Question

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.

Answer 1

The correct choice is: "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."

Reason: x = 6 does not make either denominator zero (6+2 = 8, 6−4 = 2), and 8/8 = 2/2 = 1 so the equation holds. An extraneous solution would occur only if the candidate made one (or both) denominators zero so the original expression is undefined — the student's "difference of the numerators" explanation is incorrect.