Posted by Anneliese on .
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,
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,
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