A student solves the following equation for all possible values of x: StartFraction 8 Over x + 2 EndFraction = StartFraction 2 Over x minus 4 EndFraction 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.