Subtract $$ \frac{7}{x+7}-\frac{5}{x^{2}+5 x-14} $$ and define the domain. $$ x
eq \square $$

Question

Subtract $$ \frac{7}{x+7}-\frac{5}{x^{2}+5 x-14} $$ and define the domain. $$ x 
eq \square $$

UpStudy AI · Accepted Answer

To solve the expression $$ \frac{7}{x+7}-\frac{5}{x^{2}+5x-14} $$ and define the domain, we will follow these steps:

1. **Identify the denominators**: The denominators in the expression are $$ x + 7 $$ and $$ x^2 + 5x - 14 $$.

2. **Factor the second denominator**: We need to factor $$ x^2 + 5x - 14 $$ to find its roots.

3. **Set the denominators to zero**: We will find the values of $$ x $$ that make the denominators zero, as these values will be excluded from the domain.

Let's start by factoring $$ x^2 + 5x - 14 $$.

### Step 1: Factor $$ x^2 + 5x - 14 $$

To factor $$ x^2 + 5x - 14 $$, we look for two numbers that multiply to $$-14$$ and add to $$5$$. The numbers $$7$$ and $$-2$$ satisfy this condition.

Thus, we can factor it as:
$$
x^2 + 5x - 14 = (x + 7)(x - 2)
$$

### Step 2: Set the denominators to zero

Now we set each denominator to zero to find the excluded values:

1. For $$ x + 7 = 0 $$:
   $$
   x = -7
   $$

2. For $$ (x + 7)(x - 2) = 0 $$:
   - $$ x + 7 = 0 $$ gives $$ x = -7 $$
   - $$ x - 2 = 0 $$ gives $$ x = 2 $$

### Step 3: Define the domain

The values that make the denominators zero are $$ x = -7 $$ and $$ x = 2 $$. Therefore, the domain of the expression is all real numbers except these values.

Thus, the domain can be defined as:
$$
x \neq -7, \quad x \neq 2
$$

### Final Answer

The expression is:
$$
\frac{7}{x+7}-\frac{5}{(x+7)(x-2)}
$$
And the domain is:
$$
x \neq -7, \quad x \neq 2
$$
The expression is $$ \frac{7}{x+7} - \frac{5}{(x+7)(x-2)} $$, and the domain is all real numbers except $$ x = -7 $$ and $$ x = 2 $$.

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

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