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

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

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