Correct. Adding the expressions and then factoring, you find that $\displaystyle\frac{{{2}{\left({3}{x}+{1}\right)}}}{{{x}+{3}}}+\frac{{{3}{x}+{7}}}{{{x}+{3}}}=\frac{{{2}{\left({3}{x}+{1}\right)}+{\left({3}{x}+{7}\right)}}}{{{x}+{3}}}=\frac{{{6}{x}+{2}+{3}{x}+{7}}}{{{x}+{3}}}=\usepackage{cmbright}\Huge\$ $\displaystyle\frac{{{9}{x}+{9}}}{{{x}+{3}}}=\frac{{{9}{\left({x}+{1}\right)}}}{{{x}+{3}}}\usepackage{cmbright}\Huge\$ . Only a value of $\displaystyle{x}=-{3}\usepackage{cmbright}\Huge\$ will cause the denominator to be 0.