How do you solve #3x + 2y = 7# and #x - y + 3 = 0 # using substitution?

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Answer 1 · 2017-03-31T09:55:32

#y=3 1/5,x=1/5#

#:.3x+2y=7#---------- (1)

#:.x-y+3=0#

#:.x-y=-3#---------(2)

----(2)# xx 3#:

#:.3x-3y=-9#--------(3)

-----#(1)-(3)# :

#:.3x+2y=7#
#:.3x-3y=-9#
#:. 5y=16#

#:.y=16/5#

#:.y=3 1/5#

substitute #y=3 1/5# in #(2)#

#:.x-(3 1/5)=-3#

#:.x=-3+3 1/5#

#:.x=1/5#

substitute #y=16/5,x=1/5# in #(1)#

#:.3(1/5)+2(16/5)=7#

#:.3/5+32/5=7 #

#:.35/5=7#

#:.7=7#