If `x(x+y+z) =4, y(x+y+z)=16 and z (x+y+z)=29` and x, y & z are positive numbers. Find x, y & z=?

Answer

To solve the equations given in the problem, we will follow these steps:

1. **Write down the equations:**
- \( x(x+y+z) = 4 \) (Equation 1)
- \( y(x+y+z) = 16 \) (Equation 2)
- \( z(x+y+z) = 29 \) (Equation 3)

2. **Add all three equations:**
\[ x(x+y+z) + y(x+y+z) + z(x+y+z) = 4 + 16 + 29 \]
This simplifies to:
\[ (x+y+z)(x+y+z) = 49 \]

3. **Let \( S = x+y+z \). Then we have:**
\[ S^2 = 49 \]
Taking the square root of both sides:
\[ S = \sqrt{49} = 7 \]

4. **Substituting \( S \) back into the equations:**
- From Equation 1:
\[ x \cdot 7 = 4 \implies x = \frac{4}{7} \]

- From Equation 2:
\[ y \cdot 7 = 16 \implies y = \frac{16}{7} \]

- From Equation 3:
\[ z \cdot 7 = 29 \implies z = \frac{29}{7} \]

5. **Final values:**
- \( x = \frac{4}{7} \)
- \( y = \frac{16}{7} \)
- \( z = \frac{29}{7} \)

Thus, the values of \( x, y, \) and \( z \) are:
- \( x = \frac{4}{7} \)
- \( y = \frac{16}{7} \)
- \( z = \frac{29}{7} \)