Solve the compound inequality:
$\frac{4}{2(x-3)} \geq -10$ or $4x + 1 < -1$
The given problem requires finding the set of real numbers \(x\) that satisfy at least one of the two inequalities: \(\frac{4}{2(x-3)} \geq -10\) or \(4x + 1 < -1\). We will solve each inequality individually and then take their union to determine the final solution set. Note that for the first inequality, the denominator cannot be zero, so \(x - 3 eq 0\), which implies \(x eq 3\).
To solve the first inequality, \(\frac{4}{2(x-3)} \geq -10\), we first simplify the expression by dividing the numerator and denominator by 2:
\[\frac{2}{x-3} \geq -10.\]
Next, move \(-10\) to the left side:
\[\frac{2}{x-3} + 10 \geq 0.\]
To combine these into a single rational expression, find a common denominator:
\[\frac{2 + 10(x-3)}{x-3} \geq 0 \implies \frac{2 + 10x - 30}{x-3} \geq 0 \implies \frac{10x - 28}{x-3} \geq 0.\]
The critical points where the numerator is zero or the denominator is undefined are \(x = 2.8\) and \(x = 3\). Using these points to test the intervals \((-\infty, 2.8]\), \((2.8, 3)\), and \((3, \infty)\), we find that the inequality holds for \(x \in (-\infty, 2.8] \cup (3, \infty)\).
The second inequality is \(4x + 1 < -1\). We solve for \(x\) as follows:
\[4x < -1 - 1\]
\[4x < -2\]
\[x < -0.5.\]
Thus, the solution set for the second inequality is \((-\infty, -0.5)\).
We now combine the solutions from both inequalities using a logical union:
\[\text{Solution set} = (-\infty, 2.8] \cup (3, \infty) \cup (-\infty, -0.5).\]
Observe that \((-\infty, -0.5)\) is a subset of \((-\infty, 2.8]\). Therefore, the union of these intervals is simply:
\[(-\infty, 2.8] \cup (3, \infty).\]
This interval represents all real numbers that satisfy at least one of the original conditions.
In electrical circuit design, the voltage across a component must often stay within safe operating ranges. If a circuit has two potential modes of failure, represented by inequalities, the combined safe region is the union of the regions where each mode of failure is avoided. For example, if a transistor must operate where current \(x\) satisfies \(\frac{2}{x-3} \geq -10\) or \(x < -0.5\), understanding how to combine these intervals ensures the device remains within safe limits. This approach helps engineers map out safe operating areas across different circuit configurations.