Expert: Sherman D. - 10/11/2005

Question
Find/solve following system of linear equations,

3y_1 + 2y_2 +y_4 = 6
5y_1 - 2y_2 +2y_4 = 5
-2y_1 + y_2 - y_4 = -2



WORK DONE :

I am told that the answers are y_1 = 1 and y_2= 1 and y_4 = 1.
But i dont understand how to obtain these values....
I know how to solve 2 linear systerm of equations...but how do i solve
3 linear system of equations like the one above?

please help. thanks!  

Answer
I hope you don't mind, but to make it easier for me to work, i am going to do it this way

x = y_1
y = y_2
z = y_4

3x + 2y + z = 6
5x - 2y + 2z = 5
-2x + y - z = -2

5x - 2y + 2z = 5
-2x + y - z = -2

Multiply bottom by 2

5x - 2y + 2z = 5
-4x + 2y - 2z = -4

the 2y's and 2z's cancel out

x = 1

3x + 2y + z = 6
3(1) + 2y + z = 6
3 + 2y + z = 6
2y + z = 3
z = -2y + 3

5x - 2y + 2z = 5
5(1) - 2y + 2(-2y + 3) = 5
5 - 2y - 4y + 6 = 5
-6y + 11 = 5
-6y = -6
y = 1

z = -2y + 3
z = -2(1) + 3
z = -2 + 3
z = 1

x = 1
y = 1
z = 1

or in this case

y_1 = 1
y_2 = 1
y_4 = 1

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If you notice, if you try using

3y_1 + 2y_2 + y_4 = 6
5y_1 - 2y_2 + 2y_4 = 5
and get rid of the 2y_2's, This will give you the wrong answer, because i've tried it.

or

-3y_1 + 2y_2 + y_4 = 6
-2y_1 + y_2 - y_4 = 6
and get rid of the y_4's, This will give you the wrong answer also.

so you just have to go with

5y_1 - 2y_2 +2y_4 = 5
-2y_1 + y_2 - y_4 = -2

also when you get to y_4 = -2y_2 + 3 and try substituting that in to 3y_1 + 2y_2 + y_4 = 6, it won't work out, so you have to use 5y_1 - 2y_2 + 2y_4 = 5 again. So i only have one tip, whenever you get an answer to these types of problems, you will have to make sure to check your answer, or a quicker way is to go to http://alumnus.caltech.edu/~chamness/equation/equation.html
but it will just give you the answers not the work. When one thing doesn't give you the correct answer, you will just have to use other methods to coming to the correct answer.