The number of integral solutions of the equation {x+1}+2x=4[x+1]-6 , is

Options

A. 0
B. 1
C. 2
D. 3

Explanation

To solve the equation \( \{x+1\} + 2x = 4[x+1] - 6 \), we will follow these steps:

### Step 1: Understand the Components of the Equation
The curly bracket notation \( \{x\} \) represents the fractional part of \( x \), which can be expressed as:
\[ \{x\} = x - [x] \]
where \( [x] \) is the greatest integer less than or equal to \( x \).

### Step 2: Rewrite the Equation
We can rewrite the equation using the definitions of the greatest integer function:
\[ \{x+1\} + 2x = 4[x+1] - 6 \]
Since \( [x+1] = [x] + 1 \), we can substitute this into the equation:
\[ \{x+1\} + 2x = 4([x] + 1) - 6 \]
This simplifies to:
\[ \{x+1\} + 2x = 4[x] + 4 - 6 \]
or
\[ \{x+1\} + 2x = 4[x] - 2 \]

### Step 3: Substitute the Fractional Part
Now, we know that \( \{x+1\} = x + 1 - [x+1] \):
\[ \{x+1\} = x + 1 - ([x] + 1) = x - [x] \]
Substituting this back into the equation gives:
\[ x - [x] + 2x = 4[x] - 2 \]
This simplifies to:
\[ 3x - [x] = 4[x] - 2 \]

### Step 4: Rearranging the Equation
Rearranging the equation yields:
\[ 3x + 2 = 5[x] \]
Thus, we can express \( [x] \) in terms of \( x \):
\[ [x] = \frac{3x + 2}{5} \]

### Step 5: Determine the Range of \( x \)
Since \( [x] \) is an integer, \( \frac{3x + 2}{5} \) must also be an integer. Therefore, \( 3x + 2 \) must be divisible by 5. Let \( 3x + 2 = 5k \) for some integer \( k \):
\[ 3x = 5k - 2 \quad \Rightarrow \quad x = \frac{5k - 2}{3} \]

### Step 6: Find the Integral Solutions
For \( x \) to be an integer, \( 5k - 2 \) must be divisible by 3. This leads to the congruence:
\[ 5k - 2 \equiv 0 \pmod{3} \]
Calculating \( 5 \mod 3 \) gives \( 2 \), so:
\[ 2k - 2 \equiv 0 \pmod{3} \quad \Rightarrow \quad 2k \equiv 2 \pmod{3} \quad \Rightarrow \quad k \equiv 1 \pmod{3} \]
Thus, \( k \) can be expressed as:
\[ k = 3m + 1 \quad \text{for some integer } m \]

### Step 7: Substitute Back to Find \( x \)
Substituting back for \( k \):
\[ x = \frac{5(3m + 1) - 2}{3} = \frac{15m + 5 - 2}{3} = \frac{15m + 3}{3} = 5m + 1 \]
Thus, the integral solutions for \( x \) are of the form \( x = 5m + 1 \).

### Step 8: Determine the Range of \( [x] \)
Now we need to find the values of \( m \) such that \( [x] \) remains an integer. Since \( [x] = m + 1 \), we need \( m \) to be an integer.

### Conclusion
The integral solution \( x = 1 \) corresponds to \( m = 0 \). Checking for other values of \( m \) will yield no additional integral solutions within the bounds of the greatest integer function.

Thus, the number of integral solutions of the equation is:
\[ \boxed{1} \]