Question 55978
Solve by elimination:
E1) 3x+ y+2z=6
E2)  x+ y+4z=3
E3) 2x+3y+2z=2
Eliminate the y in the E1) and E2), multiply E2) by -1 and add them together:
-1(x+y+4z)=-1(3)
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3x+y+2z=6
-x-y-4z=-3
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2x+0y-2z=3
E4) 2x-2z=3
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Eliminate the y in E2) and E3), multiply E2 by -3 and add them together:
-3(x+y+4z)=-3(3)
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-3x-3y-12z=-9
2x+3y+2z=2
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-x+0y-10z=-7
E5) -x-10z=-7
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Eliminate the x in E4) and E5), multiply  E5) by 2 and add them together:
2(-x-10z)=2(-7)
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2x-2z=3
-2x-20z=-14
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0x-22z=-11
-22z=-11
{{{-22z/-22=-11/-22}}}
z=1/2
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Sustitute z=1/2 into E5) and solve for x:
{{{-x-10(1/2)=-7}}}
-x-5=-7
-x-5+5=-7+5
-x=-2
-(-x)=-(-2)
x=2
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Substitute z=1/2 and x=2 into E1,E2, or E3 and solve for y.  I'm going with E2.
(2)+y+4(1/2)=3
2+y+2=3
y+4=3
y+4-4=3-4
y=-1
Therefore the solution is (x,y,z)=(2,-1,1/2)
I'll leave it to you to check.  If you can't do that then let me know.  I checked it with a calculator.
Happy Calculating!!!