a: Ta có: \(\left(y-\dfrac{1}{2}\right)\cdot\dfrac{5}{3}=\dfrac{7}{4}+\dfrac{1}{2}\)
\(\Leftrightarrow y-\dfrac{1}{2}=\dfrac{27}{20}\)
hay \(y=\dfrac{37}{20}\)
b: Ta có: \(\left(y+\dfrac{2}{5}\right):\dfrac{5}{3}=\dfrac{7}{5}\)
\(\Leftrightarrow y+\dfrac{2}{5}=\dfrac{7}{3}\)
hay \(y=\dfrac{29}{15}\)