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anonymous

  • 5 years ago

Can someone please help me. I need help "solving the system" 2x+y+z=9 -x-y+z=1 3x-y+z=9

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  1. anonymous
    • 5 years ago
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    Start off by adding line one and line two; and now you have at least the y disappearing.

  2. anonymous
    • 5 years ago
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    I did that. I did it with 1st and 3rd to get 5x+2z=9

  3. anonymous
    • 5 years ago
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    Then I went and did it to the 1st and 2nd. Giving me x+2z=10

  4. anonymous
    • 5 years ago
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    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

  5. anonymous
    • 5 years ago
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    ?? Thanks. sorry, but you lost me.

  6. anonymous
    • 5 years ago
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    Its not what our teacher said to do, and I know I will be required to show my work.

  7. anonymous
    • 5 years ago
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    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.

  8. anonymous
    • 5 years ago
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    I came up with (2,1,4) and it fits into the equation. Am I right?

  9. anonymous
    • 5 years ago
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    Let's test it with the second equation -x-y+z=1 -2-1+4=1 1=1 You are exactly right

  10. anonymous
    • 5 years ago
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    :)) Yay!!!!!! Thanks!

  11. anonymous
    • 5 years ago
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    You can also test them with the others to make sure it wasn't a fluke, but you are doing great.

  12. anonymous
    • 5 years ago
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    Now, if I can get the next one...lol!

  13. anonymous
    • 5 years ago
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    UGH... This one isnt working out right... Can I explain (the best I know how) what I am doing????

  14. anonymous
    • 5 years ago
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    Sorry I had signed out to watch basketball. Hope you figured it out.

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