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

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

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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AND (X+3)^2+Z^5+25
are they both part of the same system?
YEA THEY ARE

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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
IM NOT SURE IF THATS CORRECT
No. What is your second equation up there equal to?
I GOT SOMETHING LIKE (-3,0,5)
AND FOR MY FRST EQUATION I GOT (8,6,12)
No. Those answers aren't correct. If you can tell me what your second equation is equal to then I can help you.
OOOH SRRY THE SEC ONE EQUALS 25 I ACCENDENTLY PUT PLUS
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?
yea a little
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.
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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