To solve the rational inequality [tex]\(\frac{-4}{1-x} < 9\)[/tex], we will follow a set of steps to isolate the variable [tex]\(x\)[/tex] and determine the interval solution.
### Step-by-Step Solution:
1. Start with the given inequality:
[tex]\[
\frac{-4}{1-x} < 9
\][/tex]
2. Find the critical point where the denominator is zero:
[tex]\[
1 - x = 0 \implies x = 1
\][/tex]
This tells us that [tex]\(x = 1\)[/tex] is a point of discontinuity and we should consider this point when solving the inequality.
3. Isolate the variable [tex]\(x\)[/tex]:
First, we need to resolve the inequality by manipulating it algebraically:
[tex]\[
\frac{-4}{1-x} < 9
\][/tex]
Multiply both sides by the expression [tex]\((1-x)\)[/tex], noting that the sign of the inequality will depend on the sign of [tex]\((1-x)\)[/tex]. This step requires consideration of two cases:
Case 1: [tex]\(1 - x > 0\)[/tex] (i.e., [tex]\(x < 1\)[/tex])
[tex]\[
-4 < 9(1 - x)
\][/tex]
[tex]\[
-4 < 9 - 9x
\][/tex]
Add [tex]\(9x\)[/tex] to both sides:
[tex]\[
9x - 4 < 9
\][/tex]
Add 4 to both sides:
[tex]\[
9x < 13
\][/tex]
Divide by 9:
[tex]\[
x < \frac{13}{9}
\][/tex]
Case 2: [tex]\(1 - x < 0\)[/tex] (i.e., [tex]\(x > 1\)[/tex])
In this case, the inequality [tex]\(-4 < 9(1-x)\)[/tex] would be reversed if you multiply both sides by a negative number. This complicates the scenario and we should verify the feasible solution.
4. Combining the computed interval:
The first case gives us the interval:
[tex]\[
x < \frac{13}{9}
\][/tex]
We also have to consider [tex]\(x \ne 1\)[/tex] due to the undefined nature at this point. This means:
[tex]\[
x \in (-\infty, 1) \cup (1, \frac{13}{9})
\][/tex]
5. Write the solution in interval notation:
The solution set for the given inequality [tex]\(\frac{-4}{1-x} < 9\)[/tex] is as follows in interval notation:
[tex]\[
(-\infty, 1) \cup (1, \frac{13}{9})
\][/tex]
Therefore, the solution set is:
[tex]\[
(-\infty, 1) \cup (1, \frac{13}{9})
\][/tex]
This interval notation represents all values [tex]\(x\)[/tex] where the given inequality holds true, excluding the point where the function is undefined.
