algebra

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Could you show me how to do these math systems and show your work so i understand how to do them.
1)
3x+2y+4z=19
2x-y+z=3
6x+7y-z=17

2)
5x+8y+6z=14
2x+5y+z=12
-3x+6y-12z=20

3)
x^2+2xy-2y^2=-11
x^2+xy-2y^2=-9

  • algebra -

    If i remember you just take two out of the three and solve using elimination. Then plug in your found variables to the third equation and youll have your variables! Its a lot of work per problem though

  • algebra -

    look over the equations and decide which variable would be the "easier" one to eliminate.
    In the first I would choose the z

    add the last two equations:
    8x + 6y = 0 = 20
    reduce to
    4x + 3y = 10

    2nd times 4 :
    8x - 4y + 4z = 12
    3x + 2y + 4z = 19 , now subtract those two
    ----------------
    5x - 6y + 0 = -7

    now look at the y's, they would be the easier to get rid of ...

    double the first: 8x + 6y = 20
    keep the 2nd ...: 5x - 6y = -7
    add them
    13x = 13
    x = 1
    sub into 4x+3y=10
    4 + 3y = 10
    3y = 6
    y = 2

    back into the 2nd, the easier looking one ...
    2x - y + z = 3
    2 - 2 + z = 3
    z = 3

    x=1
    y=2
    z=3

    try the second the same way, I would choose to work on the z's to start with

    The last one is a lot different:
    notice that the x^2 and y^2 terms are the same.
    So, let's just subtract them ....
    xy = -2
    or
    y = -2/x
    sub that into the first:
    x^2 + 2x(-2/x) - 2(4/x^2) = -11
    x^2 - 4 - 8/x^2 = -11
    x^2 - 8/x^2 = -7
    times x^2

    x^4 - 8 = -7x^2
    x^4 + 7x^2 - 8 = 0
    (x^2 + 8)(x^2 - 1) = 0

    x^2 = 8 = 0 ----> x^2 = -8 ------> no real solution
    (or x = ± 2i√2 for complex numbers)

    or

    x^2 - 1 = 0
    x^2 = 1
    x = ± 1
    IF x= +1 , then y = -2/1 = -2 ----- (1 , -2)
    IF x = -1 , then y = -2/-1 = +2 --- (-1, 2)

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