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Algebra

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solving linear equations with 3 variables

1) x+y=1300
2) x+2z=1400
3) x+y+z=1600

this one should be so easy but i don't get which equation i should pick first?!

  • Algebra -

    It does not matter which you pick.
    However since there is one times x in each it is easy to eliminate x.

    x + y = 1300
    x + 2z = 1400
    -------------- subtract
    y - 2 z = -100

    then do the first and the third for example
    x + y = 1300
    x + y + z = 1600
    ---------------------
    subtract
    -z = -300
    or z = 300
    That was just lucky that there is no y term
    If there were we would use the two equations left with y and z to get rid of one or the other.
    now with z = 300 go back and use the one in x and z to get x
    x + 2(300) = 1400
    I think you can take it from there.

  • Algebra -

    You subtract? arent you supposed to add?

  • Algebra -

    A much easier way to do this would be:
    1) x+y=1300
    2) x+2z=1400
    3) x+y+z=1600
    Look at 1 and 2. Since x+y=1300 and x+y+z=1600, z obviously equals 300. Now substitute that back in #2.
    x+2(300)=1400, x+600=1400, x=800.
    Now substitute x and z back in #3.
    x+y+z=1600, 800+y+300=1600, 1100+y=1600, y=500

  • Algebra -

    THanks Damon and Jen! (especially Jen ;)

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