solve the inequation `(2x+4)/(x-1) ge 5` and represent this solution on the number line.

Explanation

To solve the inequation \(\frac{2x+4}{x-1} \geq 5\), we will follow these steps:

### Step 1: Rearranging the Inequation
We start with the given inequation:
\[ \frac{2x + 4}{x - 1} \geq 5 \]
We can rearrange this by moving 5 to the left side:
\[ \frac{2x + 4}{x - 1} - 5 \geq 0 \]

### Step 2: Finding a Common Denominator
To combine the terms, we need a common denominator. The common denominator is \(x - 1\):
\[ \frac{2x + 4 - 5(x - 1)}{x - 1} \geq 0 \]

### Step 3: Simplifying the Numerator
Now, simplify the numerator:
\[ 2x + 4 - 5x + 5 = -3x + 9 \]
So, we rewrite the inequation as:
\[ \frac{-3x + 9}{x - 1} \geq 0 \]

### Step 4: Factoring the Numerator
We can factor out \(-3\) from the numerator:
\[ \frac{-3(x - 3)}{x - 1} \geq 0 \]
This can be rewritten as:
\[ \frac{3(x - 3)}{x - 1} \leq 0 \]

### Step 5: Finding Critical Points
The critical points occur when the numerator or denominator is zero:
- From \(3(x - 3) = 0\) we get \(x = 3\).
- From \(x - 1 = 0\) we get \(x = 1\).

### Step 6: Testing Intervals
Now we will test the intervals determined by the critical points \(x = 1\) and \(x = 3\):
1. **Interval \((- \infty, 1)\)**: Choose \(x = 0\):
\[ \frac{3(0 - 3)}{0 - 1} = \frac{3(-3)}{-1} = 9 \quad (\text{positive}) \]
2. **Interval \((1, 3)\)**: Choose \(x = 2\):
\[ \frac{3(2 - 3)}{2 - 1} = \frac{3(-1)}{1} = -3 \quad (\text{negative}) \]
3. **Interval \((3, \infty)\)**: Choose \(x = 4\):
\[ \frac{3(4 - 3)}{4 - 1} = \frac{3(1)}{3} = 1 \quad (\text{positive}) \]

### Step 7: Determining the Solution Set
We need the regions where the expression is less than or equal to zero:
- The expression is negative in the interval \((1, 3)\).
- At the critical points, we check:
- At \(x = 1\), the expression is undefined.
- At \(x = 3\), the expression equals zero.

Thus, the solution set is:
\[ x \in [1, 3] \]

### Step 8: Representing on the Number Line
To represent this solution on the number line, we will mark the points 1 and 3. Since 1 is not included (undefined), we use an open circle at 1, and since 3 is included, we use a closed circle at 3.

### Final Answer
The solution to the inequation is:
\[ x \in [1, 3] \]

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