Answer :
To solve the inequality [tex]\(\frac{x+8}{5x-1} > 0\)[/tex], we need to determine where the expression [tex]\(\frac{x+8}{5x-1}\)[/tex] is positive. The first step is to identify the critical points where the expression [tex]\(\frac{x+8}{5x-1}\)[/tex] either becomes zero or undefined.
1. Find the zeros of the numerator and the denominator:
- The numerator [tex]\(x + 8 = 0\)[/tex] gives the zero [tex]\(x = -8\)[/tex].
- The denominator [tex]\(5x - 1 = 0\)[/tex] gives the point where the expression is undefined, which is [tex]\(x = \frac{1}{5}\)[/tex].
2. Mark the critical points on a number line and divide the number line into intervals:
- The critical points are [tex]\(x = -8\)[/tex] and [tex]\(x = \frac{1}{5}\)[/tex].
- These points divide the number line into three intervals: [tex]\((-\infty, -8)\)[/tex], [tex]\((-8, \frac{1}{5})\)[/tex], and [tex]\((\frac{1}{5}, \infty)\)[/tex].
3. Test each interval to determine where the expression is positive:
Choose a test point from each interval and substitute it into the inequality [tex]\(\frac{x+8}{5x-1} > 0\)[/tex].
- Interval [tex]\((-\infty, -8)\)[/tex], choose [tex]\(x = -9\)[/tex]:
[tex]\[ \frac{-9+8}{5(-9)-1} = \frac{-1}{-46} = \frac{1}{46} > 0 \][/tex]
Hence, the expression is positive in this interval.
- Interval [tex]\((-8, \frac{1}{5})\)[/tex], choose [tex]\(x = 0\)[/tex]:
[tex]\[ \frac{0+8}{5(0)-1} = \frac{8}{-1} = -8 < 0 \][/tex]
Hence, the expression is negative in this interval.
- Interval [tex]\((\frac{1}{5}, \infty)\)[/tex], choose [tex]\(x = 1\)[/tex]:
[tex]\[ \frac{1 + 8}{5(1) - 1} = \frac{9}{4} > 0 \][/tex]
Hence, the expression is positive in this interval.
4. Consider the critical points:
- At [tex]\(x = -8\)[/tex], the numerator is zero, so [tex]\(\frac{x+8}{5x-1} = 0\)[/tex], which is not positive.
- At [tex]\(x = \frac{1}{5}\)[/tex], the expression is undefined.
5. Combine the results:
The expression [tex]\(\frac{x+8}{5x-1}\)[/tex] is positive in the intervals where [tex]\(x\)[/tex] lies in [tex]\((-\infty, -8)\)[/tex] and [tex]\((\frac{1}{5}, \infty)\)[/tex], and it does not include the points where the numerator or denominator are zero or make the expression undefined.
Therefore, the solution to the inequality [tex]\(\frac{x+8}{5x-1} > 0\)[/tex] is:
[tex]\[ x < -8 \text{ or } x > \frac{1}{5} \][/tex]
So, the correct answer is:
[tex]\[ x < -8 \text{ or } x > \frac{1}{5} \][/tex]
1. Find the zeros of the numerator and the denominator:
- The numerator [tex]\(x + 8 = 0\)[/tex] gives the zero [tex]\(x = -8\)[/tex].
- The denominator [tex]\(5x - 1 = 0\)[/tex] gives the point where the expression is undefined, which is [tex]\(x = \frac{1}{5}\)[/tex].
2. Mark the critical points on a number line and divide the number line into intervals:
- The critical points are [tex]\(x = -8\)[/tex] and [tex]\(x = \frac{1}{5}\)[/tex].
- These points divide the number line into three intervals: [tex]\((-\infty, -8)\)[/tex], [tex]\((-8, \frac{1}{5})\)[/tex], and [tex]\((\frac{1}{5}, \infty)\)[/tex].
3. Test each interval to determine where the expression is positive:
Choose a test point from each interval and substitute it into the inequality [tex]\(\frac{x+8}{5x-1} > 0\)[/tex].
- Interval [tex]\((-\infty, -8)\)[/tex], choose [tex]\(x = -9\)[/tex]:
[tex]\[ \frac{-9+8}{5(-9)-1} = \frac{-1}{-46} = \frac{1}{46} > 0 \][/tex]
Hence, the expression is positive in this interval.
- Interval [tex]\((-8, \frac{1}{5})\)[/tex], choose [tex]\(x = 0\)[/tex]:
[tex]\[ \frac{0+8}{5(0)-1} = \frac{8}{-1} = -8 < 0 \][/tex]
Hence, the expression is negative in this interval.
- Interval [tex]\((\frac{1}{5}, \infty)\)[/tex], choose [tex]\(x = 1\)[/tex]:
[tex]\[ \frac{1 + 8}{5(1) - 1} = \frac{9}{4} > 0 \][/tex]
Hence, the expression is positive in this interval.
4. Consider the critical points:
- At [tex]\(x = -8\)[/tex], the numerator is zero, so [tex]\(\frac{x+8}{5x-1} = 0\)[/tex], which is not positive.
- At [tex]\(x = \frac{1}{5}\)[/tex], the expression is undefined.
5. Combine the results:
The expression [tex]\(\frac{x+8}{5x-1}\)[/tex] is positive in the intervals where [tex]\(x\)[/tex] lies in [tex]\((-\infty, -8)\)[/tex] and [tex]\((\frac{1}{5}, \infty)\)[/tex], and it does not include the points where the numerator or denominator are zero or make the expression undefined.
Therefore, the solution to the inequality [tex]\(\frac{x+8}{5x-1} > 0\)[/tex] is:
[tex]\[ x < -8 \text{ or } x > \frac{1}{5} \][/tex]
So, the correct answer is:
[tex]\[ x < -8 \text{ or } x > \frac{1}{5} \][/tex]
