Let's solve the inequality [tex]\(\frac{3x + 8}{x - 4} \geq 0\)[/tex].
1. Find the critical points of the inequality: These points occur where the numerator and denominator are zero.
[tex]\[
3x + 8 = 0 \implies x = -\frac{8}{3}
\][/tex]
[tex]\[
x - 4 = 0 \implies x = 4
\][/tex]
2. Determine the intervals: The critical points split the number line into three intervals:
- [tex]\((- \infty, -\frac{8}{3})\)[/tex]
- [tex]\((- \frac{8}{3}, 4)\)[/tex]
- [tex]\((4, \infty)\)[/tex]
3. Test the sign of the expression in each interval:
- For the interval [tex]\((- \infty, -\frac{8}{3})\)[/tex]:
Pick a test point, [tex]\( x = -3 \)[/tex].
[tex]\[
\frac{3(-3) + 8}{-3 - 4} = \frac{-9 + 8}{-7} = \frac{-1}{-7} > 0
\][/tex]
The expression is positive in this interval.
- For the interval [tex]\((- \frac{8}{3}, 4)\)[/tex]:
Pick a test point, [tex]\( x = 0 \)[/tex].
[tex]\[
\frac{3(0) + 8}{0 - 4} = \frac{8}{-4} = -2 < 0
\][/tex]
The expression is negative in this interval.
- For the interval [tex]\((4, \infty)\)[/tex]:
Pick a test point, [tex]\( x = 5 \)[/tex].
[tex]\[
\frac{3(5) + 8}{5 - 4} = \frac{15 + 8}{1} = 23 > 0
\][/tex]
The expression is positive in this interval.
4. Include the critical points if they satisfy the inequality:
- At [tex]\( x = -\frac{8}{3} \)[/tex]:
[tex]\[
\frac{3 \left( -\frac{8}{3} \right) + 8}{ -\frac{8}{3} - 4} = \frac{-8 + 8}{-\frac{8}{3} - 4} = \frac{0}{ -\frac{8}{3} - 4} = 0 \geq 0
\][/tex]
The expression is zero, hence it satisfies the inequality and should be included.
- At [tex]\( x = 4 \)[/tex]:
The denominator becomes zero and the expression is undefined, so it cannot be included.
5. Combine the intervals:
From the above steps, the solution to the inequality is:
[tex]\[
x \leq -\frac{8}{3} \quad \text{or} \quad x > 4
\][/tex]
Thus, the correct answer is:
[tex]\[ x \leq -\frac{8}{3} \text{ or } x > 4 \][/tex]
