Incorrect. Since the denominators are the same, you can subtract the numerators: $\displaystyle\frac{{{12}{{x}}^{{2}}}}{{{2}{x}-{3}}}-\frac{{{2}{{x}}^{{2}}+{15}{x}}}{{{2}{x}-{3}}}=\frac{{{10}{{x}}^{{2}}-{15}{x}}}{{{2}{x}-{3}}}\usepackage{cmbright}\Huge\$ . Can you simplify this expression?