Question

I need help with system A. I have the other answer already

Answer 1

`\begin{gathered} \text{Add the two systems together} \\ x-3y=3 \\ -x-3y=3 \\ \\ \text{The }x\text{ cancels out leaving us with} \\ -6y=6 \\ \\ \text{Divide both sides by }-6 \\ \frac{\cancel{-6}y}{\cancel{-6}}=(6)/(-6) \\ y=-1 \end{gathered}`

Substitute y = -1 to either one of the two systems

`\begin{gathered} x-3y=3 \\ x-3(-1)=3 \\ x+3=3 \\ x=3-3 \\ x=0 \end{gathered}`

Therefore, the solution (x,y) is ( 0 , -1 ).