Solve the equation. Enter any solution that checks as an integer or reduced fraction. Separate with commas if there is more than one solution. If there are no solutions, click the No Solution button. \( \frac{x}{x+2}+\frac{3}{x+7}=\frac{3 x-4}{x^{2}+9 x+14} \) One or more solutions:
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To solve the equation \( \frac{x}{x+2}+\frac{3}{x+7}=\frac{3 x-4}{x^{2}+9 x+14} \), we first recognize that the denominator on the right can be factored. The expression \( x^2 + 9x + 14 \) factors into \( (x + 2)(x + 7) \). Thus, we can rewrite the equation as: \[ \frac{x}{x+2} + \frac{3}{x+7} = \frac{3x - 4}{(x + 2)(x + 7)} \] To combine the left-hand side, we need a common denominator, which is \( (x + 2)(x + 7) \): \[ \frac{x(x + 7) + 3(x + 2)}{(x + 2)(x + 7)} = \frac{3x - 4}{(x + 2)(x + 7)} \] Now, simplify the numerator on the left: \[ x^2 + 7x + 3x + 6 = x^2 + 10x + 6 \] The equation now becomes: \[ \frac{x^2 + 10x + 6}{(x + 2)(x + 7)} = \frac{3x - 4}{(x + 2)(x + 7)} \] Setting the numerators equal gives: \[ x^2 + 10x + 6 = 3x - 4 \] Rearranging this, we get: \[ x^2 + 7x + 10 = 0 \] Now, factoring yields: \[ (x + 2)(x + 5) = 0 \] Thus, the solutions are \( x = -2 \) and \( x = -5 \). However, we must check for restrictions from the original equation. The denominators \( x + 2 \) and \( x + 7 \) cannot be zero, which means \( x \neq -2 \) and \( x \neq -7 \). Therefore, \( x = -2 \) is not valid. The only permissible solution is: \(-5\)
