For system 5. we multiply the first equation by -1
-2g - 3h + 8j = -10
The third equation is
-2g - 3h + 8j = 5
They both can't be true.
Answer: NO SOLUTIONS
8.
We can eliminate z by multiplying the third equation by 2 and adding to the first:
2x - 10y - 4z = -8
2x + 3y + 4z = 2
Adding,
4x - 7y = -6
Let's eliminate z from the second by multiplying it by 2, the third by 3, and adding.
10x - 4y + 6z = 0
3x - 15y - 6z = -12
13 x - 19 y = -12
Ok, we eliminate y by muliplying our first x,y equation by 19, the last by 7, and subtracting
4(19)x - 7(19)y = -6(19)
13(7)x - 19(7)y = -12(7)
(4(19) - 13(7)) x = -6(19)+12(7)
x = ( -6(19)+12(7)) / (4(19) - 13(7)) = 2
That worked out.
7y = 4x + 6 = 4(2) + 6 = 14
y = 2
4z = 2 - 2x - 3y = 2 - 2(2) - 3(2) = 2 - 4 - 6 = -8
z = -2
Answer: x=2, y=2, z=-2
Check:
2(2)+3(2)+4(-2) = 2 good
5(2) - 2(2) + 3(-2) = 0 good
2 - 5(2) - 2(-2) = -4 good