Solving Intermediate-Difficulty Systems of Non-Linear Equations

As before, I'll set these equations equal, and solve for the values of x:

Using the Quadratic Formula:

But I can't graph that negative inside the square root! What's going on here?

Lemme take a quick look at the graph:

The lines do not intersect; *that's* what's going on! Since there is no intersection on the graph, then there is no solution to the system. That is, this is an inconsistent system. My final answer is:

−y = +x² + 3x

+y = −x² − 7x + 5

0 = −4x + 5

x = 5/4 = 1.25

Graphically, this system is a straight line (from the first equation) crossing a circle centered at the origin (from the second equation):

There appear to be two solutions. I'll proceed algebraically to confirm this impression, and to get the exact values.

Since the first equation is already solved for y, I will plug "−x − 3" in for "y" in the second equation, and solve for the values of x:

When x = −4, then:

When x = 1, then:

Then my solution consists of the following two points:

I solved one of the equations (the first equation looked easier) for one of the variables (solving for "y=" looked easier), and then plugged the resulting expression back into the other equation. This gave me one equation in one variable (the variable happened to be x) — and a one-variable equation is something I know how to solve.

Once I had the solution values for x, I back-solved for the corresponding y-values. I emphasize "corresponding" because you have to keep track of which y-value goes with which x-value.

In the example above, the points (−4, −4) and (1, 1) were not solutions. Even though I came up with x = −4 and x = 1 and y = −4 and y = 1, the x = −4 did not go with the y = −4, and the x = 1 did not go with the y = 1.

Since both equations are already solved for y, I'll set them equal and solve for the values of x:

When , then the corresponding y-value is:

And when x = −4, then the corresponding y-value is:

Then my solutions for this system are at the following two points:

Taking a quick look at these equations, I see something new: the variables are multiplied together in the first equation. This might lead to some extra complexity in the algebra for solving this system.

Taking a quick look at the graph, I see that there appears to be only one solution:

Hmm... I guess I'll solve the second equation for y=, and plug the result into the first equation in place of the y:

Plugging the expression for the right-hand side above into the first of the original equations in place of the y, I get:

(Sneaking a glance at the graphed system, above, I see that the x-value for my solution appears to match the one marked intersection point. Good.)

Plugging this x-value into the second equation, I get:

(Checking the graph above, I see that this value also appears to match the marked point. Excellent!)

Then my system solution is a single point: