To solve the inequality
[tex]\[
\frac{x+4}{x-2}<1
\][/tex]
we will proceed step by step.
Step 1: Move all terms to one side to set up the inequality for combining the fractions.
First, we subtract 1 from both sides to get:
[tex]\[
\frac{x+4}{x-2} - 1 < 0
\][/tex]
Step 2: Combine the terms over a common denominator.
Rewrite the expression on the left side:
[tex]\[
\frac{x+4}{x-2} - \frac{x-2}{x-2} < 0
\][/tex]
This simplifies to:
[tex]\[
\frac{(x+4) - (x-2)}{x-2} < 0
\][/tex]
Now, combine the terms in the numerator:
[tex]\[
\frac{x + 4 - x + 2}{x-2} < 0
\][/tex]
Simplify the numerator:
[tex]\[
\frac{6}{x-2} < 0
\][/tex]
Step 3: Determine where the fraction is negative.
The fraction [tex]\(\frac{6}{x-2}\)[/tex] is negative when the denominator is negative. So, we need:
[tex]\[
x - 2 < 0
\][/tex]
Solving this inequality, we get:
[tex]\[
x < 2
\][/tex]
Step 4: Consider the critical points and check the intervals.
The inequality [tex]\(\frac{6}{x-2}\)[/tex] is undefined at [tex]\(x=2\)[/tex]. Therefore, we have a critical point at [tex]\(x=2\)[/tex], and we need to consider the intervals around this point:
1. [tex]\(x < 2\)[/tex]
2. [tex]\(x > 2\)[/tex] (which we are not interested in because [tex]\(\frac{6}{x-2}\)[/tex] would be positive)
Conclusion:
The solution to the inequality [tex]\(\frac{6}{x-2} < 0\)[/tex] is all [tex]\(x\)[/tex] values that make the expression negative, which corresponds to:
[tex]\[
x < 2
\][/tex]
In interval notation, the solution set is:
[tex]\[
(-\infty, 2)
\][/tex]
Thus, the answer is:
The solution set is [tex]\((- \infty, 2)\)[/tex].
