Answer:
(x, y, z) = (-22/13, 29/13, 6/13)
Step-by-step explanation:
Adding the first and second equations, we get ...
(3x +2y -3z) +(7x -2y +5z) = (-2) +(-14)
10x +2z = -16 . . . . collect terms
5x + z = -8 . . . . . . divide by 2 . . . [eq4]
Adding twice the second equation to the third, we get ...
2(7x -2y +5z) +(2x +4y +z) = 2(-14) +(6)
16x +11z = -22 . . . . . . [eq5]
Now, we have two equations with the variable y eliminated. We can subtract [eq5] from 11 times [eq4] to eliminate z:
11(5x +z) -(16x +11z) = 11(-8) -(-22)
39x = -66
x = -66/39 = -22/13
From [eq4], we can find z as ...
z = -8 -5x = -8 -5(-22/13) = 6/13
And from the second equation, we get ...
y = (1/2)(7x +5z +14) = (1/2)(7(-22/13) +5(6/13) +14) = 29/13
The solution is (x, y, z) = (-22/13, 29/13, 6/13).
