How do you solve the system #10y=x^2# and #x^2-6=-2y#?

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Answer 1 · 2016-07-15T07:26:18

We have two solutions #(-sqrt5,1/2)# and #(sqrt5,1/2)#

As #10y=x^2# and #x^2-6=-2y#, putting value of #x^2# from first equation into second we get

#10y-6=-2y# or

#10y+2y+6-6=-2y+2y+6# or

#12y=6# or

#y=6/12=1/2#.

Hence, #x^2=10×1/2=5# or

#x^2-5=0# or

#x^2-(sqrt5)^2=0# or

#(x-sqrt5)(x+sqrt5)=0#

Hence, either #x-sqrt5=0# i.e. #x=sqrt5#

or #x+sqrt5=0# i.e. #x=-sqrt5#

Hence we have two solutions #(-sqrt5,1/2)# and #(sqrt5,1/2)#