If ` x /(y+z) = y/(z + x) = z /(x + y)` , then find the value of each fraction.

Answer

To solve the equation \( \frac{x}{y+z} = \frac{y}{z+x} = \frac{z}{x+y} \), we will follow these steps:

### Step 1: Set the common value
Let \( k \) be the common value of the fractions. Therefore, we can write:
\[ \frac{x}{y+z} = k, \quad \frac{y}{z+x} = k, \quad \frac{z}{x+y} = k \]

### Step 2: Express \( x \), \( y \), and \( z \) in terms of \( k \)
From the first equation, we can express \( x \):
\[ x = k(y + z) \]
From the second equation, we can express \( y \):
\[ y = k(z + x) \]
From the third equation, we can express \( z \):
\[ z = k(x + y) \]

### Step 3: Substitute \( x \), \( y \), and \( z \)
Now, we will substitute \( x \), \( y \), and \( z \) into one of the equations. Let's substitute \( x \) and \( y \) into the equation for \( z \):
\[ z = k(k(z + k(y + z))) \]
Substituting \( y = k(z + k(y + z)) \) into the equation will lead to a complex expression. Instead, we can use the equality of the fractions to derive a simpler relation.

### Step 4: Cross-multiply the first two fractions
Cross-multiplying \( \frac{x}{y+z} = \frac{y}{z+x} \):
\[ x(z + x) = y(y + z) \]
This simplifies to:
\[ xz + x^2 = y^2 + yz \]

### Step 5: Rearranging the equation
Rearranging gives:
\[ x^2 - y^2 + xz - yz = 0 \]
Factoring this, we get:
\[ (x - y)(x + y) + z(x - y) = 0 \]
Factoring out \( (x - y) \):
\[ (x - y)(x + y + z) = 0 \]

### Step 6: Solve for \( x \) and \( y \)
This gives us two cases:
1. \( x - y = 0 \) which implies \( x = y \)
2. \( x + y + z = 0 \)

### Step 7: Analyze the case \( x = y \)
If \( x = y \), we can substitute back into the original fractions:
\[ \frac{x}{y+z} = \frac{x}{x+z} = \frac{z}{x+x} \]
This leads to:
\[ \frac{x}{x+z} = \frac{z}{2x} \]
Cross-multiplying gives:
\[ 2x^2 = z(x + z) \]
This implies a relationship between \( x \), \( y \), and \( z \).

### Step 8: Conclude the values of \( x \), \( y \), and \( z \)
By symmetry and the nature of the equations, we find that:
\[ x = y = z \]
Thus, substituting back into the original fractions, we find:
\[ \frac{x}{y+z} = \frac{x}{2x} = \frac{1}{2} \]
So, each fraction equals \( \frac{1}{2} \).

### Final Answer
The value of each fraction is \( \frac{1}{2} \).

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