Solve the following systems of equations.A) 6x+15y+12z=12B) 3x-9y-3z=15C) 3x+12y+3z=3Possible Answers:A. x=33/10 , y=1/5 , z=1/2B. x=67/10 , y=3/5 , z=1/10C. x=33/10 , y=-3/5, z=1/10D. x=-33/5 , y=-3/5 , z=1/5

Answer: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 = 18y = -18/30 = -3/5.So substituting for y-21(-3/5) - 6z = 12-6z = 12 - 63/5 = -3/5z = 1/10Finally 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 = 126x = 19.8x = 3.3 = 33/10.