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anonymous
 4 years ago
Solve for x, y, z:
x+y+z=12
6x2yz=16
3x+4y+2z=28
anonymous
 4 years ago
Solve for x, y, z: x+y+z=12 6x2yz=16 3x+4y+2z=28

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0what are you stuck on?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0solve the first equation for x and substitute into the last two equations... then solve those two equations for y and z

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I don't get any of it.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0"solve the first equation for x" move all the terms without 'x' in them to one side of the equal sign

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I know THAT. x=yz+12

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0"substitute into the last two equations." 6x2yz=16  > 6(yz+12) 2yz=16 3x+4y+2z=28 > 3(yz+12) 4y+2z=28

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0do some basic algebra to simplify... what are you left with?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0two equations with two unknowns...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0solve them the same way we just 'solved' for x...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0solve one equation for z or y.... substitute into the other equation.
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