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anonymous

  • 4 years ago

Solve the following system of equations. 2x – 3y + z = –1,,,, 3x + 2y + 2z = –1,,,, x – y – 3z = –4

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  1. anonymous
    • 4 years ago
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    please help i dont know how to do this

  2. y2o2
    • 4 years ago
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    add all of them, 2x-3y+z = -1 + 3x+2y+2z = -1 + x-y-3z = -4 ------------------- 6x-2y+0z = -6 ====>[1] (divide by 2) 3x-y = -3 y = 3x+3 Substitute in any two equations and you'll get the answer

  3. anonymous
    • 4 years ago
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    2x-3y+z=-1...................1 3x-2y+2z=-1..................2 x=y+3z-4.........................3 sub 3 in 1,2 2(y+3z-4)-3y+z=-1..............1 3(y+3z-4)-2y+2z=-1.............2 2y+6z-8-3y+z=-1 then it -y+ 7z =7............1 3y+9z-12-2y+2z=-1 then it y+11z=-1...........2 let we take eq 2 y=-1-11z sub this in eq 1 -(-1-11z)+7z=7..........1 1+11z+7z=7 18 z=6 z= 1/3 sub this in eq 2 y=-1-(11/3)=-4.666 sub y and z in eq 3 x=-4.666+(3/3)-4 =-7.666 you are welcome

  4. anonymous
    • 4 years ago
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    so what is the answer

  5. anonymous
    • 4 years ago
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    A.(–1, 0, 1) B.(1, 2, 1) C.(1, 1, 4) D.(–1, –1, –2)

  6. anonymous
    • 4 years ago
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    x=-7.6666 , y=-4.666 , z=1/3

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