Question 798952: How do I solve this ordered triple system of equations?
3x-2y+3z=-27 (1)
4x+y-2z=5 (2)
2x+4y-z=13 (3) Answer by Finavon(81) (Show Source):
You can put this solution on YOUR website! (2) - 2*(3) gives
4x+y-2z -2*(2x+4y-z) = 5 -2*13
4x+y-2z-4x-8y+2z=5-26
-7y=-21
so y=3
(1) + 3*(3) gives
3x-2y+3z +3*(2x+4y-z) =-27 +3*13
3x-2y+3z+6x+12y-3z=-27+39
9x+10y=12 but y=3
so 9x+30=12
and 9x = 12 - 30 = -18
so x=-2
put these values in (3)
2*(-2) + 4*3 -z = 13
add (z-13) to both sides
-4 + 12 -z +z-13 = 13+ z-13
-5 = z
x= -2; y= 3 ; z= -5
Check:
3*(-2) -2*3 + 3*(-5) = -6-6-15 = -27
4*(-2) +3 -2*(-5)= -8+3+10 = 5
2*(-2) +4*3 -(-5)= -4+12+5 = 13
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