# Example 04 - Simultaneous Non-Linear Equations of Three Unknowns

**Problem**

Solve for *x*, *y*, and *z* from the following system of equations.

$x(y + z) = 12$ → Equation (1)

$y(x + z) = 6$ → Equation (2)

$z(x + y) = 10$ → Equation (3)

**Solution**

$x(y + z) + y(x + z) + z(x + y) = 12 + 6 + 10$

$(xy + xz) + (xy + yz) + (xz + yz) = 28$

$2xy + 2yz + 2xz = 28$

$xy + yz + xz = 14$ → Equation (4)

Equation (4) - Equation (3)

$(xy + yz + xz) - z(x + y) = 14 - 10$

$(xy + yz + xz) - (xz + yz) = 4$

$xy = 4$ → Equation (5)

Equation (4) - Equation (1)

$(xy + yz + xz) - x(y + z) = 14 - 12$

$(xy + yz + xz) - (xy + xz) = 2$

$yz = 2$ → Equation (6)

Equation (4) - Equation (2)

$(xy + yz + xz) - y(x + z) = 14 - 6$

$(xy + yz + xz) - (xy + yz) = 8$

$xz = 8$ → Equation (7)

Multiply equations (5), (6), and (7)

$(xy)(yz)(xz) = 4(2)(8)$

$x^2 y^2 z^2 = 64$

$(xyz)^2 = 64$

$xyz = \pm 8$ → Equation (8)

Divide Equation (6) from Equation (8)

$\dfrac{xyz}{yz} = \dfrac{\pm 8}{2}$

$x = \pm 4$ *answer*

Divide Equation (7) from Equation (8)

$\dfrac{xyz}{xz} = \dfrac{\pm 8}{8}$

$y = \pm 1$ *answer*

Divide Equation (5) from Equation (8)

$\dfrac{xyz}{xy} = \dfrac{\pm 8}{4}$

$z = \pm 2$ *answer*