Answer:
(5, 3, 3)
Explanations:
The given system of equations is:
x − 2y + 3z = 8.......(1)
3y + z = 12.............(2)
−2x + 2z = −4.........(3)
Make x the subject of the formula in equation (1):
x = 2y - 3z + 8...........(4)
Substitute equation (4) into equation (3):
−2x + 2z = −4
-2(2y - 3z + 8) + 2z = -4
-4y + 6z - 16 + 2z = -4
-4y + 8z = -4 + 16
-4y + 8z = 12...............(5)
Make z the subject of the formula in equation (2):
z = 12 - 3y...........(6)
Substitute equation (6) into equation (5):
-4y + 8z = 12
-4y + 8(12 - 3y) = 12
-4y + 96 - 24y = 12
-4y - 24y = 12 - 96
-28y = -84
y = -84/-28
y = 3
Substitute y = 3 into equation (6):
z = 12 - 3y
z = 12 - 3(3)
z = 12 - 9
z = 3
Substitute y = 3 and z = 3 into equation (4)
x = 2y - 3z + 8
x = 2(3) - 3(3) + 8
x = 6 - 9 + 8
x = 5
Therefore the solution (x, y, z) to the system of equations is (5, 3, 3)