How do you solve the system #y = x +4# and #2x + 3y = 22# by substitution?

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Answer 1 · 2015-05-21T11:18:02

You are given #y=color(red)(x+4)#, so substitute this for the value of #y# in
#2x+3y = 22#

#2x + 3color(red)((x+4)) = 22#

#2x+3x+12 =22#

#5x= 10#

#x=color(blue)(2)#

Now substitute this value for #x# back into #y=x+4#
#y=color(blue)(2)+4#

#y=6#