How do you solve #2x + y = 7# and #y - x = 1 # using substitution?

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Answer 1 · 2018-03-08T16:57:51

#therefore# #color(darkred)(x=2 and y=3#

#2x+y=7#

#y=7-2x#

Given that,

#y-x=1#

Substituting #y=7-2x#

#7-2x-x=1#

#7-3x=1#

#-3x=1-7#

#-3x=-6#

#x=(-6)/-3#

#color(magenta)(x=2#

#y=7-2x#

#y=7-2xx2#

#y=7-4#

#color(magenta)(y=3#

#therefore# #color(darkred)(x=2 and y=3#

~Hope this helps! :)

Answer 2 · 2018-03-08T17:59:52

#y=3,x=2#

#2x+y=7------(1)#
#-x+y=1--------(2)#

#(1)-(2)#

#3x=6#

#x=6/3#

#x=2#

substitute #x=2# in #(2)#

#:.-(2)+y=1#

#:.y=1+2#

#:.y=3#
~~~~~~~~~~~
check:-

substitute #y=3 and x=2 # in (1)

#:.2(2)+(3)=7#

#:.4+3=7#