# How do you solve (1) #x + 2y + z = 1#, (2) #2x - 3y - z = 6#, (3) #3x + 5y + 4z = 5#, (4) #4x + y + z = 8#?

##### 1 Answer

Aug 6, 2016

#### Explanation:

Given:

(1):#color(white)(X)x+2y+z=1#

(2):#color(white)(X)2x-3y-z=6#

(3):#color(white)(X)3x+5y+4z=5#

(4):#color(white)(X)4x+y+z=8#

First note that we have

Note that: **(1)** + **(1)** + **(2)** is **(4)**. So there's our expected redundancy. We can ignore **(4)** and work with the first three equations.

Subtract **(1)** from **(2)** to get:

(5):#color(white)(X)-7y-3z=4#

Subtract **(1)** from **(3)** to get:

(6):#color(white)(X)-y+z=2#

Add **(6)** to **(5)** to get:

#-10y = 10#

So

Hence from **(6)** we have

Then from **(1)** we have

So: