First you have to solve one of the two equations for one variable. In the present case you can see that equation $\mathrm{I}\backslash mathrm\{I\}$ is already solved for $yy$.

$\backslash ;$Plug $y=3x+4y=3x+4$ from equation $\mathrm{I}\backslash mathrm\{I\}$ into $\mathrm{I}\mathrm{I}\backslash mathrm\{II\}$

$\mathrm{I}\mathrm{I}4\cdot (3x+4)-3x=9\backslash mathrm\{II\}\; \backslash quad\; 4\backslash cdot(3x\; +\; 4)\; -3x\; =\; 9$

Multiply out the parenthesis and solve for $xx$.

$\backslash ;$Plug $x=-\frac{7}{9}x=-\backslash frac\; 79$ from $\mathrm{I}\mathrm{I}\backslash mathrm\{II\}$ into $\mathrm{I}\backslash mathrm\{I\}$

And finally write down the solution set.

$\mathbb{L}=\left\{(-{\displaystyle \frac{7}{9}}\mid {\displaystyle \frac{5}{3}})\right\}\backslash mathbb\{L\}=\backslash left\backslash \{\backslash left(-\backslash dfrac\{7\}\{9\}\backslash left|\backslash dfrac\{5\}\{3\}\backslash right.\backslash right)\backslash right\backslash \}$.

First you have to solve one of the two equations for $ss$ or $tt$. It is, for instance, a good idea to solve equation $\mathrm{I}\mathrm{I}\backslash mathrm\{II\}$ for $tt$.

Solve equation for $tt$.

$\backslash ;$Plug $t=-2-4st=-2-4s$ from $\mathrm{I}\mathrm{I}\backslash mathrm\{II\}$ into $\mathrm{I}\backslash mathrm\{I\}$

$\phantom{\mathrm{I}}\mathrm{I}3s-4\cdot (-2-4s)=4\backslash hphantom\{\backslash mathrm\{I\}\}\backslash mathrm\{I\}\; \backslash quad\; 3s\; -\; 4\backslash cdot(-2\; -\; 4s)\; =\; 4$

Multiply out the parenthesis and solve for $ss$.

Substitute $s=-\frac{4}{19}s\; =\; -\backslash frac\{4\}\{19\}$ into the transformed second equation to determine $tt$.

And finally, write down the solution set.

$\mathbb{L}=\{(s\mathrm{\mid }t)=(-\frac{4}{19}\mathrm{\mid }-\frac{22}{19})\}\backslash mathbb\{L\}=\backslash left\backslash \{(s|t)=\backslash left(-\backslash frac\{4\}\{19\}|-\backslash frac\{22\}\{19\}\backslash right)\backslash right\backslash \}$.