Answer:
[tex]\large\boxed{y=\dfrac{32}{9}}[/tex]
Step-by-step explanation:
[tex]\underline{+\left\{\begin{array}{ccc}x+3y=1\\2x-3y=-30\end{array}\right}\qquad\text{add both sides of the equations}\\.\qquad3x=-29\qquad\text{divide both sides by 3}\\.\qquad x=-\dfrac{29}{3}\\\\\text{put the value of x to the first equation}\\\\-\dfrac{29}{3}+3y=1\qquad\text{multiply both sides by 3}\\\\3\!\!\!\!\diagup^1\cdot\left(-\dfrac{29}{3\!\!\!\!\diagup_1}\right)+(3)(3y)=(3)(1)\\\\-29+9y=3\qquad\text{add 29 to both sides}\\\\9y=32\qquad\text{divide both sides by 9}\\\\y=\dfrac{32}{9}[/tex]
