SOLUTION: I've been trying to figure out how to do this problem but I can't seem to get it right. The problem is... Solve the system using any algebraic method {{{2x+y+2z=7}}} {{{2x-y+2z

Algebra ->  Linear-equations -> SOLUTION: I've been trying to figure out how to do this problem but I can't seem to get it right. The problem is... Solve the system using any algebraic method {{{2x+y+2z=7}}} {{{2x-y+2z      Log On


   

Question 53993This question is from textbook McDougal Littel Algebra 2
: I've been trying to figure out how to do this problem but I can't seem to get it right. The problem is...
Solve the system using any algebraic method
2x%2By%2B2z=7
2x-y%2B2z=1
5x%2By%2B5z=13
I tried the problem a couple of times but I still couldn't get the answer right.
This is what I tried:
2x%2By%2B2z=7
2x-y%2B2z=1
and that cancelled out the y value and left me with 4x+4z=8. Then I combined 2x-y%2B2z=1
5x%2By%2B5z=13
that cancelled out the y value again and left 7x+7z=14 and from there I got stuck. This question is from textbook McDougal Littel Algebra 2

Found 2 solutions by stanbon, checkley71:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Your procedure is good. You now have two
equations that say x+z=2
These equations are dependent. The system
of equations does not have a unique solution.
Cheers,
Stan H.

Answer by checkley71(8403) About Me  (Show Source):
You can put this solution on YOUR website!
2X+Y+2Z=7
2X-Y+2Z=1 A BETTER MOVE HERE IS TO SUBTRACT THE SECOND EQUATION FROM THE FIRST
-------------
2Y=6 OR Y=6/2 OR Y=3 2X+3+2Z=7 OR 2X+2Z=4 & 2X-3+2Z=1 OR 2X+2Z=4 OR X+Z=2
THEN X & Z CAN EQUAL 1 & 1 OR X & Z CAN EQUAL 3 & -1 OR X & Z CAN EQUAL
4 & -2 OR ANY OTHER COMBINATION THAT=2