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anonymous

  • 5 years ago

OKSY HOW DO I FIND (X,Y,Z).... 3X+4Y+2Z=24

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  1. anonymous
    • 5 years ago
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    AND (X+3)^2+Z^5+25

  2. anonymous
    • 5 years ago
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    are they both part of the same system?

  3. anonymous
    • 5 years ago
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    YEA THEY ARE

  4. anonymous
    • 5 years ago
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    WELL WHAT I WAS DOING WAS SOLVE INDIVIDUALY LIKE 3X=24 AND THEN DIVIDING IT TO BOTH SIDES WICH GAVE ME 8 SO X=8 AND THEN I DID THAT TO Y AND Z

  5. anonymous
    • 5 years ago
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    IM NOT SURE IF THATS CORRECT

  6. anonymous
    • 5 years ago
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    No. What is your second equation up there equal to?

  7. anonymous
    • 5 years ago
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    I GOT SOMETHING LIKE (-3,0,5)

  8. anonymous
    • 5 years ago
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    AND FOR MY FRST EQUATION I GOT (8,6,12)

  9. anonymous
    • 5 years ago
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    No. Those answers aren't correct. If you can tell me what your second equation is equal to then I can help you.

  10. anonymous
    • 5 years ago
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    OOOH SRRY THE SEC ONE EQUALS 25 I ACCENDENTLY PUT PLUS

  11. anonymous
    • 5 years ago
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    No problem. Solve your first equation for x... x=8-(4/3)y-(2/3)z...and so... x+3=11-(4/3)y-(2/3)z... do you follow?

  12. anonymous
    • 5 years ago
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    yea a little

  13. anonymous
    • 5 years ago
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    well, this gets kind of messy, so it's important that you can get that far, because you also solve it for z & substitute each into your second equation.

  14. anonymous
    • 5 years ago
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    HEHE OKAY THNX SO MUCH I JUS HAD TO LOOK AT IT A COUPLE OF TIMES TO UNDERSTAND IT THANK YU

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