Algebra

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1. I need help solving the system by substitution
-x-y-z=-8
-4x+4y+5z=7
2x+2z=4

  • Algebra -

    get rid of y so you can use the third eqn

    y = 8-x-z
    so
    -4x + 4(8-x-z) + 5z = 7
    -8 x + z + 32 = 7
    -8x + z = -25 and we know
    +2x +2z = 4

    -16x +2z =-50
    + 2x +2z = 4
    ---------------
    -18 x = -54
    x = 3
    now go back

  • Algebra -

    let's clean them up a bit first
    -x-y-z=-8 ---> x + y + z = 8 , #1
    2x+2z=4 ----> x + z = 2 , #3

    how about subtracting those two:
    y = 6 , well that is a good start

    let's put that into #1
    x + 6 + z = 8
    x + z = 2 , but that is what the original #3 said, so we have nothing new.

    but let's change that to z = 2-x and put that data into #2

    -4x + 24 + 5(2-x) = 7
    -4x + 24 + 10 - 5x = 7
    -9x = -27
    x = 3
    then z = 2-3 = -1

    x = 3, y = 6, z = -1

  • Algebra -

    well, we both used elimination at one point or another, going back a bit we had

    -8x + z = -25 and we know
    +2x +2z = 4

    so use z = 8x -25 to substitute in the last one
    2x +2(8x-25) = 4
    18 x -50 = 4
    18 x = 54
    x = 3 again, whew !

  • Algebra -

    You said to use substitution but there is no practical reason not to use elimination and I did so automatically for the second half of the problem, sorry.

  • Algebra -

    Thank you very much Reiny for showing me the proper steps. and Thank you too Damon for trying to help me.

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