How do you solve this system of equations: #x+ y + z = 3 and x + y - z = 1 and x - y - z = - 1#?

3 Answers
Manasvini Nittala
Oct 23, 2017

here the answer is #x=y=z=1#

Explanation:

#x+y+z=3#
#x+y-z=1#
adding these 2 equations we get #2(x+y)=4# #rArr(x+y)=2#
#x-y-z=-1#
#x+y+z=3#
adding these 2 equations we get #2x=2 rArrx=1#
put #x=1# in #x+y=2# we get #y=1#
now put #x=y=1# in #x+y+z=3 rArr1+1+z=3rArr2+z=3rArrz=1#
thus it is the final answer

Alan P.
Oct 23, 2017

#(x,y,z)=(1,1,1)#

Explanation:

Given
[1]#color(white)("XXX")x+y+z=3#
[2]#color(white)("XXX")x+y-z=1#
[3]#color(white)("XXX")x-y-z=-1#

Adding [1] and [3]
[4]#color(white)("XXX")2x=2color(white)("xxx")rarrx=1#

Subtracting [2] from [1]
[5]#color(white)("XXX")2z=2color(white)("xxx")rarrz=1#

Substituting #1# for #x# (from [4])
and #1# for #z# (from [5])
into [1]
[6]#color(white)("XXX")1+y+1=3color(white)("xxx")rarr y=1#

sjc
Oct 23, 2017

#x=1,y=1,z=1#

Explanation:

we solve by elimination method

#x+y+z=3--(1)#

#x+y-z=1--(2)#

#x-y-z=-1---(3)#

#(1)+(3)#

#2x+0+0=2#

#:.x=1#

#(1)rarr1+y+z=3#

#=>y+z=2--(1a)#

#(2)rarr1+y-z=1#

#=>y-z=0--(2a)#

#(1a)+(2a)#

#2y=2=>y=1#

#:.z=2-1=1#

#x=1,y=1,z=1#

test for consistency

#(3)#

#LHS=1+1-1=0-1=-1=RHS#

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