To solve the inequality \(\frac{14x}{x+1} - \frac{9x-30}{x-4} < 0\), we will follow these steps:
### Step 1: Find a common denominator
The common denominator for the fractions is \((x + 1)(x - 4)\). We will rewrite the inequality with this common denominator.
\[
\frac{14x(x - 4) - (9x - 30)(x + 1)}{(x + 1)(x - 4)} < 0
\]
### Step 2: Expand the numerator
Now, we will expand the numerator:
1. Expand \(14x(x - 4)\):
\[
14x^2 - 56x
\]
2. Expand \((9x - 30)(x + 1)\):
\[
9x^2 + 9x - 30x - 30 = 9x^2 - 21x - 30
\]
Now, substituting back into the inequality:
\[
\frac{14x^2 - 56x - (9x^2 - 21x - 30)}{(x + 1)(x - 4)} < 0
\]
### Step 3: Combine like terms in the numerator
Combine the terms in the numerator:
\[
14x^2 - 56x - 9x^2 + 21x + 30 = 5x^2 - 35x + 30
\]
So we have:
\[
\frac{5x^2 - 35x + 30}{(x + 1)(x - 4)} < 0
\]
### Step 4: Factor the numerator
Now we will factor the quadratic \(5x^2 - 35x + 30\):
1. Factor out the common factor of 5:
\[
5(x^2 - 7x + 6)
\]
2. Factor the quadratic:
\[
5(x - 6)(x - 1)
\]
So the inequality becomes:
\[
\frac{5(x - 6)(x - 1)}{(x + 1)(x - 4)} < 0
\]
### Step 5: Determine the critical points
The critical points are where the expression is equal to zero or undefined:
1. From the numerator: \(x - 6 = 0 \Rightarrow x = 6\) and \(x - 1 = 0 \Rightarrow x = 1\)
2. From the denominator: \(x + 1 = 0 \Rightarrow x = -1\) and \(x - 4 = 0 \Rightarrow x = 4\)
The critical points are \(-1, 1, 4, 6\).
### Step 6: Test intervals
We will test the intervals defined by these critical points:
1. \( (-\infty, -1) \)
2. \( (-1, 1) \)
3. \( (1, 4) \)
4. \( (4, 6) \)
5. \( (6, \infty) \)
Choose test points from each interval:
- For \(x = -2\) in \((- \infty, -1)\):
\[
\frac{5(-2 - 6)(-2 - 1)}{(-2 + 1)(-2 - 4)} = \frac{5(-8)(-3)}{(-1)(-6)} > 0
\]
- For \(x = 0\) in \((-1, 1)\):
\[
\frac{5(0 - 6)(0 - 1)}{(0 + 1)(0 - 4)} = \frac{5(-6)(-1)}{(1)(-4)} < 0
\]
- For \(x = 2\) in \((1, 4)\):
\[
\frac{5(2 - 6)(2 - 1)}{(2 + 1)(2 - 4)} = \frac{5(-4)(1)}{(3)(-2)} > 0
\]
- For \(x = 5\) in \((4, 6)\):
\[
\frac{5(5 - 6)(5 - 1)}{(5 + 1)(5 - 4)} = \frac{5(-1)(4)}{(6)(1)} < 0
\]
- For \(x = 7\) in \((6, \infty)\):
\[
\frac{5(7 - 6)(7 - 1)}{(7 + 1)(7 - 4)} = \frac{5(1)(6)}{(8)(3)} > 0
\]
### Step 7: Conclusion
The intervals where the expression is less than zero are:
1. \((-1, 1)\)
2. \((4, 6)\)
Thus, the solution to the inequality is:
\[
x \in (-1, 1) \cup (4, 6)
\]
