Subtract \( \frac{7}{x+7}-\frac{5}{x^{2}+5 x-14} \) and define the domain. \( x \neq \square \)

To subtract the fractions \( \frac{7}{x+7} - \frac{5}{x^2 + 5x - 14} \), we first need to factor the second denominator \( x^2 + 5x - 14 \). Factoring gives us: \[ x^2 + 5x - 14 = (x+7)(x-2) \] Now we can rewrite the expression: \[ \frac{7}{x+7} - \frac{5}{(x+7)(x-2)} \] To subtract these fractions, we need a common denominator, which is \( (x+7)(x-2) \). Rewriting the first fraction to have this common denominator: \[ \frac{7(x-2)}{(x+7)(x-2)} - \frac{5}{(x+7)(x-2)} = \frac{7(x-2) - 5}{(x+7)(x-2)} \] Now simplifying the numerator: \[ 7(x-2) - 5 = 7x - 14 - 5 = 7x - 19 \] Thus, the combined fraction is: \[ \frac{7x - 19}{(x+7)(x-2)} \] Next, we need to define the domain of this expression. The function is undefined when the denominator equals zero. Therefore, we set: \[ (x + 7)(x - 2) = 0 \] This gives us the points: \[ x + 7 = 0 \quad \Rightarrow \quad x = -7 \\ x - 2 = 0 \quad \Rightarrow \quad x = 2 \] So the domain is all real numbers except where the denominator is zero, leading us to \( x \neq -7 \) and \( x \neq 2 \). In conclusion, we have: \[ \frac{7x - 19}{(x+7)(x-2)} \quad \text{with the domain: } x \neq -7, \; x \neq 2 \]