How to find the intersection of the line y=x+3 with the ellipse (x^2/9) + (y^2/4)=1?

We know that the intersection of the graphs consists of the common points of the graphs, these common points being found by solving the equations of the graphs, simultaneously.

In our case, we'll substitute in the equation of the ellipse, the unknown y, by the expression from the equation of the line.

(x^2/9) + [(x+3)^2/4]=1

The common denominator of the 2 ratios is 36, so we'll multiply the first ratio by 4 and the second, by 9.

4x^2 + 9(x+3)^2=36

4x^2 + 9x^2 + 54x + 81 - 36 = 0

13x^2 + 54x + 45 = 0

Now we'll use the formula of the quadratic equations for finding the solutions.

x1 = [-54+sqrt(54^2 - 4*13*45)]/2*13

x1 = (-54 + sqrt(2916 - 2340))/26

x1 = (-54 + sqrt(576))/26

x1 = (-54+24)/26

x1 = -30/26

x1 = -15/13, so y1 = x1+3 = -15/13 + 3 = 24/13

x2 = (-54-24)/26

x2 = -78/26

x2 = -3, so y2 = -3+3=0

So, the intersection points are:

(-15/13, 24/13) and  (-3,0)