Answer :
To solve the inequality [tex]\(\frac{x+8}{5x-1} > 0\)[/tex], we need to determine the values of [tex]\(x\)[/tex] that make the expression positive. Let's break it down and solve it step-by-step:
1. Identify the critical points:
Firstly, find the points where the numerator and the denominator are zero. These points divide the number line into intervals.
- The numerator [tex]\(x + 8 = 0 \Rightarrow x = -8\)[/tex]
- The denominator [tex]\(5x - 1 = 0 \Rightarrow x = \frac{1}{5}\)[/tex]
2. Intervals to consider:
The critical points divide the [tex]\(x\)[/tex]-axis into three intervals:
[tex]\[ (-\infty, -8), \quad (-8, \frac{1}{5}), \quad (\frac{1}{5}, +\infty) \][/tex]
3. Determine the sign of the expression in each interval:
- For [tex]\(x \in (-\infty, -8)\)[/tex]:
Both the numerator [tex]\(x + 8\)[/tex] and the denominator [tex]\(5x - 1\)[/tex] are negative. A negative divided by a negative is positive.
- For [tex]\(x \in (-8, \frac{1}{5})\)[/tex]:
The numerator [tex]\(x + 8\)[/tex] is positive while the denominator [tex]\(5x - 1\)[/tex] is negative. A positive divided by a negative is negative.
- For [tex]\(x \in (\frac{1}{5}, +\infty)\)[/tex]:
Both the numerator [tex]\(x + 8\)[/tex] and the denominator [tex]\(5x - 1\)[/tex] are positive. A positive divided by a positive is positive.
4. Combine the intervals where the expression is positive:
- We found the expression [tex]\(\frac{x+8}{5x-1} > 0\)[/tex] in the intervals [tex]\((- \infty, -8)\)[/tex] and [tex]\((\frac{1}{5}, + \infty)\)[/tex].
5. Exclude the critical points where the expression is undefined or zero:
- The critical point [tex]\(x = -8\)[/tex] makes the numerator zero, which results in the expression being zero, not positive.
- The critical point [tex]\(x = \frac{1}{5}\)[/tex] makes the denominator zero, which makes the expression undefined.
Thus, the solution of [tex]\(\frac{x+8}{5x-1} > 0\)[/tex] is:
[tex]\[ (-\infty, -8) \cup \left(\frac{1}{5}, +\infty\right) \][/tex]
This corresponds to [tex]\(x < -8\)[/tex] or [tex]\(x > \frac{1}{5}\)[/tex].
Therefore, the correct choice is:
[tex]\[ \boxed{x < -8 \text{ or } x > \frac{1}{5}} \][/tex]
1. Identify the critical points:
Firstly, find the points where the numerator and the denominator are zero. These points divide the number line into intervals.
- The numerator [tex]\(x + 8 = 0 \Rightarrow x = -8\)[/tex]
- The denominator [tex]\(5x - 1 = 0 \Rightarrow x = \frac{1}{5}\)[/tex]
2. Intervals to consider:
The critical points divide the [tex]\(x\)[/tex]-axis into three intervals:
[tex]\[ (-\infty, -8), \quad (-8, \frac{1}{5}), \quad (\frac{1}{5}, +\infty) \][/tex]
3. Determine the sign of the expression in each interval:
- For [tex]\(x \in (-\infty, -8)\)[/tex]:
Both the numerator [tex]\(x + 8\)[/tex] and the denominator [tex]\(5x - 1\)[/tex] are negative. A negative divided by a negative is positive.
- For [tex]\(x \in (-8, \frac{1}{5})\)[/tex]:
The numerator [tex]\(x + 8\)[/tex] is positive while the denominator [tex]\(5x - 1\)[/tex] is negative. A positive divided by a negative is negative.
- For [tex]\(x \in (\frac{1}{5}, +\infty)\)[/tex]:
Both the numerator [tex]\(x + 8\)[/tex] and the denominator [tex]\(5x - 1\)[/tex] are positive. A positive divided by a positive is positive.
4. Combine the intervals where the expression is positive:
- We found the expression [tex]\(\frac{x+8}{5x-1} > 0\)[/tex] in the intervals [tex]\((- \infty, -8)\)[/tex] and [tex]\((\frac{1}{5}, + \infty)\)[/tex].
5. Exclude the critical points where the expression is undefined or zero:
- The critical point [tex]\(x = -8\)[/tex] makes the numerator zero, which results in the expression being zero, not positive.
- The critical point [tex]\(x = \frac{1}{5}\)[/tex] makes the denominator zero, which makes the expression undefined.
Thus, the solution of [tex]\(\frac{x+8}{5x-1} > 0\)[/tex] is:
[tex]\[ (-\infty, -8) \cup \left(\frac{1}{5}, +\infty\right) \][/tex]
This corresponds to [tex]\(x < -8\)[/tex] or [tex]\(x > \frac{1}{5}\)[/tex].
Therefore, the correct choice is:
[tex]\[ \boxed{x < -8 \text{ or } x > \frac{1}{5}} \][/tex]
