Solve #{(y^2+x=12),(x^2+y=12):}# ?

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Answer 1 · 2016-10-30T00:02:59

#((x= 1/2 (1 + 3 sqrt[5]), y= 1/2 (1 - 3 sqrt[5])),(x= 1/2 (1 - 3 sqrt[5]), y= 1/2 (1 + 3 sqrt[5])))#

With

#{(y^2+x=12),(x^2+y=12):}#

subtracting both sides

#y^2-x^2+x-y=0# or

#(y^2-x^2)/(y-x)=1=x+y#

Now solving

#{(y^2+x=12),(x+y=1):}#

we have the solutions

#((x= 1/2 (1 + 3 sqrt[5]), y= 1/2 (1 - 3 sqrt[5])),(x= 1/2 (1 - 3 sqrt[5]), y= 1/2 (1 + 3 sqrt[5])))#