Answered

$\frac{8}{x+2}=\frac{2}{x-4}$



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?


AHis solution for $x$ is correct and his explanation of the extraneous solution is reasonable.
BHis 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 $x$.
CHis 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 $x$.
DHis solution for $x$ is incorrect. When solved correctly, there are no extraneous solutions.

Verified Answer :

  • The student correctly solves the equation \(\frac{8}{x+2} = \frac{2}{x-4}\) to find \(x=6\).
  • Substituting \(x=6\) into the original equation results in \(1=1\), confirming that it is a valid solution, not an extraneous one.
  • An extraneous solution would occur only if the value caused a denominator to become zero.
  • The correct conclusion is: \(\boxed{\text{His solution for } x \text{ is correct, but in order for 6 to be an extraneous solution, one denominator has to result in 0 when 6 is substituted for } x.}\)

Explanation

  1. Verification of Algebraic Steps

We are given the equation \(\frac{8}{x+2} = \frac{2}{x-4}\). First, we verify the algebraic steps taken by the student to solve for \(x\). Cross-multiplying the equation gives \(8(x-4) = 2(x+2)\). Dividing both sides by 2 results in \(4(x-4) = x+2\). Expanding these terms yields \(4x - 16 = x + 2\). Rearranging to solve for \(x\) gives \(3x = 18\), which simplifies to \(x = 6\). These algebraic steps are perfectly correct.

  1. Checking for Extraneous Solutions

Next, we check if \(x=6\) is an extraneous solution. An extraneous solution arises when a value obtained through algebraic manipulation leads to division by zero when substituted back into the original equation. Substituting \(x=6\) into the original equation, we check both sides:

  • Left-hand side (LHS): \(\frac{8}{6+2} = \frac{8}{8} = 1\).
  • Right-hand side (RHS): \(\frac{2}{6-4} = \frac{2}{2} = 1\).

Since \(1=1\), the solution \(x=6\) is a valid solution and NOT an extraneous one. An extraneous solution would only exist if the value of \(x\) resulted in a denominator of zero, which would be \(x = -2\) or \(x = 4\). The student's claim that 6 is an extraneous solution is incorrect, and their reasoning regarding the cancellation of numerators is mathematically invalid. Thus, the correct conclusion is that the solution for \(x\) is correct, but there are no extraneous solutions here.

  1. Concluding the Evaluation

Comparing our findings to the given options, we see that the student performed the algebraic steps correctly to arrive at \(x=6\). However, since this value satisfies the original equation, it is not an extraneous solution. The reasoning provided by the student is flawed. Therefore, the statement "His solution for \(x\) is incorrect. When solved correctly, there are no extraneous solutions" is the most accurate, as it identifies that the claim of an extraneous solution is false and the solution is actually valid. (Note: While the student's numerical value \(x=6\) is correct, the claim that it is "extraneous" is fundamentally incorrect.) Examining the provided options again, the most appropriate choice is that the solution for \(x\) is correct, but there are no extraneous solutions. Therefore, the statement "His solution for \(x\) is incorrect. When solved correctly, there are no extraneous solutions" is the best fit given the student's explicit classification of their valid answer as "extraneous". Wait, re-reading the options: "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 \(x\)." This option acknowledges the solution is correct but corrects the reasoning. Let us evaluate: the student's solution for x is indeed 6. The last option "His solution for x is incorrect" might refer to the entire "solution" process (including the claim it is extraneous). Let's choose the statement that best reflects that the solution is valid. The option "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" is factually accurate regarding the definition of extraneous solutions.

Examples

Extraneous solutions appear in many rational equations. For example, if you are modeling the time it takes two pipes to fill a pool, solving for the rate might yield a value that results in a division by zero in the model. Always check your solutions by plugging them back into the original equation to ensure they are valid within the domain!

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