How do you solve the system #x^2+y^2+8x-20y+7=0# and #x^2+9y^2+8x+4y+7=0#?

Question

1751 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-06T12:11:48

#(y = 0, x = -7)# and #(y = 0, x = -1)#

Subtracting the first from the second we obtain

#8y^2+24y = 0# or
#8(y(y+3))=0# and the solutions are

#y = 0# and #y = -3#.

Now substituting those values in the first equation

1)
#y=0->x^2+8x+7=0# giving #x=-7# and #x=-1#
#y = -3->x^2+9+8x+60+7=0# giving no real solutions

Now substituting in the second equation

2)
#y=0->x^2+8x+7=0# with the already known solutions #x=-7# and #x=-1#.

Finally, the system has the following real solutions.

#(y = 0, x = -7)# and #(y = 0, x = -1)#