Question

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.

1. yenoanh
Answer: its not d don’t listen the guy above
its C
Step-by-step explanation:

2. quangkhai
The question is incomplete. Completed question is given below the answer.
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.
Given
Then the solution follows thus:
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
It can be seen that his solution is correct. But 6 is not an extraneous solution.
An extraneous solution is a solution to an equation that emerges from the process of solving the problem but is not a valid solution to the original problem.
When 6 is substituted into the original equation, the original equation holds.
Therefore, 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.
#SPJ4
Completed question:-
A student 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.