We start with a system of 3 linear equations.
\[ \begin{align} 3x + 2y & = 1 \\ x - y & = 2 \\ 4x + 2y & = 2 \\ \end{align} \]
We’ll first subtract the first from the third row, to eliminate a \(y\) variable. This gives us:
\[ \begin{align} x - y & = 2 \\ x & = 1 \\ \end{align} \]
We can plug our newfound value of \(x\) into the other equation, which gives us:
\[ \begin{align} 1 - y & = 2 \\ \end{align} \]
We subtract 1 from both sides to get:
\[ \begin{align} - y & = 1 \\ \end{align} \]
And multiply both sides by \(-1\) to obtain our solution: \[ \begin{align} y & = -1 \\ \end{align} \]
So, the solution to the system of linear equations
\[ \begin{align} 3x + 2y & = 1 \\ x - y & = 2 \\ 4x + 2y & = 2 \\ \end{align} \]
equals:
\[ \begin{align} x & = 1 \\ y & = - 1 \end{align} \]
Which we can verify by plugging our answers back in to the original system:
\[ \begin{align} 3(1) + 2(-1) & = 3 - 2 = 1 \\ 1 - (-1) & = 1 + 1 = 2 \\ 4(1) + 2(-1) & = 4 -2 = 2 \\ \end{align} \]