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

• Solve the following system:

y = 2x² + 3x + 4
y = x² + 2x + 3

As before, I'll set these equations equal, and solve for the values of x:		2x2 + 3x + 4 = x2 + 2x + 3 x2 + x + 1 = 0
Using the Quadratic Formula:
But I cannot 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.

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 into the first equation, and solve for the values of the first variable. Here are some additional examples:  Copyright © 2006-2008 Elizabeth Stapel All Rights Reserved

• Solve the following system:

y = –x – 3
x² + y² = 17

Graphically, this system is a straight line crossing a circle centered at the origin:

There appear to be two solutions. I'll proceed algebraically 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:

x² + y² = 17
x² + (–x – 3)² = 17
x² + (–x – 3)(–x – 3) = 17
x² + (x² + 6x + 9) = 17
2x² + 6x + 9 = 17
2x² + 6x – 8 = 0
x² + 3x – 4 = 0
(x + 4)(x – 1) = 0
x = –4, x = 1

When x = –4,  y = –x – 3 = –(–4) – 3 = 4 – 3 = 1

When x = 1,  y = –x – 3 = –(1) – 3 = –4

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

Note the procedure: I solved one of the equations (it was my choice which to solve) for one of the variables (again, this was a matter of choice), and then plugged the resulting expression back into the other equation. This gave me one equation in one variable, which is something we know how to solve. Once I had the 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, (–4, –4) and (1, 1) are not solutions. Even though I came up with x = –4 and 1 and y = –4 and 1, x = –4 did not go with y = –4, and x = 1 did not go with y = 1. You must match the x-values and y-values correctly!

• Solve the following system of equations:

y = (¹/₂)x – 5
y = x² + 2x – 15

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

(¹/₂)x – 5 = x² + 2x – 15
    x – 10 = 2x² + 4x – 30
            0 = 2x² + 3x – 20
            0 = (2x – 5)(x + 4)

x = ⁵/₂, x = –4

When x = ⁵/₂:

y = (¹/₂)x – 5 = (¹/₂)(⁵/₂) – 5 = ⁵/₄ – ²⁰/₄ = – ¹⁵/₄ = –3.75

When x = –4:

y = (¹/₂)x – 5 = (¹/₂)(–4) – 5 = –2 – 5 = –7

Then the solution is the points ( ⁵/₂, ^–15/₄ ) and (–4, –7).

Graphically, the above system looks like this:

• 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:

x + y = 2
y = –x + 2

Then:

xy = 1
x(–x + 2) = 1
–x² + 2x = 1
–x² + 2x – 1 = 0
x² – 2x + 1 = 0
(x – 1)(x – 1) = 0
x = 1

Then:

x + y = 2
(1) + y = 2
y = 1

Then the solution is the point (1, 1).

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