Let’s say we want to solve the following system of equations using substitution:
\begin{equation*}
\begin{cases}y \amp = 2x-1\\ 3 \amp= y+x\end{cases}
\end{equation*}
From the first equation, we can see that \(y\) is equal to (so it’s the same as) \(2x-1\text{.}\) Since that’s the same as the \(y\) that’s in the second equation, we can plug in \(2x-1\) in for that \(y\) and then solve the equation we have left:
\begin{align*}
3 \amp= y+x\\
3 \amp = (2x-1) + x\\
3 \amp=3x-1\\
4 \amp= 3x\\
\frac{4}{3} \amp= x
\end{align*}
Now that we figured out what \(x\) is, don’t forget that a solution is a point, so we still need to figure out what \(y\) is. To do that, we plug our \(x=\frac{4}{3}\) into either of the starting equations and solve for \(y\text{:}\)
\begin{equation*}
y = 2x-1 = 2\left(\frac{4}{3}\right)-1 = \frac{8}{3} - 1 = \frac{8}{3} - \frac{3}{3} = \frac{5}{3}
\end{equation*}