How do you solve the system of equations #x+ y + z = 82#, #x - y - z = 72#, and #x - y + z = 24#?

Question

669 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-04T07:49:18

#x = 77#, #y=29#, and #z=-24# or the ordered triple #(77,29,-24)#

To make it easier to discuss I'm going to number the equations:
(1) #x+y+z=82#
(2) #x-y-z=72#
(3) #x-y+z=24#

For these equations adding equations (1) and (2) will eliminate #y# and #z#:

(1) + (2):

#(x+y+z=82)# + #(x-y-z=72)#

Results in:

#2x = 154#

divide both sides by 2:

#x = 77#

Adding (1) and (3) will eliminate #y# and leave us with #x# and #z#.

(1) + (3):

#(x+y+z=82)# + #(x-y+z=24)#

results in:

#2x + 2z = 106#

divide both sides by 2:

#x+z = 53#

We already found that #x = 77# so we'll substitute that value:

#77+z=53#

Subtract 77 from both sides:

#z = -24#

Now we know #x = 77# and #z=-24# so we can substitute both of those values into (1) and solve for #y#:

#x+y+z=82\rightarrow 77+y+(-24) = 82#

Combine like terms on the left:

#53+y=82#

Subtract 53 from both sides:

#y = 29#

So our solution is #x = 77#, #y=29#, and #z=-24# or the ordered triple #(77,29,-24)#