How do you solve #y=5-x# and #1=4x+3x# using substitution?

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Answer 1 · 2017-01-25T13:13:59

#y=4 6/7, x= 1/7#

#y=5-x,1=4x+3x#

#:.y=5-x#------(1)

#:.7x=1#

#x= 1/7#

substitute #x= 1/7# in-----(1)

#:.y=5-(1/7)#

#:.y=35/7-1/7#

#:.y=34/7#

#:.y=4 6/7#

check:

substitute #y= 34/7, x=1/7 # in--------(1)

#:.34/7=35/7-1/7#

#:.34/7=34/7#