Posted by Anneliese on Wednesday, February 25, 2009 at 5:19pm.
Find all solutions to the following system of equations:
x + 3y + 2z = 6
3x + y + 5z = 29
2x  3y + z = 14
This just did not work out for me. Here is my work:
I first solved Equation #1 for x.
x = 3y  2z + 6
Then, I did substitution, and replaced x for 3y  2z + 6 in equations #2 and 3.
3(3y  2z +6) + y + 5z = 29
2(3y  2z +6)  3y + z = 14
That is the same as:
10y + 11z = 47
and
3y + 5z = 26
Would elimination be the next step?
Thanks, Liese

Algebra  Reiny, Wednesday, February 25, 2009 at 5:25pm
why not use elimination right from the beginning
look at the y's
If you add the first and the last, they are gone
If you add 3times the second to the last, they are gone.
now you have 2 equations in x and z
not bad....

Question  Anneliese, Wednesday, February 25, 2009 at 5:37pm
I finished it up like I said with elimination and got y = 3. Is this correct?
If not, I'll try it the way you suggested with elimination the whole way through.
Thanks, Liese
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