Because $\displaystyle{{x}}^{{2}}+{x}-{12}={\left({x}+{4}\right)}{\left({x}-{3}\right)}\usepackage{cmbright}\Huge\$ , this is the least common denominator.

$\displaystyle\frac{{{3}{x}+{2}}}{{{x}+{4}}}+\frac{{{x}-{1}}}{{{{x}}^{{2}}+{x}-{12}}}=\frac{{{\left({3}{x}+{2}\right)}{\left({x}-{3}\right)}}}{{{\left({x}+{4}\right)}{\left({x}-{3}\right)}}}+\frac{{{x}-{1}}}{{{\left({x}+{4}\right)}{\left({x}-{3}\right)}}}\usepackage{cmbright}\Huge\$

$\displaystyle=\frac{{{3}{{x}}^{{2}}-{7}{x}-{6}}}{{{\left({x}+{4}\right)}{\left({x}-{3}\right)}}}+\frac{{{x}-{1}}}{{{\left({x}+{4}\right)}{\left({x}-{3}\right)}}}=\frac{{{3}{{x}}^{{2}}-{6}{x}-{7}}}{{{{x}}^{{2}}+{x}-{12}}}\usepackage{cmbright}\Huge\$