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

1 Answer
Nov 21, 2015

Answer:

x=1
y=0
z=10

Explanation:

Write all the equations

#3*x + z = 13# equation 1
#2*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*x + z -(2*y + z) = 13-10# (#equation 1- equation 2#)
#3*x - 2*y = 3# (#equation 1- equation 2#)

Add equation 3 two times

#3*x - 2*y + 2*(x+y)= 3 + 2*1#
#5*x =5#
#x = 1#

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

#2*y+z=10#
#2*0+z=10#
#z=10#

Check so that this works with the equation system,