C - Solving Systems of Non-Linear Equations Lesson

Solving Systems of Non-Linear Equations 

In previous courses, you learned how to solve systems of equations using graphing, substitution and elimination. In fact, in the last module, you used matrices to solve systems of equations. Recall, the solution(s) to a system of equations is(are) the set of values that make the equations true.  

Let's start by solving a system graphically:

$y=-3x\\ x^2+y^2=3$

Step One: Analyze these two functions - we have a line and a circle. Visualize the possible number of intersections on the coordinate plane.   Notice we could have no solutions, one solution, or two solutions.

            

Step Two: So, let's start by graphing each of the equations we've been given.  

     graph of  y=-3x                  graph of  x² + y² = 3

                 

Graph the system on the same coordinate plane and determine the points of intersection.

 

Step 3: Well, we can see that we have two points of intersection, but the coordinates are not easy to determine from the graph - so let's try solving algebraically.

Solve the system algebraically using the substitution method.

$y=-3x\\ x^2+y^2=3$

Step	Math
Substitute for y in the 2nd equation	$x^2+\left(-3x\right)^2=3$
Apply the Power Rule for Exponents	$x^2+9x^2=3$
Combine like terms	$10x^2=3$
Isolate x2	$x^2=\frac{3}{10}$
Take the square root on both sides and rationalize	$x=\pm\sqrt[]{\frac{3}{10}}\pm\frac{\sqrt[]{30}}{10}$
Substitute x back in to solve for y	$x=-3\left(\frac{\sqrt[]{30}}{10}\right)=\frac{-3\sqrt[]{30}}{10}\\ x=-3\left(-\frac{\sqrt[]{30}}{10}\right)=\frac{3\sqrt[]{30}}{10}$
So, our ordered pairs are	$\left(\frac{\sqrt[]{30}}{10},\:-\frac{3\sqrt[]{30}}{10}\right)\approx\left(0.55,\:-1.64\right)\\ \left(-\frac{\sqrt[]{30}}{10},\:\frac{3\sqrt[]{30}}{10}\right)\approx\left(-0.55,\:1.64\right)$
Looking back at our system of equations, we can see that those are the approximate solutions.

 

Watch these videos to try more examples.

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