Exercises: systems with two variables

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

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

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

Multiply out the parenthesis and solve for $x$.

$\begin{array}{crcll}\mathrm{II}& 12x+16-3x& =& 9& |-16\\ & 9x& =& -7& |:9\\ & x& =& -\frac{7}{9}& \end{array}$

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

$\begin{array}{crcl}\mathrm{I}& 3\cdot (-{\displaystyle \frac{7}{9}})+4& =& y\\ & y& =& \frac{5}{3}\end{array}$

And finally write down the solution set.

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

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

Solve equation for $t$.

$\begin{array}{crcll}\mathrm{II}& 4s+t& =& -2& |-4s\\ & t& =& -2-4s& \end{array}$

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

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

Multiply out the parenthesis and solve for $s$.

$\begin{array}{crcll}\mathrm{I}& 3s+8+16s& =& 4& |-8\\ & 19s& =& -4& |:19\\ & s& =& -\frac{4}{19}& \end{array}$

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

$\begin{array}{crcl}\mathrm{II}& t& =& -2-4\cdot (-\frac{4}{19})\\ & t& =& -\frac{22}{19}\end{array}$

And finally, write down the solution set.

$𝕃=\{(s|t)=(-\frac{4}{19}|-\frac{22}{19})\}$.