How do you solve the system #x^2+y^2=9# and #x-3y=3#?

This is an interesting system of equations problem because it yields more than one solution per variable.

Why this happens is something we can analyze right now. The first equation, is the standard form for a circle with radius #3#. The second is a slightly messy equation for a line. Cleaned up, it would look like this:
#y = 1/3 x - 1#

So naturally if we consider that a solution to this system will be a point where the line and the circle intersect, we should not be surprised to learn that there will be two solutions. One when the line enters the circle, and another when it leaves. See this graph:
graph{(x^2 + y^2 - 9)((1/3)x -1-y)=0 [-10, 10, -5, 5]}

First we start by manipulating the second equation:
#x - 3y = 3#
#x = 3 + 3y#

We can insert this directly into the first equation to solve for #y#:
#x^2 + y^2 = 9#
#(3 + 3y)^2 + y^2 = 9#
#9 + 18y + 9y^2 + y^2 = 9#
#18y + 10y^2 = 0#
#y(9 + 5y) = 0#

Obviously this equation has two solutions. One for #y = 0# and another for #9+5y = 0# which means #y = -9/5#.

Now we can solve for the #x# at each of these #y# values.
If #y=0#:
#x - 3*0 = 3#
#x = 3#

If #y = -9/5#:
#x + 3*(9/5) = 3#
#x + 27/5 = 15/5#
#x = -12/5#

So our two solutions are the points: #(3,0)# and #(-12/5 , -9/5)#. If you look back to the graph, you can see that these clearly correspond to the two points at which the line crossed the circle.