Can you help with a math problem with Systems?

Question

solve using a method of your choice:

x^2-y^2-4x+6y-4=0

x^2+y^2-4x-6y+12=0

Answer 1

x² - y² - 4x + 6y - 4 =0 .............(1)

x² + y² - 4x - 6y + 12 = 0 ..........(2)

2x² - 8x + 8 = 0 .......(1) + (2)

x² - 4x + 4 = 0

(x - 2)(x -2) =0

therefore x = 2.

when x = 2,

2² + y² - 8 - 6y + 12 = 0

y² - 6y + 8 = 0

(y - 2)(y - 4) = 0

therefore y = 2 or y = 4

Answer 2

add these two equations

==>

2x^2-8x+8=0,

x=2

==>

y^2-6y+8=0,

y=2 or 4

Answer 3

If you add the two equations together, you'll notice that the y terms drop out, leaving you with one equation in terms of just x.  Factor and solve this, then use the x values to find the values of y.

Answer 4

x=2;y=6