Solve the inequality algebraically.
$\frac{x+4}{x-11} \leq 1$
The problem asks for the set of all real values of \(x\) that satisfy the inequality:
\[\frac{x+4}{x-11} \leq 1\]
First, we note that the denominator cannot be zero, which implies \(x-11
eq 0\), so \(x
eq 11\). We will keep this constraint in mind throughout the solution.
To solve the inequality, we move all terms to one side to compare the expression with zero. Subtracting 1 from both sides gives:
\[\frac{x+4}{x-11} - 1 \leq 0\]
Next, we find a common denominator to combine the terms into a single rational expression:
\[\frac{x+4}{x-11} - \frac{x-11}{x-11} \leq 0\]
\[\frac{(x+4) - (x-11)}{x-11} \leq 0\]
\[\frac{x+4 - x + 11}{x-11} \leq 0\]
\[\frac{15}{x-11} \leq 0\]
We now examine the inequality \(\frac{15}{x-11} \leq 0\). Since the numerator is the positive constant 15, the sign of the entire fraction depends solely on the denominator \(x-11\).
For the quotient to be less than or equal to zero, the denominator must be negative (it cannot be zero, as division by zero is undefined). Thus, we require:
\[x - 11 < 0\]
Solving this for \(x\), we find:
\[x < 11\]
This inequality naturally satisfies the restriction \(x
eq 11\) determined at the beginning. Thus, all real numbers less than 11 are solutions to the given inequality. The solution in interval notation is \((-\infty, 11)\).
This type of algebraic inequality appears when analyzing thresholds in various systems. For example, suppose you are designing a storage tank where the concentration of a chemical is modeled by the function \(f(x) = \frac{x+4}{x-11}\) based on a parameter \(x\). If safety regulations mandate that the concentration must not exceed a value of 1, finding the range of safe operation involves solving the inequality \(\frac{x+4}{x-11} \leq 1\). This calculation ensures the system remains within safe limits.