Example 04- Simultaneous equations

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simultaneous equations

three equations, 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

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Add the three equations
$x(y + z) + y(x + z) + z(x + y) = 12 + 6 + 10$

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

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

$xy + yz + xz = 14$ → 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

‹ Example 03 - Simultaneous equations up Smallest number for given remainders ›

Submitted by Romel Verterra on June 15, 2012 - 10:47am

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Example 04- Simultaneous equations

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Example 04- Simultaneous equations

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Algebra

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