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Answer

D.

Explanation

To find the correct equation Sean could have solved, we need to consider how an extraneous solution arises in solving rational equations. An extraneous solution is one that is found during the process of solving the equation but is not a valid solution to the original equation. This often occurs when solving rational equations because the process of solving them can introduce solutions that make the original denominators zero, which is not allowed.In rational equations, the key is to check for values that make the denominators zero, as these values are not permissible solutions. Let's analyze each option:A. 10/(x^2-1)=5/(3x-3): Here, x = 1 makes the denominator 3x - 3 zero, which is not allowed. Thus, 1 is an extraneous solution. This equation fits the condition as 1 is not a solution of the original equation but appears as a solution when cross multiplying.B. (x+2)/(x+3)=6x/8: In this equation, x = 1 does not make any denominator zero. Therefore, 1 cannot be an extraneous solution in this case.C. (4x-4)/(x+6)=(x-1)/10: Here, x = 1 makes the numerator 4x - 4 zero, not the denominator. The focus should be on the denominator for extraneous solutions, so this option is not suitable.D. 4/(x-1)=(x+2)/10: In this equation, x = 1 makes the denominator x - 1 zero. However, 1 is not a solution of the equation derived from cross multiplying, hence it is an extraneous solution.The correct answer must be the one where 1 is an extraneous solution, meaning it should make a denominator zero in the original equation but should not be a solution of the equation derived from cross multiplying.Option D fits these criteria, making it the correct choice.

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