uzman1243
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Solving Equations with Gaussian Elimination: Troubleshooting Mistakes
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uzman1243 said:Homework Statement
When solving 2 equations using gaussian elimination, can we divide one equation by the other?
Can you help me find where I went wrong?
Homework Equations
4x + 2y = 14
2x-y=1
The Attempt at a Solution
The answer to your question is "Never."
For simple row operations, you are permitted to:
1. Swap two rows with one another.
2. Multiply a row by a non-zero constant.
3. Add one row multiplied by a non-zero constant to another row.
http://en.wikipedia.org/wiki/Gaussian_elimination
And when multiplying a row by a non-zero constant, you must multiply all of the terms in that row (including the right-hand side) by the same non-zero constant.
It appears you have several mistakes in your attached elimination exercise. I would suggest that
you start over from the beginning.
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uzman1243 said:Homework Statement
When solving 2 equations using gaussian elimination, can we divide one equation by the other?
Can you help me find where I went wrong?
Homework Equations
4x + 2y = 14
2x-y=1
The Attempt at a Solution
No, you cannot divide equations like that. You can add (or subtract) multiples of one equation onto another, or you can multiply all terms in an equation by a common constant.
So, in the above, you can multiply the second equation by 2 to get 4x - 2y = 2 and then subtract that from the first equation, to get 4x - 4x + 2y + 2y = 14 - 2 --> 4y = 12. Or you can add twice the second equation to the first to get 4x + 2y + 4x - 2y = 14 + 2 --> 8x = 16.
However, fundamentally, what Gaussian elimination is really doing (although it is not always presented as such) is to use one equation to solve for some variable in terms of the others, then to substitute that expression into the other equations, giving a smaller set of equations in which one of the variables has been eliminated. So, we could solve for y from the second equation: y = 2x - 1. Now substitute that into the first equation: 4x + 2(2x-1) = 14, or 8x = 16, as before.