If `(x)/(2x+y+z) =(y)/(x+2y+z) =(z)/(x+y+2z)` then each terms is equal to

Options

A. `(1)/(4)`
B. `(1)/(3)`
C. `(1)/(2)`
D. None of these

Answer

To solve the equation \(\frac{x}{2x+y+z} = \frac{y}{x+2y+z} = \frac{z}{x+y+2z}\), we will set each term equal to a constant \(k\).

### Step-by-step Solution:

1. **Set each fraction equal to \(k\)**:
\[ \frac{x}{2x+y+z} = k \implies x = k(2x+y+z) \quad \text{(Equation 1)} \]
\[ \frac{y}{x+2y+z} = k \implies y = k(x+2y+z) \quad \text{(Equation 2)} \]
\[ \frac{z}{x+y+2z} = k \implies z = k(x+y+2z) \quad \text{(Equation 3)} \]

2. **Rearranging the equations**:
- From Equation 1:
\[ x = k(2x + y + z) \implies x - 2kx - ky - kz = 0 \implies x(1 - 2k) = ky + kz \implies x = \frac{ky + kz}{1 - 2k} \quad \text{(Equation 4)} \]
- From Equation 2:
\[ y = k(x + 2y + z) \implies y - kx - 2ky - kz = 0 \implies y(1 - 2k) = kx + kz \implies y = \frac{kx + kz}{1 - 2k} \quad \text{(Equation 5)} \]
- From Equation 3:
\[ z = k(x + y + 2z) \implies z - kx - ky - 2kz = 0 \implies z(1 + 2k) = kx + ky \implies z = \frac{kx + ky}{1 + 2k} \quad \text{(Equation 6)} \]

3. **Adding the equations**:
\[ x + y + z = \frac{ky + kz}{1 - 2k} + \frac{kx + kz}{1 - 2k} + \frac{kx + ky}{1 + 2k} \]
Combine the fractions:
\[ x + y + z = k\left(\frac{(y + z) + (x + z) + (x + y)}{1 - 2k}\right) \]
Simplifying gives:
\[ x + y + z = k\left(\frac{2x + 2y + 2z}{1 - 2k}\right) = \frac{2k(x + y + z)}{1 - 2k} \]

4. **Setting up the equation**:
\[ (1 - 2k)(x + y + z) = 2k(x + y + z) \]
If \(x + y + z \neq 0\), we can divide both sides by \(x + y + z\):
\[ 1 - 2k = 2k \implies 1 = 4k \implies k = \frac{1}{4} \]

### Final Result:
Thus, each term is equal to \(\frac{1}{4}\).