How do you solve #y= x+3# and #x-4y=0 # using substitution?

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Answer 1 · 2017-08-22T06:57:11

#x = -4#

#y = -1#

#y = x+3" "(1)#

#x-4y = 0" "(2)#

Substitute the value of #y# in equation #(2)#. We get

#x-4(x+3) = 0#

#x-4x-12 = 0#

#-3x = 12#

#x = -4#

Substitute the value in equation #(1)#. We get

#y = -4+3#

#y = -1#