# SOLUTION: x+2y-z=6 -x+y+3z=5 x+2y+2z=9 Im not sure what its called exactly what were working on, but you have to solve for x,y, and z in all of the equations.

Algebra ->  Algebra  -> Coordinate Systems and Linear Equations -> SOLUTION: x+2y-z=6 -x+y+3z=5 x+2y+2z=9 Im not sure what its called exactly what were working on, but you have to solve for x,y, and z in all of the equations.       Log On

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 Click here to see ALL problems on Linear-systems Question 526644: x+2y-z=6 -x+y+3z=5 x+2y+2z=9 Im not sure what its called exactly what were working on, but you have to solve for x,y, and z in all of the equations. Answer by KMST(1868)   (Show Source): You can put this solution on YOUR website!It's a system of linear equations. You have 3 pieces of information (the 3 equations), and 3 variables to be found. An easier system would look like a staircase, like this: x=1 -x+y=2 x+2y-z=6 That would be easier because the first equation gives you the value for one of the variable; plugging that value in the second equation will lead you to find the value for another variable, and substituting the values found in the third equation will allow you to find the remaining variable. You can transform your system by combining equations. We pick one of the equations that we will leave untouched, and we'll make combinations of the chosen one with the other two. -x+y+3z=5 will be kept Adding up x+2y-z=6 and -x+y+3z=5 yields 3y+2z=11 Adding up x+2y+2z=9 and -x+y+3z=5 yields 3y+5z=14 We will continue to work with the equivalent system below -x+y+3z=5 (which will be kept) 3y+2z=11 3y+5z=14 With the two last equations we will pick one to be kept untouched and make a combination of the two to replace the other Keeping 3y+5z=14, we multiply 3y+2z=11 times (-1) to get -3y-2z=-11 and add 3y+5z=14 yielding 3z=3 (which is the same as z=1). Now we have 3z=3 (or z=1) 3y+5z=14 -x+y+3z=5 Can you solve that one?