How do you solve #y= 1/x# and #x+5y=6# using substitution?

1 Answer
Jul 27, 2016

Answer:

#(x,y)=color(green)(""(1,1)) " or " color(green)(""(5,1/5))#

Explanation:

Given
[1]#color(white)("XXX")y=1/x#
[2]#color(white)("XXX")x+5y=6#

Using [1] we can substitute #1/x# for #y# in [2]
[3]#color(white)("XXX")x+5/x=6#

Multiplying both sides by #x#
[4]#color(white)("XXX")x^2+5=6x#

Rearranging into standard quadratic form:
[5]#color(white)("XXX")x^2-6x+5=0#

Factoring
[6]#color(white)("XXX")(x-1)(x-5)=0#

Allowing the two solutions:
#{:(x=1,color(white)("XX"),x=5),(rarr y=1/1=1,,rarr y=1/5):}#