Question

a student solves the following equation for all possible values of x.\\(\frac{8}{x + 2}=\frac{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?\
\\(\bigcirc\\) his solution for x is correct and his explanation of the extraneous solution is reasonable.\
\\(\bigcirc\\) 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.\
\\(\bigcirc\\) 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.\
\\(\bigcirc\\) his solution for x is incorrect. when solved correctly, there are no extraneous solutions.

Explanation:

1. First, verify the student's solution for $x$:

• Cross-multiplying $\frac{8}{x+2}=\frac{2}{x-4}$ gives $8(x-4)=2(x+2)$, which simplifies correctly through the student's steps to $x=6$.
• Substitute $x=6$ back into the original equation: $\frac{8}{6+2}=\frac{8}{8}=1$, and $\frac{2}{6-4}=\frac{2}{2}=1$. The equation holds true, so $x=6$ is a valid solution, not extraneous.

1. An extraneous solution is one that arises from solving the equation but makes the original equation undefined (i.e., causes a denominator to equal 0). For $x=6$, denominators $x+2=8$ and $x-4=2$, neither is 0, so it is not extraneous.
2. Evaluate the options: The student's calculation of $x=6$ is correct, but their reasoning for it being extraneous is wrong. The correct understanding is that extraneous solutions make denominators 0, which $x=6$ does not do, so there are no extraneous solutions here.

Answer:

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