Question: solve x+2y-z=12,3/2x+2/3y-1/2z=-7,1/4x+2/3y-1/2z=1.
The three simultaneous equations are,
x+2y-z=12 ------------------------------- (1)
3/2x+2/3y-1/2z=-7 -------------------- (2)
1/4x+2/3y-1/2z=1 -------------------- (3)
Multiply (2) by 6, and (3) by12 to get rid of fractions.
x + 2y - z = 12 ------------------------- (4)
9x + 4y - 3z = -42 -------------------- (5)
3x + 8y - 6z = 12 -------------------- (6)
Substitute x = 12 - 2y + z, from (4), into (5) and (6).
9(12 - 2y + z) + 4y - 3z = -42 -------------------- (5)
3(12 - 2y + z) + 8y - 6z = 12 -------------------- (6)
These equations simplify to
-14y + 6z = -150 ----------------------------------- (7)
2y - 3z = -24 --------------------------------------- (8)
Add 2*(8) to (7).
-10y = -198
y = 99/5
Substitute for y = 99/5 into (8).
198/5 - 3z = -24
66/5 - z = -8
z = 66/5 + 8 = (66+40)/5 = 106/5
z = 1065/5
Substitute for y = 99/5 and z = 106/5 into (1).
x + 198/5 - 106/5 = 12
x + 92/5 = 12
x = 12 - 92/5 = (60 - 92)/5 = -32/5
x = -32/5
Answer: x = -32/5, y = 99/5, z = 106/5