Questions: Classify the solutions of 3/(x+5)+1/5=2/(x+5) as extraneous or non-extraneous. x=-5; extraneous x=-5; non-extraneous x=-10; extraneous x=-10; non-extraneous

Solution Steps

To classify the solutions of the given equation as extraneous or non-extraneous, we need to solve the equation and then check if the solutions satisfy the original equation. An extraneous solution is one that does not satisfy the original equation.

1. Combine the fractions on the left-hand side of the equation.
2. Set the combined fraction equal to the right-hand side.
3. Solve for \( x \).
4. Check if the solutions satisfy the original equation to determine if they are extraneous or non-extraneous.

We start with the equation: \[ \frac{3}{x+5} + \frac{1}{5} = \frac{2}{x+5} \] Combine the fractions on the left-hand side: \[ \frac{3}{x+5} + \frac{1}{5} = \frac{3 + 0.2(x+5)}{x+5} = \frac{3 + 0.2x + 1}{x+5} = \frac{0.2x + 4}{x+5} \] So the equation becomes: \[ \frac{0.2x + 4}{x+5} = \frac{2}{x+5} \]

Since the denominators are the same, we can equate the numerators: \[ 0.2x + 4 = 2 \] Solving for \( x \): \[ 0.2x = 2 - 4 \] \[ 0.2x = -2 \] \[ x = \frac{-2}{0.2} = -10 \]

We need to check if \( x = -10 \) satisfies the original equation: \[ \frac{3}{-10+5} + \frac{1}{5} = \frac{2}{-10+5} \] \[ \frac{3}{-5} + \frac{1}{5} = \frac{2}{-5} \] \[ -0.6 + 0.2 = -0.4 \] \[ -0.4 = -0.4 \] Since \( x = -10 \) satisfies the original equation, it is a non-extraneous solution.

Final Answer