Solving Simple Systems of Nonlinear Equations

Since I am looking for the intersection points, I am therefore looking for the points where the equations overlap, where they share the same values. That is, I am trying to find any spots where y = x² equals y = 8 − x²:

The algebra comes in when I manipulate useful bits of this last equation. I can pick out whichever parts I like. (They're all equal, after all — at least at the intersection points, but the intersection points are the only points that I care about anyway!) So I can pick out any of the following:

Each of these sub-equations is true, but only the last one is usefully new and different:

I can solve this for the x-values that make the equation true:

Then the solutions to the original system will occur when x = −2 and when x = +2.

What are the corresponding y-values? To find them, I plug the x-values back in to either of the two original equations. (It doesn't matter which one I pick because I only care about the points where the equations spit out the same values. So I can pick whichever equation I like better.)

I'll plug the x-values into the first equation, because it's the simpler of the two:

Then the solutions (as we already knew, from looking at the graph earlier) are at:

I can solve this in the same manner as I did on the previous problem. The solution to the system will be any point(s) that the lines share; that is, any point(s) where the x-value and corresponding y-value for y = x² + 3x + 2 is the same as the x-value and corresponding y-value for y = 2x + 3; that is, where the lines overlap or intersect; that is, where y = x² + 3x + 2 equals y = 2x + 3.

Looking at a quick graph of the system:

...I can see that there appear to be solutions at around (x, y) = (−1.5, −0.25) and (x, y) = (0.5, 4.25). But I cannot assume that this is the answer! The picture can give me a good idea, but only the algebra can give me the actual answer.

(If my algebra leads me to an answer that is nowhere close to the intersections on the graph, then I know that I need to check my work, and check what I plugged into the graphing calculator.)

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I'll set the right-hand sides of the two equations equal to each other, and solve:

Using the Quadratic Formula gives me:

Then I have one solution (from the "minus;" in the "±" in front of the radical) that is:

...which has a corresponding y-value of:

The other solution (from the "plus" in the "±" in front of the square root) is:

....which gives me a y-value of:

So the solutions are:

(x, y) = (−1.62, −0.24), (0.62, 4.24)