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June 26th, 2017, 08:33 PM   #1
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Help with Solving a System of Equations with a Unique Solution

I'm a bit confused by the Echelon Method. The directions in my example say to solve a system of equations with a unique solution. From the reading I gather that a unique solution is one in which the line of two graphs intersect at one point. The two equations listed are:

3x + 10y = 115

11x + 4y = 95

The solution states:

We first use transformation 3 to eliminate the x-term from the equation (2). We multiply equation (1) by 11 and add the results to -3 times equation (2)

Transformation 3 says we replace any equation by a nonzero multiple of that equation plus a nonzero multiple of any equation. What exactly does that mean?

The solution continues algebraically:

11(3x + 10y) = 11 * 115 --> 33x + 110y = 1265

-3(11x + 4y) = -3 * 95 --> . -33x - 12y = -285

__________________

98y = 980

What just happened? I see the 33 and -33 cancel each other out and we get 33 in the first equation by 11*3 and -33 in the second equation by -3*11, but why did we have to multiply by 11 and -3 in the first place?

In the second part of solving this problem the solution states that the first equation remains unchanged and the new system becomes:

3x + 10y = 115

98y = 980

Why did the first equation remain unchanged?

Then it says to use transformation 2, which is "multiply both sides of an equation by any nonzero real number", to make the coefficient of the first term in each row equal to 1. Hence we must multiply equation 1 by 1/3 and equation 3 by 1/98.

The confusion is in transformation 2 it doesn't say anything about making the coefficient 1, so how do we know this? It states it in my solution but not in the transformation. Am I missing something?
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June 26th, 2017, 09:01 PM   #2
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FYI, this thread should be in the Algebra forum, not Abstract Algebra.

Quote:
Originally Posted by geekchick View Post
I'm a bit confused by the Echelon Method. The directions in my example say to solve a system of equations with a unique solution. From the reading I gather that a unique solution is one in which the line of two graphs intersect at one point. The two equations listed are:

3x + 10y = 115

11x + 4y = 95

The solution states:

We first use transformation 3 to eliminate the x-term from the equation (2). We multiply equation (1) by 11 and add the results to -3 times equation (2)

Transformation 3 says we replace any equation by a nonzero multiple of that equation plus a nonzero multiple of any equation. What exactly does that mean?

The solution continues algebraically:

11(3x + 10y) = 11 * 115 --> 33x + 110y = 1265

-3(11x + 4y) = -3 * 95 --> . -33x - 12y = -285

__________________

98y = 980

What just happened? I see the 33 and -33 cancel each other out and we get 33 in the first equation by 11*3 and -33 in the second equation by -3*11, but why did we have to multiply by 11 and -3 in the first place?
Because 33x and -33x cancel each other out, and we want them to cancel each other out - that allows us to remove the x-term.

Quote:
In the second part of solving this problem the solution states that the first equation remains unchanged and the new system becomes:

3x + 10y = 115

98y = 980

Why did the first equation remain unchanged?
Transformation 3 was applied only to the second equation with the 'help' of the first equation. It was not applied to the first equation.

Quote:
Then it says to use transformation 2, which is "multiply both sides of an equation by any nonzero real number", to make the coefficient of the first term in each row equal to 1. Hence we must multiply equation 1 by 1/3 and equation 3 by 1/98.

The confusion is in transformation 2 it doesn't say anything about making the coefficient 1, so how do we know this? It states it in my solution but not in the transformation. Am I missing something?
We know this because we want the coefficients to be 1 - this will help us find x and y. The statements of the transformations just tell you what a valid transformation is - that is, after you make the transformations, the resulting system of equations remains equivalent to the original - it doesn't tell you what a good transformation is.
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June 26th, 2017, 09:15 PM   #3
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Thank you. How do I move my thread to algebra?
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June 26th, 2017, 09:32 PM   #4
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Quote:
Originally Posted by geekchick View Post
Thank you. How do I move my thread to algebra?
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You don't have to - a mod will do it for you soon
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