The given system of equations is expressed as
- x + 2y = - 3
x - 4y - z = 4
2x - y + 3z = 0
From the first equation,
x = 2y + 3
We would substitute x = 2y + 3 into the second and third equation.
Substituting into the second equation, it becomes
2y + 3 - 4y - z = 4
2y - 4y - z = 4 - 3
- 2y - z = 1 fourth equation
Substituting into the third equation, it becomes
2(2y + 3) - y + 3z = 0
4y + 6 - y + 3z = 0
4y - y + 3z = 0 - 6
3y + 3z = - 6 fifth equation
From the fourth equation,
z = - 2y - 1
Substituting z = - 2y - 1 into the fifth equation, it becomes
3y + 3(- 2y - 1) = - 6
3y - 6y - 3 = - 6
3y - 6y = - 6 + 3
- 3y = - 3
y = - 3/ - 3
y = 1
Substituting y = 1 into x = 2y + 3, it becomes
x = 2 * 1 + 3
x = 5
Substituting y = 1 into z = - 2y - 1,
z = - 2 * 1 - 1
z = - 2 - 1
z = - 3
Therefore,
x = 5, y = 1, z = - 3