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

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Answer 1 · 2016-07-01T16:18:10

#x=4/3# or #x=-1#

To solve #y=4/x# and #y=3x-1# using substitution, just substitute value of #y# from first equation into second, which leads to

#4/x=3x-1# or multiplying each by #x# (we assome #x!=0#, we get

#4=3x^2-x# or

#3x^2-x-4=0# or

#3x^2+3x-4x-4=0# or

#3x(x+1)-4(x+1)=0# or

#(3x-4)(x+1)=0#, which means that #3x-4=0# or #x+1=0#

i.e. #x=4/3# or #x=-1#