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AEB047

  • 3 years ago

Solve for x, y, z: x+y+z=12 6x-2y-z=16 3x+4y+2z=28

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  1. AEB047
    • 3 years ago
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    HELP!!!!!

  2. _irawk_
    • 3 years ago
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    what are you stuck on?

  3. Algebraic!
    • 3 years ago
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    solve the first equation for x and substitute into the last two equations... then solve those two equations for y and z

  4. AEB047
    • 3 years ago
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    I don't get any of it.

  5. Algebraic!
    • 3 years ago
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    "solve the first equation for x" move all the terms without 'x' in them to one side of the equal sign

  6. Algebraic!
    • 3 years ago
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    can you do that?

  7. AEB047
    • 3 years ago
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    I know THAT. x=-y-z+12

  8. Algebraic!
    • 3 years ago
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    great.

  9. AEB047
    • 3 years ago
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    6(-y-z+12)-2y-z=16

  10. Algebraic!
    • 3 years ago
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    "substitute into the last two equations." 6x-2y-z=16 - > 6(-y-z+12) -2y-z=16 3x+4y+2z=28 -> 3(-y-z+12) 4y+2z=28

  11. Algebraic!
    • 3 years ago
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    yep

  12. Algebraic!
    • 3 years ago
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    do some basic algebra to simplify... what are you left with?

  13. Algebraic!
    • 3 years ago
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    two equations with two unknowns...

  14. Algebraic!
    • 3 years ago
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    solve them the same way we just 'solved' for x...

  15. Algebraic!
    • 3 years ago
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    solve one equation for z or y.... substitute into the other equation.

  16. AEB047
    • 3 years ago
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    what's16-72?

  17. Algebraic!
    • 3 years ago
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    good one.

  18. AEB047
    • 3 years ago
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    never mind.

  19. Arhin
    • 3 years ago
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    x=4 y=0 z=8

  20. jishan
    • 3 years ago
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    x=4,y=0 & z=8

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