alexwing285
contestada

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.

A student solves the following equation for all possible values of x His solution is as follows Step 1 8x 4 2x 2 Step 2 4x 4 x 2 Step 3 4x 16 x 2 Step 4 3x 18 S class=

Respuesta :

nobillionaireNobley
Given  [tex] \frac{8}{x+2} = \frac{2}{x-4} [/tex]
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.

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.

Otras preguntas