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Start off by adding line one and line two; and now you have at least the y disappearing.
I did that. I did it with 1st and 3rd to get 5x+2z=9
Then I went and did it to the 1st and 2nd. Giving me x+2z=10
OK, you have infinitely many choices and your classmates may go about a different way. in x+2z=10 Let z=0. It is your equation you can do what ever you want. So when z=0 in your eq x=10. So x=10, z=0, punch those values in the third eq and solve for y
?? Thanks. sorry, but you lost me.
Its not what our teacher said to do, and I know I will be required to show my work.
The proof is you can plug in the values of x,y,z and see if it equals what is on the other side. Math has a lot of different methods. Tell me specifically what you don't understand.
I came up with (2,1,4) and it fits into the equation. Am I right?
Let's test it with the second equation -x-y+z=1 -2-1+4=1 1=1 You are exactly right
:)) Yay!!!!!! Thanks!
You can also test them with the others to make sure it wasn't a fluke, but you are doing great.
Now, if I can get the next one...lol!
UGH... This one isnt working out right... Can I explain (the best I know how) what I am doing????
Sorry I had signed out to watch basketball. Hope you figured it out.