[tex]\begin{gathered} 2\cdot\frac{2}{5}-3\cdot\frac{1}{2}=\frac{(2\cdot5)+2}{5}-\frac{(3\cdot2)+1}{2} \\ =\frac{10+2}{5}-\frac{7}{2} \\ =\frac{12}{5}-\frac{7}{2} \end{gathered}[/tex][tex]\begin{gathered} \text{ Now, we remember that } \\ \frac{a}{b}-\frac{c}{d}=\frac{(a\cdot d)-(b\cdot c)}{b\cdot d} \end{gathered}[/tex][tex]\begin{gathered} \text{ Thus we get } \\ \frac{12}{5}-\frac{7}{2}=\frac{(12\cdot2)-(5\cdot7)}{5\cdot2} \\ =\frac{24-35}{10} \\ =-\frac{11}{10} \end{gathered}[/tex][tex]\begin{gathered} \text{ So, we finally have} \\ 2\cdot\frac{2}{5}-3\cdot\frac{1}{2}=\frac{-11}{10} \end{gathered}[/tex]
