Astudent solves the following equation for all possible values of x:

his solution is as follows:
step 1: 8(x – 4) = 2(x + 2)
step 2: 4(x – 4) = (x + 2)
step 3: 4x – 16 = x + 2
step 4: 3x = 18
step 5: x = 6
he determines that 6 is an extraneous solution because the difference of the numerators is 6, so the 6s cancel to 0.
which best describes the reasonableness of the student’s solution?
his solution for x is correct and his explanation of the extraneous solution is reasonable.
his solution for x is correct, but in order for 6 to be an extraneous solution, both denominators have to result in 0 when 6 is substituted for x.
his solution for x is correct, but in order for 6 to be an extraneous solution, one denominator has to result in 0 when 6 is substituted for x.
his solution for x is incorrect. when solved correctly, there are no extraneous solutions.