Solve : ` (x)/(6)ge (8-3x)/(2) +4`

Explanation

To solve the inequality \( \frac{x}{6} \geq \frac{8 - 3x}{2} + 4 \), we will follow these steps:

### Step 1: Rewrite the inequality
Start with the original inequality:
\[ \frac{x}{6} \geq \frac{8 - 3x}{2} + 4 \]

### Step 2: Move all terms to one side
Subtract \( \frac{8 - 3x}{2} + 4 \) from both sides:
\[ \frac{x}{6} - \left(\frac{8 - 3x}{2} + 4\right) \geq 0 \]

### Step 3: Simplify the right side
Combine the terms on the right side. First, rewrite \( 4 \) as \( \frac{24}{6} \) to have a common denominator:
\[ \frac{x}{6} - \left(\frac{8 - 3x}{2} + \frac{24}{6}\right) \geq 0 \]
Now, find a common denominator for \( \frac{8 - 3x}{2} \):
\[ \frac{8 - 3x}{2} = \frac{3(8 - 3x)}{6} = \frac{24 - 9x}{6} \]
So the inequality becomes:
\[ \frac{x}{6} - \left(\frac{24 - 9x}{6} + \frac{24}{6}\right) \geq 0 \]

### Step 4: Combine the fractions
Combine the fractions:
\[ \frac{x}{6} - \frac{24 - 9x + 24}{6} \geq 0 \]
This simplifies to:
\[ \frac{x - (48 - 9x)}{6} \geq 0 \]
\[ \frac{x + 9x - 48}{6} \geq 0 \]
\[ \frac{10x - 48}{6} \geq 0 \]

### Step 5: Multiply both sides by 6
Since 6 is positive, we can multiply both sides without changing the inequality:
\[ 10x - 48 \geq 0 \]

### Step 6: Solve for x
Add 48 to both sides:
\[ 10x \geq 48 \]
Now divide by 10:
\[ x \geq \frac{48}{10} \]
\[ x \geq 4.8 \]

### Step 7: Write the solution in interval notation
The solution can be expressed in interval notation as:
\[ x \in [4.8, \infty) \]

### Final Answer:
Thus, the solution to the inequality is:
\[ x \geq 4.8 \quad \text{or} \quad x \in [4.8, \infty) \]