ect the correct answer. lanie used cross multiplication to correctly solve a rational equation. She found one valid solution and one extraneous solution. If 3 is the extraneous ution, which equation could she have solved? A. The equation is $$ \frac{4}{x-3}=\frac{x}{10} $$ because. 3 makes a denominator equal zero and is not a solution of the equation derived from cross multiplying. B. The equation is $$ \frac{x-3}{4}=\frac{2 x-6}{4 x} $$ because 3 makes a numerator equal zero and is a solution of the equation derived from cross multiplying. C. The equation is $$ \frac{8}{x^{2}-9}=\frac{5}{2 x-6} $$ because 3 makes a denominator equal zero and is a solution of the equation derived from cross multiplying. D. The equation is $$ \frac{8}{x+3}=\frac{12}{4 x-3} $$ because 3 is a solution of both the original equation and the equation derived from cross multiplying.

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UpStudy AI · Accepted Answer

To determine which equation could have been solved by Lanie, we need to analyze each option based on the information provided: Lanie found one valid solution and one extraneous solution, with 3 being the extraneous solution. An extraneous solution typically arises when a value makes a denominator zero or when it does not satisfy the original equation after cross-multiplying.

Let's evaluate each option:

### Option A: 
$$
\frac{4}{x-3}=\frac{x}{10}
$$
- If $$ x = 3 $$, the left side becomes $$ \frac{4}{3-3} = \frac{4}{0} $$, which is undefined. 
- Therefore, 3 is not a valid solution, and it is indeed an extraneous solution because it makes the denominator zero.

### Option B:
$$
\frac{x-3}{4}=\frac{2x-6}{4x}
$$
- If $$ x = 3 $$, the left side becomes $$ \frac{3-3}{4} = \frac{0}{4} = 0 $$.
- The right side becomes $$ \frac{2(3)-6}{4(3)} = \frac{0}{12} = 0 $$.
- Here, 3 is a valid solution, not an extraneous one.

### Option C:
$$
\frac{8}{x^{2}-9}=\frac{5}{2x-6}
$$
- If $$ x = 3 $$, the left side becomes $$ \frac{8}{3^{2}-9} = \frac{8}{9-9} = \frac{8}{0} $$, which is undefined.
- Therefore, 3 is not a valid solution, and it is an extraneous solution because it makes the denominator zero.

### Option D:
$$
\frac{8}{x+3}=\frac{12}{4x-3}
$$
- If $$ x = 3 $$, the left side becomes $$ \frac{8}{3+3} = \frac{8}{6} = \frac{4}{3} $$.
- The right side becomes $$ \frac{12}{4(3)-3} = \frac{12}{12-3} = \frac{12}{9} = \frac{4}{3} $$.
- Here, 3 is a valid solution, not an extraneous one.

### Conclusion:
The only option where 3 is an extraneous solution (making a denominator zero) is **Option A** and **Option C**. However, since we are looking for the equation that Lanie could have solved, the correct answer is:

**A. The equation is $$ \frac{4}{x-3}=\frac{x}{10} $$ because 3 makes a denominator equal zero and is not a solution of the equation derived from cross multiplying.**
The correct equation Lanie could have solved is $$ \frac{4}{x-3}=\frac{x}{10} $$, as 3 makes the denominator zero and is not a valid solution after cross-multiplying.

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