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Algebra 2

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Solving systems of linear equations in three variables.
Ok so i don't really totally get these problems. Can you help me solve this one? and explain everything please?

2x+3y=6+z
x-2y=-1-z
3x+y=-1+3z

  • Algebra 2 - ,

    put them in standard form (x,y,z)

    2x+3y-z=6
    x-2y+z=-1
    3x+y-3z=-1

    You can solve these a number of ways, I will show you one.
    add 1st and second equations
    3x+y=5
    now multiply equation 2 by three, then add equations 2 and 3.
    6x-5y=-4 check that.
    now on these two remaining equations, multiply the first by 2, then subtract.
    7y=14
    y=2 work back up to find x, then z.

  • Algebra 2 - ,

    Ok, this helps a lot, but hold on, i don't get how you got 7y=14? because when i multiply 3x+y=5 by (-2) i get -6x-2y=8. And then when combining that with 6x-5y=-4 i get -7y=4? how did you get 7y=14???

  • Algebra 2 - ,

    3x+y=5 multiply -2, your way
    6x-5y=-4 to get

    -6x-2y=-10
    6x-5y=-4 add to get

    -7y=-14
    y=2

  • Algebra 2 - ,

    Oh wait, i forgot to multiply the 5 by -2! (idk how i got 8, but watever its not that important) so right you would get y=2.
    ok i think i get it now! THANKS!

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