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Answer 1 · 2018-02-08T11:54:04

See below.

#{(y=sqrt(x+2)),((y+x)(y-x)=0):}#

or squaring the first

#{(y^2=x+2),(y^2-x^2=0):}#

or

#x+2=x^2 rArr x^2-x-2=0# which solving gives

#x = {-1,2}# and then the solution pairs

#(-1,1)# and #(2,2)#