Find x such that ` 3x-3 ge x +5 and (1)/(x -1) ge (1)/(8)`

Answer

To solve the inequalities \( 3x - 3 \geq x + 5 \) and \( \frac{1}{x - 1} \geq \frac{1}{8} \), we will break it down into steps.

### Step 1: Solve the first inequality \( 3x - 3 \geq x + 5 \)

1. Start by rearranging the inequality:
\[ 3x - 3 - x \geq 5 \]
This simplifies to:
\[ 2x - 3 \geq 5 \]

2. Next, add 3 to both sides:
\[ 2x \geq 8 \]

3. Now, divide both sides by 2:
\[ x \geq 4 \]

### Step 2: Solve the second inequality \( \frac{1}{x - 1} \geq \frac{1}{8} \)

1. Rearranging gives:
\[ \frac{1}{x - 1} - \frac{1}{8} \geq 0 \]

2. Finding a common denominator (which is \( 8(x - 1) \)):
\[ \frac{8 - (x - 1)}{8(x - 1)} \geq 0 \]
This simplifies to:
\[ \frac{9 - x}{8(x - 1)} \geq 0 \]

3. Now, we need to determine the critical points where the expression is zero or undefined:
- The numerator \( 9 - x = 0 \) gives \( x = 9 \).
- The denominator \( 8(x - 1) = 0 \) gives \( x = 1 \) (undefined).

4. We will analyze the sign of the expression \( \frac{9 - x}{8(x - 1)} \) in the intervals determined by the critical points \( x = 1 \) and \( x = 9 \):
- For \( x < 1 \): Both numerator and denominator are positive, so the expression is positive.
- For \( 1 < x < 9 \): The numerator is positive, and the denominator is negative, so the expression is negative.
- For \( x = 9 \): The expression equals zero.
- For \( x > 9 \): Both numerator and denominator are negative, so the expression is positive.

5. Thus, the solution for this inequality is:
\[ x \in (1, 9] \]

### Step 3: Find the intersection of the two solutions

1. From the first inequality, we have \( x \geq 4 \).
2. From the second inequality, we have \( x \in (1, 9] \).

3. The intersection of these two sets is:
\[ x \in [4, 9] \]

### Final Solution

The solution to the given inequalities is:
\[ x \in [4, 9] \]