How do find x= y= z= w=?

-2x - 1y + 2z + 2w = -15

-2x - 2y + 1z + 0w = -13

8 Answers

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  • 1 decade ago
    Favorite Answer

    You cannot find all of the variables. You need one equation for each variable, you have 2 equations and 4 variables.

    You could, at best, find each variable in terms of 2 of the other variables.

  • 1 decade ago

    -2x - 1y + 2z + 2w = -15

    -2x - 2y + 1z + 0w = -13

    There are too many variables and not enough equations to give a unique answer

    Here is a possible answer

    Let w = -1

    Now:

    -2x - 1y + 2z - 2 = -15

    -2x - 1y + 2z = -13 = -2x - 2y + 1z

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

    2z - y = z - 2y

    z = -y

    -2x - 2y + 1z + 0w = -13

    -2x + 2z + 1z + 0w = -13

    -2x + 3z + 0w = -13

    -2x - 1y + 2z + 2w = -15

    -2x + 1z + 2z + 2w = -15

    -2x + 3z + 2w = -15

    and as we decided w was -1 earlier

    -2x + 3z = -13

    So we could decide z was -5 and x was -1

    (-2 * -1) + (3 * -5) = -13

    and y = -z so y = 5

    x = -1

    y = 5

    z = -5

    w = -1

    Once again, this is only one possible solution

    Another possible one, from observation is:

    x = 4

    y = 1

    z = -3

    w = 0

    -2x - 1y + 2z + 2w = -15

    -8 - 1 - 6 + 0 = -15

    -2x - 2y + 1z + 0w = -13

    -8 - 2 - 3 + 0 = -13

    So there are at least two different, but correct combinations. So proof by contradiction that there is not a unique solution to this problem

  • DanE
    Lv 7
    1 decade ago

    -2x - 1y + 2z + 2w = -15

    substitute (because x = y = z = w, I can put an x where these variables were)

    -2x - 1x + 2x + 2x = -15

    simplify

    -3x + 4x = -15

    x = -15

  • Since there are four variables, there will be four answers. Use the transitive postulate for this.

    Have fun plugging and chugging!

    Source(s): C_3252.ew
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  • 1 decade ago

    ROTFLMSFAO!!!!!!!!!!!!!

    You really are totally lost. Quite wasting bandwidth and go back to 1'st year Algebra. And this time, *learn* something instead of just fumbling around and squeeking through the class.

    Doug

  • Anonymous
    1 decade ago

    [ matrix ]

    _ _ _ _

    | -2 -1 2 2 | ^(-1) X | -15 | = no possible answer

    |_-2 -2 1 0 _| |_-13 _|

  • Anonymous
    1 decade ago

    You cant solve that there are too many variables and not enought information

  • Anonymous
    1 decade ago

    -(-2x-y+2z+2w=-15)

    -2x-2y+z+0w=-13

    ==============

    2x+y-2z-2w=15

    -2x-2y+z+0w=-13

    ==============

    2(y-2z-2w=15)

    -2y+z+0w=-13

    ==============

    2y-4z-4w=30

    -2y+z+0w=-13

    ==============

    -4z-4w=30

    4(z+0w=-13)

    ==============

    -4z-4w=30

    4z+0w=-52

    =========

    -4w=-22

    w=22/4=5.5

    4z+0(5.5)=-52

    4z=-52

    z=-13

    -2x-y+2(-13)+2(5.5)=-15

    -2x-2y+(-13)+0(5.5)=-13

    ==================

    -2x-y-26+11=-15

    -2x-2y-13+0=-13

    ============

    -(-2x-y-15=-15)

    -2x-2y-13=-13

    ============

    2x+y+15=15

    -2x-2y-13=-13

    ============

    2x+y=0

    -2x-2y=0

    =======

    -y=0

    y=0

    -2x-0+2(-13)+2(5.5)=-15

    -2x-0-26+11=-15

    -2x-15=-15

    -2x=0

    x=0

    (0,0,-13,5.5)

    Check:

    -2(0)-1(0)+2(-13)+2(5.5)=-15

    -26+11=-15

    -15=-15

    -2(0)-2(0)+(-13)+0(5.5)=-13

    -13=-13

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