How do you solve #x+y+z=73, -2x+2y=8, x-z=-9#?

Question

1927 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-21T04:31:16

This is a system of linear equations.Let's label them as

eq1 : #x+y+z=73#
eq2 : #-2x+2y=8#
eq3 : #x-z=-9#

Now eq2 can be written as #-x+y=4# or #y=4+x# (divide by #2#)

and eq3 can be written as #z=x+9#

Now we replace the values of #y# and #z# in eq1 hence

#x+(4+x)+(x+9)=73#

#3x+13=73#

#3x=60#

#x=20#

Now replace the value of #x# in eq2 and eq3 to get the values of #y# and #z# hence

#y=4+x=4+20=24#

#z=x+9=20+9=29#

Finally the solution of the system is #(x,y,z)=(20,24,29)#