How do you solve the system #xy = 9#, #x^2 + y^2 = 82#?

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Answer 1 · 2015-11-27T22:27:09

#(-1, -9)# and #(1,9)#
#(-9, -1)# and #(9, 1)#

This system can be solve by substitution

Eq 1: #xy= 9#

#=># Solve for #x#

#(xy)/y= 9/y#

#=> color(red)(x= 9/y)# Substitute into Eq.2

Eq 2: #x^2 +y^2= 82#

#color(red)((9/y)^2) +y^2 = 82 #

#=># #(9/y)(9/y)+y^2 = 82#

#=>81/y^2 +y^2 = 82#

#=>color(blue)(y^2)(81/y^2+y^2=82)#

#=>81+y^4 =82y^2#

#=>y^4 -82y^2+81=0#

Factor #(y^2-81)(y^2-1)= 0#

Set each factor equal to #0#
a) #( y^2-81)= 0 #
#y^2 = 81#
#y= -9 , 9#

b) #y^2 -1 = 0 #
#y^2 = 1#
#y= -1, 1#

When #y= -1#
#x= 9/(-1)= -9# ; #(-9, -1)#
When #y= 1#
#x= 9/(10= 9# ; #(9, 1)#

When #y= -9#
#x= 9/(-9) = -1# ; #(-1, -9)#

When #y= 9#
#x= 9/(9) = 1# ; #(1, 9)#