Respuesta :
[tex]\huge\boxed{\left \{ {{x+y=3} \atop {3x-y=5}} \right.}[/tex]
We can solve this problem by testing the solution against each system of equations. We'll do this by substituting the point into each equation for each system.
System 1 - [tex]\Large\textbf{X}[/tex]
[tex]\begin{array}{c|c}\textbf{First Equation}&\textbf{Second Equation}\\\cline{1-2}\begin{aligned}2x+y&=5\\2(2)+1&=5\\4+1&=5\\5&=5\\&\checkmark\end{aligned}&\begin{aligned}-2x+y&=2\\-2(2)+1&=2\\-4+1&=2\\-3&=2\\&\text{X}\end{aligned}\end{array}[/tex]
System 2 - [tex]\Large\textbf{X}[/tex]
[tex]\begin{array}{c|c}\textbf{First Equation}&\textbf{Second Equation}\\\cline{1-2}\begin{aligned}-2x-y&=-5\\-2(2)-1&=-5\\-4-1&=-5\\-5&=-5\\&\checkmark\end{aligned}&\begin{aligned}2x-y&=2\\2(2)-1&=2\\4-1&=2\\3&=2\\&\text{X}\end{aligned}\end{array}[/tex]
System 3 - [tex]\Large\textbf{X}[/tex]
[tex]\begin{array}{c|c}\textbf{First Equation}&\textbf{Second Equation}\\\cline{1-2}\begin{aligned}-x+y&=3\\-2+1&=3\\-1&=3\\&\text{X}\end{aligned}&\begin{aligned}-3x+y&=-5\\-3(2)+1&=-5\\-6+1&=-5\\-5&=-5\\&\checkmark\end{aligned}\end{array}[/tex]
System 4 - [tex]\Large\checkmark[/tex]
[tex]\begin{array}{c|c}\textbf{First Equation}&\textbf{Second Equation}\\\cline{1-2}\begin{aligned}x+y&=3\\2+1&=3\\3&=3\\&\checkmark\end{aligned}&\begin{aligned}3x-y&=5\\3(2)-1&=5\\6-1&=5\\5&=5\\&\checkmark\end{aligned}\end{array}[/tex]
Since both equations in the final system are true when solved with [tex](2, 1)[/tex], the answer is the last system.