## Intermediate Algebra (12th Edition)

We need at least two points to graph a line. To find the point $(3,y)$, we plug in $x=3$ and solve for $y$: $2x-3y=0$ $2*3-3y=0$ $6-3y=0$ $3y=6$ $y=2$ Similarly, to find the point $(x,4)$, we plug in $y=4$ and solve for $x$: $2x-3y=0$ $2x-3*4=0$ $2x=12$ $x=6$ Thus the points $(3,2)$ and $(6,4)$ lie on the graph of $2x-3y=0$.