How do you solve the system 3x + z = 13, 2y + z = 10, x + y = 1?

Nov 21, 2015

x=1
y=0
z=10

Explanation:

Write all the equations

$3 \cdot x + z = 13$ equation 1
$2 \cdot y + z = 10$ equation 2
$x + y = 1$ equation 3

The goal is to have at least one equation with only x, y or z

We subtract equation 1 with equation 2 we get:

$3 \cdot x + z - \left(2 \cdot y + z\right) = 13 - 10$ ($e q u a t i o n 1 - e q u a t i o n 2$)
$3 \cdot x - 2 \cdot y = 3$ ($e q u a t i o n 1 - e q u a t i o n 2$)

Add equation 3 two times

$3 \cdot x - 2 \cdot y + 2 \cdot \left(x + y\right) = 3 + 2 \cdot 1$
$5 \cdot x = 5$
$x = 1$

$x + y = 1$ equation 3
$y = 1 - x = 1 - 1 = 0$
$y = 0$

$2 \cdot y + z = 10$
$2 \cdot 0 + z = 10$
$z = 10$

Check so that this works with the equation system,