Systems
of Non-Linear Equations: Intermediate-Difficulty
Systems(page
4 of 6)

The non-linear systems we've solved so far have been one quadratic equation and one linear equation, which graphed as a parabola and a straight line, respectively. Moving up in difficulty, we come to solving systems of two quadratic equations, which will graph as two parabolas; and similarly messy systems.

Solve the following
system:

y
= 2x^{2} + 3x + 4 y = x^{2}
+ 2x + 3

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

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

Take a look at
the graph:

The lines do not intersect.
Since there is no intersection, then there is no solution. That is,
this is an inconsistent system. My final answer is: no
solution: inconsistent system.

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In general, the method
of solution for
general systems of equations is to solve one of the equations (you choose
which) for one of the variables (again, you choose which). Then you plug
the resulting expression into the other
equation for the chosen variable, and solve for the values of the other
variable. Then you plug those solutions back into the first equation,
and solve for the values of the first variable. Here are some additional
examples:
Copyright 2002-2011 Elizabeth Stapel All Rights Reserved

Solve the following
system:

y
= x 3 x^{2} + y^{2} =
17

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:

Then the
solution consists of the points (4,
1) and (1,
4).

Note the procedure: 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 correspondingy-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) are not solutions.
Even though I came up with x
= 4 and 1
and y
= 4 and 1,
the x
= 4 did not go with
the y
= 4, and the x
= 1 did not go with
the y
= 1.

Warning: You must
match the x-values
and y-values
correctly. Be careful!

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Solve the following
system of equations:

y
= (^{1}/_{2})x 5 y = x^{2}
+ 2x 15

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

Then the solutions are
the points (
^{5}/_{2}, ^{15}/_{4} )
and (4, 7).

Graphically, the
above system looks like this:

The intersection points on the graph appear to be good matches for the numerical solutions I got via algebra, confirming that I've done the exercise correctly.

Solve the following
system of equations:

xy
= 1 x + y = 2

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

I guess I'll solve the
second equation for y,
and plug the result into the first equation:

Stapel, Elizabeth.
"Systems of Non-Linear Equations: Intermediate Systems."
Purplemath. Available from https://www.purplemath.com/modules/syseqgen4.htm.
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