To determine the number of valid and extraneous solutions for the equation:
[tex]\[
\frac{2}{x+2} + \frac{1}{10} = \frac{3}{x+3}
\][/tex]
we need to perform a series of steps:
1. Identify potential extraneous solutions: These occur when the denominators of fractions become zero, because division by zero is undefined. For this equation, extraneous solutions could occur when [tex]\(x+2 = 0\)[/tex] or [tex]\(x+3 = 0\)[/tex].
So, [tex]\(x = -2\)[/tex] and [tex]\(x = -3\)[/tex] are potential extraneous solutions.
2. Solve the equation: Solving the given equation typically involves finding a common denominator and simplifying the equation to solve for [tex]\(x\)[/tex]. This process will yield potential solutions.
3. Analyze the solutions:
- Check if the solutions are valid: Substituting the solutions back into the equation to ensure they do not cause any denominators to be zero (i.e., they are not [tex]\(x = -2\)[/tex] or [tex]\(x = -3\)[/tex]).
- Check if the solutions are extraneous: Solutions that cause any denominator in the original equation to be zero are extraneous solutions.
After performing the analysis, we find:
- There are two solutions to the equation.
- None of these solutions are extraneous (i.e., neither solution makes the denominators zero).
Thus, the correct statement describing the solutions to the equation is:
A. The equation has two valid solutions and no extraneous solutions.
