How do you solve the system #-x^2+y^2+10=0# and #-3y^2+x+1=0#?

Question

1683 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-18T08:24:17

#{x = 1/6 (1 + sqrt[373]), y = -1/3 sqrt[1/2 (7 + sqrt[373])]}# and
#{x = 1/6 (1 + sqrt[373]), y = 1/3 sqrt[1/2 (7 + sqrt[373])]}#

Suming both sides of

#3(-x^2+y^2+10)=0# and
#-3y^2+x+1=0#

we obtain

#-3x^2+x+31=0#

solving for #x#

#x=1/6 (1 pm sqrt[373])#

but

#y^2=(x+1)/3#

so

#y = pmsqrt((1+1/6 (1 + sqrt[373]))/3)#

so the solutions are

#{x = 1/6 (1 + sqrt[373]), y = -1/3 sqrt[1/2 (7 + sqrt[373])]}# and
#{x = 1/6 (1 + sqrt[373]), y = 1/3 sqrt[1/2 (7 + sqrt[373])]}#