Incorrect. A common denominator of $\displaystyle\frac{{{5}{{x}}^{{2}}}}{{{8}{x}{y}}}-\frac{{{3}{y}}}{{{10}{{x}}^{{3}}}}\usepackage{cmbright}\Huge\$ is $\displaystyle{40}{{x}}^{{3}}{y}\usepackage{cmbright}\Huge\$ . Does this help you simplify the expression?