The complete set of values of 'x' which satisfy the inequations: `5x+2<3x+8` and `(x+2)/(x-1)<4` is :

Options

A. `(-oo, 1)`
B. (2, 3)
C. `(-oo, 3)`
D. `(-oo, 1)cup(2, 3)`

Explanation

To solve the given inequalities and find the complete set of values of \( x \) that satisfy both inequalities, we will proceed step by step.

### Step 1: Solve the first inequality

The first inequality is:

\[ 5x + 2 < 3x + 8 \]

**Rearranging the inequality:**

1. Subtract \( 3x \) from both sides:

\[ 5x - 3x + 2 < 8 \]

This simplifies to:

\[ 2x + 2 < 8 \]

2. Next, subtract \( 2 \) from both sides:

\[ 2x < 6 \]

3. Finally, divide both sides by \( 2 \):

\[ x < 3 \]

### Step 2: Solve the second inequality

The second inequality is:

\[ \frac{x + 2}{x - 1} < 4 \]

**Rearranging the inequality:**

1. Move \( 4 \) to the left side:

\[ \frac{x + 2}{x - 1} - 4 < 0 \]

2. To combine the fractions, rewrite \( 4 \) as \( \frac{4(x - 1)}{x - 1} \):

\[ \frac{x + 2 - 4(x - 1)}{x - 1} < 0 \]

3. Simplifying the numerator:

\[ x + 2 - 4x + 4 = -3x + 6 \]

So we have:

\[ \frac{-3x + 6}{x - 1} < 0 \]

4. Factoring out \(-3\):

\[ \frac{-3(x - 2)}{x - 1} < 0 \]

5. Dividing both sides by \(-3\) (remember to flip the inequality sign):

\[ \frac{x - 2}{x - 1} > 0 \]

### Step 3: Determine the critical points

The critical points from the inequality \( \frac{x - 2}{x - 1} > 0 \) are:

- \( x = 2 \) (numerator is zero)
- \( x = 1 \) (denominator is zero)

### Step 4: Test intervals

We will test the intervals determined by these critical points: \( (-\infty, 1) \), \( (1, 2) \), and \( (2, \infty) \).

1. **Interval \( (-\infty, 1) \)**: Choose \( x = 0 \):
\[ \frac{0 - 2}{0 - 1} = \frac{-2}{-1} = 2 > 0 \quad \text{(True)} \]

2. **Interval \( (1, 2) \)**: Choose \( x = 1.5 \):
\[ \frac{1.5 - 2}{1.5 - 1} = \frac{-0.5}{0.5} = -1 < 0 \quad \text{(False)} \]

3. **Interval \( (2, \infty) \)**: Choose \( x = 3 \):
\[ \frac{3 - 2}{3 - 1} = \frac{1}{2} > 0 \quad \text{(True)} \]

### Step 5: Combine the results

From the first inequality, we found:

\[ x < 3 \]

From the second inequality, we found:

\[ x \in (-\infty, 1) \cup (2, \infty) \]

### Step 6: Find the intersection

Now we need to find the intersection of the two results:

1. From the first inequality: \( (-\infty, 3) \)
2. From the second inequality: \( (-\infty, 1) \cup (2, \infty) \)

The intersection is:

\[ (-\infty, 1) \cup (2, 3) \]

### Final Answer

Thus, the complete set of values of \( x \) that satisfy both inequalities is:

\[ \boxed{(-\infty, 1) \cup (2, 3)} \]