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

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Answer 1 · 2016-03-27T05:42:02

#(-1, -4)#
#(4, 1)#

We have two equations
#y=4/x" "#first equation
#y=x-3" "#second equation

Substitute first equation into the second

#y=x-3" "#second equation
#4/x=x-3" "#second equation

Multiply both sides of the equation by x

#x*(4/x)=x(x-3)#

#cancelx*(4/cancelx)=x(x-3)#

#4=x(x-3)#

#4=x^2-3x#

We have a quadratic equation

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

We can solve this by factoring method

#x^2-3x-4=0#
#(x+1 )(x-4)=0#

Equate each factor to 0 to find the roots

First factor

#(x+1 )=0#

#x=-1# first root
and #y=-4#

Second factor

#x-4=0#

#x=4# second root
and #y=1#

God bless....I hope the explanation is useful.