Solve this system of equations.
\begin{align*}
x + y \amp = 1 \\
x + z \amp = 0 \\
y + z \amp = 2
\end{align*}
Solution 1.
\begin{align*}
x + y \amp = 1 \amp \amp \amp \amp\\
x \amp 1 - y \amp \amp \amp \amp\\
1 - y + z \amp = 0 \amp \amp \amp \amp\\
y \amp = 1 + z \amp \amp \amp \amp\\
1 + z + z \amp = 2 \amp \amp \amp \amp\\
2z \amp = 1 \amp \amp \amp \amp\\
z \amp = \frac{1}{2}
\amp y \amp = \frac{3}{2}
\amp x \amp = \frac{-1}{2}
\end{align*}
Solution 2.
I can solve for \(x\) in the first equation.
\begin{equation*}
x = 1 - y
\end{equation*}
Then I put this expression for \(x\) into the second equation.
\begin{equation*}
1 - y + z = 0
\end{equation*}
Now I can solve for \(y\) in this second equation.
\begin{equation*}
y = 1 + z
\end{equation*}
I can put this expression for \(y\) into the third equation.
\begin{equation*}
1 + z + z = 2
\end{equation*}
Now \(z\) is the only variable, so I can simply solve for it.
\begin{equation*}
z = \frac{1}{2}
\end{equation*}
I can use this value for \(z\) to determine \(y\) using a previous expression.
\begin{equation*}
y = 1 + z = \frac{3}{2}
\end{equation*}
Finally, I can use this value for \(y\) to determine \(x\) using the first equation.
\begin{equation*}
x = 1 - y = \frac{-1}{2}
\end{equation*}
That gives me the full solution.
\begin{align*}
\amp x = \frac{-1}{2} \amp \amp y = \frac{3}{2} \amp \amp
z = \frac{1}{2}
\end{align*}
