Question

Solve the system of equations using any method. 2X – Y + Z = – 2, 6X + 3Y – 4Z = 8, – 3X + 2Y +3Z = – 6

Answer 1

Answer:

Z = - 2, Y = 0, X = 0

Step-by-step explanation:

2X – Y + Z = – 2

6X + 3Y – 4Z = 8

– 3X + 2Y +3Z = – 6

isolate "X" in the first equation

2X - Y + Z = - 2 => isolate "2X"

2X = - 2 - Z + Y => divide either side by 2

X = - 1 - Z/2 + Y/2

substitute this value of X in the second two equations (applying the substitution method here)

6(- 1 - Z/2 + Y/2) + 3Y - 4Z = 8,

6Y - 7Z - 6 = 8

- 3(- 1 - Z/2 + Y/2) + 2Y +3Z = - 6,

Y + 9Z + 6/2 = - 6

isolate Y in the second equation (6Y - 7Z - 6 = 8) and substitute in the third equation (Y + 9Z + 6/2 = - 6)

6Y - 7Z - 6 = 8 => isolate 6Y

6Y = 14 + 7Z => divide either side by 6

Y = 7Z + 14/6 => substitute in second equation

7Z + 14/6 + 9Z + 6/2 = - 6 => solve for Z

61Z + 50/12 = - 6, Z = - 2

substitute the value of Z into the equations "Y = 7Z + 14/6" and using the value of Y and Z, substitute into "X = - 1 - Z/2 + Y/2"

Y = (7(- 2) + 14)/6 = 0 / 6

= 0

X = - 1 - (- 2)/2 + 0/2 = - 1 - (-1) + 0

= - 1 + 1 + 0 = 0