A student solves the following equation for all possible values of x: 8/x+2 = 2/x-4 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?

a) His solution for x is correct and his explanation of the extraneous solution is reasonable.

b) 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.

c) 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.

b) His solution for x is incorrect. When solved correctly, there are no extraneous solutions.

Question

Jean Mullins

Mathematics

4 April, 19:31

A student solves the following equation for all possible values of x: 8/x+2 = 2/x-4 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?

a) His solution for x is correct and his explanation of the extraneous solution is reasonable.

b) 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.

c) 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.

b) His solution for x is incorrect. When solved correctly, there are no extraneous solutions.

+4

Answers (1)

Know the Answer?

Evan Ponce · Answer 1

For this case we have the following expression:

8 / x + 2 = 2 / x-4

The steps to solve it are:

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

We observe that the solution is correct.

Also, x = 6 does not make any denominator equal to zero.

Therefore, it is not a extraneus solution.

Answer:

c) 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.