How do you solve #2x+y=0# and #x-y=1# using substitution?

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Answer 1 · 2016-06-16T21:49:27

#x =1/3 and y = -2/3#

Write one of the equations with a single variable as the subject.
This could be :
#y = -2x " or " x = 1 + y " or " y = x-1#

I would use the fact that both equations can be written with y as the subject.

As # y = y# it follows that

#x - 1 = -2x#
#" "3x = 1#

#x = 1/3#

Now that we know the value of #x#, substitute into one of the equations to find the value for #y#

#y = -2 xx 1/3 = -2/3#

Answer 2 · 2016-06-16T21:55:25

#2x+y=0.......(i)#
#x-y=1.........(ii)#

From #(ii)# put #y=x-1# in #(i)#.

#2x+(x-1)=0#
#implies 3x=1#
#implies x=1/3#

Now,put #x=1/3# in either #(i)# or #(ii)# to find the value of #y#. I am going to put #x=1/3# in #(i)# but you can put in #(ii)# for practice.

#2(1/3)+y=0#
#implies y=-2/3#

#Solution# #Set={(1/3,-2/3)}#