Answer
No solution
Work Step by Step
The second equation: $-3x+y-2z=6$
$\rightarrow y=3x+2z+6$
Substitute for x in the first equation:
$4x - 8y + 2z = 10$
$4x - 8(3x+2z+6) + 2z = 10$
$4x-24x-16z-48+2z=10$
$-20x-14z=58$ (1)
Then continue to substitute for x in the third equation:
$2x - 4y + z =8$
$2x - 4(3x+2z+6) + z =8$
$2x-12x-8z-24+z=8$
$-10x-7z=32$ (2)
From (1) and (2) we get the system of equations:
$-20x-14z=58$
$-10x-7z=32$
Multiply both sides of the second equation by $-2$:
$-20x-14z=58$
$20x+14z=-64$
_____________________
$0x+0z=-6$
Hence, there is no solution for this system of equations.
