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Systems of Non-Linear Equations:
     Solving Simple Systems (page 3 of 6)

To find the exact solution to a system of equations, you must use algebra. Let's look at that first system again:

• Solve the following system algebraically:

y = x²
y = 8 – x²

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:

y = x² = 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:

y = x²
y = 8 – x²
y = y
x² = 8 – x²

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

x² = 8 – x²

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

x² = 8 – x²
2x² = 8
x² = 4
x = –2, +2

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:

x = –2:

y = x²
y = (–2)² = 4

x = +2:

y = x²
y = (+2)² = 4

Then the solutions (as we already knew) are (x, y) = (–2, 4) and (2, 4).

In this case, the solutions were "neat" values; no fractions or decimals. But solutions will not always be neat, so, while the pictures can be very useful for giving you a "feel" for what is going on, graphing is not as accurate as doing the algebra. Warning: Students are often taught nowadays to "round" absolutely everything, and are thus implicitly taught that all answers will be "neat" answers. But this is wrong; don't fall for it. For instance:

• Solve the following system:

y = x² + 3x + 2
y = 2x + 3

I can solve this in the same manner as we 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.  Copyright © 2006-2009 Elizabeth Stapel All Rights Reserved

Looking at the 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.
I'll set the equations equal, and solve:		x2 + 3x + 2 = 2x + 3 x2 + x – 1 = 0
Using the Quadratic Formula gives me:
Then I have one solution:
...which has a corresponding  y-value of:
The other solution (from the "±" in front of the square root) is:
....which gives me a y-value of:
So the solutions are:

For purposes of graphing, the approximate solutions are:

(x, y) = (–1.62, –0.24) and (0.62, 4.24).

In other words, while our guess from the picture was close, it was not entirely correct. (However, if the algebra had given me answers that are far afield of these picture-based guesses, I would have been able to safely assume that I had messed up the math somewhere. In this way, the graph can be helpful for checking your work.)

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