When we have a system of linear equations where one of the equations is of the form $y = \text{[stuff]}$ or $x=\text{[stuff]}$, we can solve it algebraically by using a technique called substitution. The basic idea is to replace a variable with an expression that it is equal to (so the expression is like a substitute for the variable). For example, let's start with the system:

$\displaystyle \begin{cases} y = 5x\\2x - y = 9 \end{cases}$

Because we know that $y = 5x$, we can substitute $5x$ for $y$ in the equation $2x - y = 9$,

$\displaystyle 2x - (5x) = 9$

and then solve the equation for $x$,

$\displaystyle x =\text -3$

We can find $y$ using either equation. Using the first one, $y = 5 \boldcdot \text-3$.
So $(\text-3,\text -15)$ is the solution to this system.

We can verify this by looking at the graphs of the equations in the system:

Sure enough! They intersect at $(\text-3, \text-15)$.

<p>Graph of two lines, origin O, with grid. Horizontal axis, x, scale negative 10 to 10, by 2’s. Vertical axis, y, scale negative 30 to 30, by 10’s. One line is labeled as y equals 5 x. Another line is labeled as 2 x minus y equals 9. The lines intersect at negative 3 comma negative 15. </p>  

We didn't know it at the time, but we were actually using substitution in the last lesson as well. In that lesson, we looked at the system

$\begin{cases} y = 2x + 6 \\ y  = \text-3x - 4 \end{cases}$

We substituted $2x+6$ for $y$ into the second equation to get $2x+6=\text-3x-4$. Go back and check for yourself!