Solve the following systems of equations.

A) 6x+15y+12z=12

B) 3x-9y-3z=15

C) 3x+12y+3z=3

Possible Answers:

A. x=33/10, y=1/5, z=1/2

B. x=67/10, y=3/5, z=1/10

C. x=33/10, y=-3/5, z=1/10

D. x=-33/5, y=-3/5, z=1/5

C. x=33/10, y=-3/5, z=1/10.

Step-by-step explanation:

6x+15y+12z=12 (A)

3x-9y-3z=15 (B)

3x+12y+3z=3 (C)

Subtract B - C to eliminate the term in x:

-21y - 6z = 12 (D) Now multiply B by 2:

6x - 18y - 6z = 30 (E) A - E will eliminate x:

33y + 18z = - 18 (F)

Now solve equations (D) and (F):

-21y - 6z = 12 Multiply this by 3:

-63y - 18z = 36 (G)

G + F will eliminate z:

-30y = 18

y = - 18/30 = - 3/5.

So substituting for y

-21 (-3/5) - 6z = 12

-6z = 12 - 63/5 = - 3/5

z = 1/10

Finally we substitute the values of y and z in equation A to find x:

6x+15 (-3/5) + 12 (1/10) = 12

6x - 9 + 1.2 = 12

6x = 19.8

x = 3.3 = 33/10.