[tex]\left\{\begin{array}{ccc}\dfrac{s}{2}-3t=-1&\text{multiply both sides by 4}\\3t-2s=3\end{array}\right\\\underline{+\left\{\begin{array}{ccc}2s-12t=-4\\3t-2s=3\end{array}\right}\qquad\text{add both sides of the equations}\\.\qquad\qquad -9t=-1\qquad\text{divide both sides by (-9)}\\.\qqud\qquad\boxed{t=\dfrac{1}{9}}\\\\\text{Put the value of t to the second equation}\\\\3\left(\dfrac{1}{9}\right)-2s=3\\\\\dfrac{1}{3}-2s=3\qquad\text{subtract}\ \dfrac{1}{3}\ \text{from both sides}[/tex]
[tex]-2s=2\dfrac{2}{3}\qquad\text{divide both sides by (-2)}\\\\s=\dfrac{8}{3}\cdot\left(-\dfrac{1}{2}\right)\\\\\boxed{s=-\dfrac{4}{3}}\\\\Answer:\ \boxed{\left(-\dfrac{4}{3},\ \dfrac{1}{9}\right)}[/tex]
