How do you solve the system #x^2-4y^2-20x-64y-172=0# and #4x^2+y^2-80x+16y+400=0#?

Question

2205 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-11T11:45:29

#((x=6,y=-8),(x=14,y=-8))#

Calling #x2=x^2# and #y2=y^2# we have the system

#{ (x2 - 4 y2 - 20 x - 64 y - 172 = 0), (4 x2 + y2 - 80 x + 16 y + 400 = 0) :}#

we obtain

#((x2 = 4 (5 x-21)), (y2 = -(64 +16 y)))#

Solving now

#x^2-20 x+84 = 0# and
#y^2+16 y +64 = 0#

we obtain

#((x=6),(x=14))#

and a double root

#y = -8#
The solutions are

#((x=6,y=-8),(x=14,y=-8))#