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Answer :
(i) We have: \(-\frac{2}3×\frac{3}5+\frac{5}2-\frac{3}5×\frac{1}6 \)
\(= -\frac{2}3×\frac{3}5-\frac{3}5×\frac{1}6+\frac{5}2 (By \quad regrouping) \)
\(= \frac{3}5×(-\frac{2}3-\frac{1}6)+\frac{5}2 (by\quad distributive\quad property) \)
\(= \frac{3}5×(\frac{-2×2}{3×2}-\frac{1×1}{6×1})+\frac{5}2 \)
\(= \frac{3}5×(\frac{-4}{6}-\frac{1}{6})+\frac{5}2= \frac{3}5×(\frac{-4-1}{6})+\frac{5}2 \)
\(= \frac{3}5×(\frac{-5}{6})+\frac{5}2 \)
\(= -\frac{3×5}{5×6}+\frac{5}2 \)
\(= -\frac{1}2 +\frac{5}2=(\frac{-1+5}2)=\frac{4}2 =2 \)
so, the value obtained is =2.
(ii) we have:\(\frac{2}5×(-\frac{3}7)-\frac{1}6×\frac{3}2+\frac{1}{14}×\frac{2}5 \)
\(= \frac{2}5×(-\frac{3}7)+\frac{1}{14}×\frac{2}5 -\frac{1}6×\frac{3}2 (by \quad regrouping)\)
\(= \frac{2}5×(\frac{-3}7+\frac{1}{14})-\frac{1}6×\frac{3}2 (using \quad distributive \quad property)\)
\(= \frac{2}5×(\frac{-3×2}{7×2}+\frac{1×1}{14×1})-\frac{1}6×\frac{3}2 \)
\(= \frac{2}5×(\frac{-6}{14}+\frac{1}{14})-\frac{1}6×\frac{3}2\)
\(= \frac{2}5×(\frac{-6+1}{14})-\frac{3}{12}\)
\(= \frac{2}5×(\frac{-5}{14})-\frac{1}{4}\)
\(= -\frac{1}7-\frac{1}{4}= \frac{-1×4}{7×4}-\frac{1×7}{4×7}\)
\(= \frac{-4}{28}-\frac{7}{28}=\frac{-4-7}{28}=\frac{-11}{28}\)
Thus, the value obtained is \(\frac{-11}{28} \).