Answer
(1, -1, 1, 2)
Work Step by Step
The question asks to find the solution to the system of equations.
Given:
1. $x + z + 2w = 6$
2. $y - 2z = -3$
3. $x + 2y - z = -2$
4. $2x + y + 3z - 2w = 0$
Substitute $ y = -3 + 2z $ (from equation 2)
1. $x + z + 2w = 6$
3. $x + 2(-3 + 2z) - z = -2$
4. $2x + (-3 + 2z) + 3z - 2w = 0$
1. $x + z + 2w = 6$
3. $x + 3z = 4$
4. $2x + 5z - 2w = 3$
Substitute $x = 4 - 3z$ (from equation 3)
1. $(4 - 3z) + z + 2w = 6$
4. $2(4-3z) + 5z - 2w = 3$
$-2z + 2w = 2$
+ ($-z - 2w = -5$)
$-3z = -3$
$z = 1$
Solve for other variables
$-2(1) + 2w = 2$
$2w = 4$
$w = 2$
$x = 4-3(1)$
$x = 1$
$y = -3 + 2(1)$
$y = -1$
Solution: (1, -1, 1, 2)
