \red ( - \frac{4}{3} \red) + 0.8$ $\tt remove \: unnecessary \: parenthless$ $\tt ⬇⬇⬇⬇⬇$ $4 \div 2 \not = - \frac{4}{3} + 0.8$ $\tt convert \: decimal \: no. \: into \: a \: \underline \orange{fraction.}$ $\tt ⬇⬇⬇⬇⬇$ $2 \not = - \frac{4}{3} + \frac{4}{5}$ $2 \not = \red{ - \frac{4}{3} + \frac{4}{5} }$ $\tt{calculate \: sum}$ $2 \not= \red{ \frac{ - 8}{15} }$ $\tt{here \: l.h.s \: not \: equal \: to \: r.h.s}$ HENCE VERIFIED