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

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

Algebra
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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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HELP!!!!!
what are you stuck on?
solve the first equation for x and substitute into the last two equations... then solve those two equations for y and z

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Other answers:

I don't get any of it.
"solve the first equation for x" move all the terms without 'x' in them to one side of the equal sign
can you do that?
I know THAT. x=-y-z+12
great.
6(-y-z+12)-2y-z=16
"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
yep
do some basic algebra to simplify... what are you left with?
two equations with two unknowns...
solve them the same way we just 'solved' for x...
solve one equation for z or y.... substitute into the other equation.
what's16-72?
good one.
never mind.
x=4 y=0 z=8
x=4,y=0 & z=8

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