## "Comparasion Method"

Quadratic equations and inequalities, variation equations, function notation, systems of equations, etc.
SplitWindowVette
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### "Comparasion Method"

I know how to solve systems of equations by graphing and substituting, but not with the comparision method, can someone help?

stapel_eliz
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I've never heard of the "comparison" method. How does that work? Can you give an example from your book, or maybe a link? (All I could find is a PDF worksheet that makes it look like "comparison" is just a form of "substitution", where both equations are already solved for one of the variables.)

Thank you!

SplitWindowVette
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Joined: Wed Dec 02, 2009 7:34 pm
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### Re: Re:

I've never heard of the "comparison" method. How does that work? Can you give an example from your book, or maybe a link? (All I could find is a PDF worksheet that makes it look like "comparison" is just a form of "substitution", where both equations are already solved for one of the variables.)

Thank you!
This is the text from my book:
"Another way to eliminate a variable from a system is sometimes referred to as the comparison method. Solve for the same variable in each equation; then form one equation by setting those two expressions equal to each other. The idea is that if two quantities are equal to the same thing, then they must equal each other.

We will solve the same system that we just did but use this method so that you can compare the two solutions. Both are correct solutions; the different method just gives you more options when you wish to solve a system algebraically.

Example:
Solve the following system of equations by the comparison method.

2x + 6y + 3 = 0
x - 4y - 9 = 0

Solution:
Make y the subject of each equation; then solve each equation:

2x + 6y + 3 = 0
6y = -2y - 3
$y = \frac{-2x - 3}{6}$

x - 4y - 9 = 0
-4y = -x + 9
$y = \frac{-x + 9}{-4}$
Combine the two results into a single equation of the results and find the value of x:
$\frac{-2x - 3}{6} = \frac{-x + 9}{-4}$

$8x + 12 = -6x + 54$

$14x + 12 = 54$

$14x = 42$

$x = \frac{42}{14} or 3$

Substitute 3 in for x into one of the original equations and solve for y:
2(3) + 6y + 3 = 0
6 + 6y + 3 = 0
6y + 9 = 0
6y = -9
y = - $\frac{3}{2}$

The checks were shown above."

Can you explain that in simpler terms?

stapel_eliz
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Okay, so your book is using the same definition as the document at the link. It's substitution, but you solve for "y=" or "x=" in both of the equations, instead of just one. Then you set the other side of the two equations equal to each other.

In other words, instead of getting "y=" for one equation and plugging that in for "y" in the other, you get "y=" in both equations, and then plug the other side of the "y=" from the one equation in for "y" in the other.

It's a form of substitution, and it can generate messy fractions more easily than will "regular" substitution.

SplitWindowVette
Posts: 13
Joined: Wed Dec 02, 2009 7:34 pm
Contact:

### Re: Re:

Okay, so your book is using the same definition as the document at the link. It's substitution, but you solve for "y=" or "x=" in both of the equations, instead of just one. Then you set the other side of the two equations equal to each other.

In other words, instead of getting "y=" for one equation and plugging that in for "y" in the other, you get "y=" in both equations, and then plug the other side of the "y=" from the one equation in for "y" in the other.

It's a form of substitution, and it can generate messy fractions more easily than will "regular" substitution.
Thanks, I thought that regular substitution was alot easier, but my book told me to use the comparision method for some of the problems.