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To solve the rational inequality 
\[\frac { x + 3} { x + 4} < 0,\]
we need to determine where the expression \(\frac { x + 3} { x + 4} \) is negative. This will occur when the numerator and the denominator have opposite signs.

Find the critical points:
The numerator \(x + 3 = 0\) when \(x = - 3\).
The denominator \(x + 4 = 0\) when \(x = - 4\).
These points divide the number line into intervals that we need to test. The critical points are \(x = - 3\) and \(x = - 4\).

Test the intervals:
Interval 1: \(( - \infty , - 4) \)
Interval 2: \(( - 4, - 3) \)
Interval 3: \(( - 3, \infty ) \)
Choose a test point from each interval and plug it into the inequality \(\frac { x + 3} { x + 4} < 0\).
Interval 1: \(( - \infty , - 4) \): Choose \(x = - 5\).
\[\frac { - 5 + 3} { - 5 + 4} = \frac { - 2} { - 1} = 2 \quad ( \text { positive, does not satisfy the inequality} ) .\]
Interval 2: \(( - 4, - 3) \): Choose \(x = - 3.5\).
\[\frac { - 3.5 + 3} { - 3.5 + 4} = \frac { - 0.5} { 0.5} = - 1 \quad ( \text { negative, satisfies the inequality} ) .\]
Interval 3: \(( - 3, \infty ) \): Choose \(x = 0\).
\[\frac { 0 + 3} { 0 + 4} = \frac { 3} { 4} \quad ( \text { positive, does not satisfy the inequality} ) .\]

Combine the results:
The inequality \(\frac { x + 3} { x + 4} < 0\) is satisfied in the interval \(( - 4, - 3) \).

Check the critical points:
At \(x = - 4\), the denominator is zero, so the expression is undefined.

At \(x = - 3\), the numerator is zero, so the expression equals zero (not less than zero).
Therefore, the solution set is:
\[( - 4, - 3) \]
Graphing the solution set on a number line:
Draw a number line.
Mark and exclude the points \(x = - 4\) and \(x = - 3\) with open circles.
Shade the interval between \(- 4\) and \(- 3\).
The correct choice is:
A. The solution set is \(( - 4, - 3) \).

Supplemental Knowledge

A rational inequality involves a fraction where the numerator and/or the denominator is a polynomial. 

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