System of equations problem?

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−x+y+z=2,4x−3y−z=−3, x+y+z=8

3 Answers

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  • 6 months ago

    ❶: −x+y=2-z, ❷: x+y=8-z, and ❸: z=3+4x−3y

    From ❶ and ❷:

    ❹: x = ((2-z)×1-1×(8-z))/(-1×1-1×1) = 3 (𝑥 𝑠𝑜𝑙𝑣𝑒𝑑)

    ❺: y = (-1×(8-z)-(2-z)×1)/(-1×1-1×1) = 5-z

    From ❸, ❹, and ❺:

    ❻: z = 3+4×3−3(5-z) = 3z = 0 (𝑧 𝑠𝑜𝑙𝑣𝑒𝑑)

    From ❺ and ❻:

    ❼: y = 5-0 = 5 (𝑦 𝑠𝑜𝑙𝑣𝑒𝑑)

    Solution: x=3, y=5, z=0

  • Philip
    Lv 6
    6 months ago

     -x  +y+ z =  2...(1).

    4x -3y - z = -3...(2).

      x  +y + z =  8...(3).

    [(1)+(3)]--->  y  + z = 5...(4).

    [(2)+4(1)]--> y +3z = 5...(5).

    [(5)-(4)]-----> 2z = 0, ie., z = 0...(6).

    [(4)-(6)]---> y = 5.

    (3)---> x = 8 - (y+z) = (8-5) = 3.

    Then (x,y,z) = (3,5,0).

  • Ash
    Lv 7
    6 months ago

    −x+y+z=2 ..................(1)

    4x−3y−z=−3  .............(2)

    x+y+z=8   ...................(3)

    Add (1) and (3)

    −x+y+z + x+y+z = 2+8

    2y+2z = 10

    y+z = 5 .............(4)

    Add (2) to 4 times (1)

    4x−3y−z + 4(−x+y+z) = -3 + 4(2)

    4x−3y−z -4x + 4y + 4z = 5

    y + 3z = 5 ...........(5)

    Subtract (4) from (5)

    y + 3z - (y+z) = 5 - 5

    y + 3z -y - z = 0

    2z = 0

    z = 0

    plug in (4)

    y + 0 = 5

    y = 5

    plug value of y and z in (1)

    -x + 5 + 0 = 2

    -x = -3

    x = 3

    x = 3, y = 5, z = 0

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