Algebra II

Solve the system by substitution.

{-x - y - z = -8

{ -4x + 4y + 5z = 7

{ 2x + 2z = 4

Solve the system by elimination.

{-2x + 2y + 3z = 0

{-2x - y + z = -3

{2x+ 3y+ 3z = 5

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  1. Using the first equation, -x=y+z-8
    and substituting that into the others, you have
    4(y+z-8)+4y+5z = 7
    -2(y+z-8)+2z = 4
    or,
    8y+9z = 39
    -2y = 12
    and the rest is easy

    For the elimination, start by subtracting row1 from row2, and adding row1 to row3. That gives you

    -2x+2y+3z = 0
    0 -3y-2z = -3
    0 +5y+6z = 5
    Now work with row 2 to get rid of the y in row3, and then you have z, and you can then find y and x.

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  2. So what the answer? I’m still confused. How would I do the full equations?

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  3. Steve showed you how to do it with elimination. You got z = -6. Now go back and get x and y
    for the second one try the Gauss/Jordan link I gave you for the procedure.
    http://www.gregthatcher.com/Mathematics/GaussJordan.aspx

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  4. I mean he used substitution first. That gave z = -6. Then the link will help with elimination.

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  5. I got 6 from that first part not -6.. and I'm confused on what to do after you get the value of z not before ?? Can yall actually help or no cus so far u just confused my g right here and gave us the wrong answer AND didn't actually help with the rest of it

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