\(10-\frac{9}{16}=\frac{160}{16}-\frac{9}{16}=\frac{151}{16}=9\frac{7}{16}\)
\(\frac{2}{3}-\left(\frac{1}{6}+\frac{1}{8}\right)=\frac{2}{3}-\left(\frac{8}{48}-\frac{6}{48}\right)=\frac{2}{3}-\frac{1}{24}=\frac{16}{24}-\frac{1}{24}=\frac{15}{24}\)
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