Solve the following system of equations 6x-y+z=4 4x+2y-3z=-8 x-3y+2z=12 Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice A. There is one solution. The solution is (type integers or simplified fractions ) B. There are infinitely many solutions. The solutions are ( (square ,square ,z) (Type integers or simplified fractions) C. There is no solution

To solve the system of equations, we can use the method of substitution or elimination. Let's use the elimination method.<br /><br />First, let's eliminate the variable y by multiplying the first equation by 2 and adding it to the second equation:<br /><br />$12x - 2y + 2z = 8$<br />$4x + 2y - 3z = -8$<br /><br />Adding these two equations, we get:<br /><br />$16x - z = 0$<br /><br />Simplifying, we have:<br /><br />$z = 16x$<br /><br />Now, let's substitute this value of z into the third equation:<br /><br />$x - 3y + 2(16x) = 12$<br /><br />Simplifying, we get:<br /><br />$33x - 3y = 12$<br /><br />Now, let's solve for y in terms of x:<br /><br />$-3y = -33x + 12$<br />$y = 11x - 4$<br /><br />So, the solution to the system of equations is:<br /><br />$x = x$<br />$y = 11x - 4$<br />$z = 16x$<br /><br />Therefore, the correct choice is B. There are infinitely many solutions. The solutions are $(x, 11x - 4, 16x)$.