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March 4, 2015

March 4, 2015

Posted by **Anneliese** on Wednesday, February 25, 2009 at 5:19pm.

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 5:25pmwhy 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 5:37pmI 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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