From the second equation, \(\displaystyle{2}{x}={3}{y}-{1}\), so \(\displaystyle{x}=\frac{{{3}{y}-{1}}}{{2}}.\)

Substitute for x in the first equation:

\(\displaystyle{3}\frac{{{3}{y}-{1}}}{{2}}+{4}{y}={7}\), multiply each side by 2:

\(\displaystyle{3}{\left({3}{y}-{1}\right)}+{8}{y}={14},{9}{y}-{3}+{8}{y}={14},{17}{y}={17},{y}={1},\) therefore \(\displaystyle{x}=\frac{{{3}-{1}}}{{2}}=\frac{{2}}{{2}}={1}\).

Solution: x=1, y=1.

Substitute for x in the first equation:

\(\displaystyle{3}\frac{{{3}{y}-{1}}}{{2}}+{4}{y}={7}\), multiply each side by 2:

\(\displaystyle{3}{\left({3}{y}-{1}\right)}+{8}{y}={14},{9}{y}-{3}+{8}{y}={14},{17}{y}={17},{y}={1},\) therefore \(\displaystyle{x}=\frac{{{3}-{1}}}{{2}}=\frac{{2}}{{2}}={1}\).

Solution: x=1, y=1.