Please solve this question-

(2x+1)/(7x-1)>5 and (x+7)/(x-8)>2

Find x?

To solve the inequalities \( \frac{2x+1}{7x-1} > 5 \) and \( \frac{x+7}{x-8} > 2 \), we'll tackle them one at a time, breaking down the steps clearly to find the values of \( x \) that satisfy each inequality.

First Inequality: \( \frac{2x+1}{7x-1} > 5 \)

Let's start with the first inequality:

1. Begin by isolating the fraction:

We can rewrite the inequality as:

\( \frac{2x+1}{7x-1} - 5 > 0 \)

2. To combine the terms, we need a common denominator:

Rewriting 5 as \( \frac{5(7x-1)}{7x-1} \) gives us:

\( \frac{2x+1 - 5(7x-1)}{7x-1} > 0 \)

3. Simplifying the numerator:

• Expand: \( 2x + 1 - (35x - 5) = 2x + 1 - 35x + 5 = -33x + 6 \)

4. Now, we have:

\( \frac{-33x + 6}{7x - 1} > 0 \)

5. Identify where the fraction changes sign:

• Set the numerator to zero: \( -33x + 6 = 0 \) gives \( x = \frac{6}{33} = \frac{2}{11} \)
• Set the denominator to zero: \( 7x - 1 = 0 \) gives \( x = \frac{1}{7} \)

6. To find the intervals, check the signs of the expression in the intervals determined by \( x = \frac{2}{11} \) and \( x = \frac{1}{7} \). The critical points are \( x = \frac{2}{11} \) and \( x = \frac{1}{7} \).

7. Testing the intervals:

• For \( x < \frac{1}{7} \), for example, \( x = 0 \): \( \frac{6}{-1} < 0 \) (not valid)
• For \( \frac{1}{7} < x < \frac{2}{11} \), for example, \( x = \frac{1}{8} \): \( \frac{6 - 33 \cdot \frac{1}{8}}{7 \cdot \frac{1}{8} - 1} \) is positive (valid)
• For \( x > \frac{2}{11} \), for example, \( x = 1 \): \( \frac{-27}{6} < 0 \) (not valid)

Thus, the solution for the first inequality is:

\( \frac{1}{7} < x < \frac{2}{11} \)

Second Inequality: \( \frac{x+7}{x-8} > 2 \)

Now let's move on to the second inequality:

1. Again, we rearrange the inequality:

\( \frac{x+7}{x-8} - 2 > 0 \)

2. Rewrite 2 as \( \frac{2(x-8)}{x-8} \):

\( \frac{x + 7 - 2(x - 8)}{x - 8} > 0 \)

3. Simplifying the numerator:

• Expand: \( x + 7 - 2x + 16 = -x + 23 \)

4. This gives us:

\( \frac{-x + 23}{x - 8} > 0 \)

5. Set the numerator and denominator to zero:

• Numerator: \( -x + 23 = 0 \) gives \( x = 23 \)
• Denominator: \( x - 8 = 0 \) gives \( x = 8 \)

6. Check the intervals defined by these critical points:

• For \( x < 8 \), for example, \( x = 0 \): \( \frac{23}{-8} < 0 \) (not valid)
• For \( 8 < x < 23 \), for example, \( x = 10 \): \( \frac{13}{2} > 0 \) (valid)
• For \( x > 23 \), for example, \( x = 25 \): \( \frac{-2}{17} < 0 \) (not valid)

The valid solution for the second inequality is:

\( 8 < x < 23 \)

Combining the Results

Now, we have two sets of solutions:

• From the first inequality: \( \frac{1}{7} < x < \frac{2}{11} \)
• From the second inequality: \( 8 < x < 23 \)

Since there is no overlap between the intervals \( \frac{1}{7} < x < \frac{2}{11} \) and \( 8 < x < 23 \), there are no values of \( x \) that satisfy both inequalities simultaneously. Therefore, the final answer is that there are no solutions for \( x \) that satisfy both inequalities. If you have any further questions or need clarification on any steps, feel free to ask!