(TEN POINTS!!!!) Intersection points for conic problems

-x^2+y^2-36=0
x^2+y^2-32y+156=0

and the second one is
x^2+y^2=17
y=x+3

On these systems, use addition to eliminate, or use substitution.
1) this one is an hyperbola and a circle.
-x^2+y^2. -36=0
x^2+y^2-32y+156=0
---------------------------
2y^2 -32y +120=0
Y^2 -16y +60= 0
(y-6)(y-10)=0
Y=6 or 10

Plug back in to find x
Y=6: -x^2 +36 -36= 0
X=0
(0,6)

Y= 10: -x^2 +100-36= 0
X^2 = 64
X=+/-8
(8,10) and (-8,10)

(So there are three points of intersection)

2) and the second one is a circle and a line.
x^2+y^2=17
y=x+3

Use substitution: x^2 + (x+3)^2 = 17
X^2+ x^2 +6x + 9= 17

2x^2 + 6x -8= 0
X^2 +3x -4=0

(x+4)(x-1)=0
X=-4 or 1
Y=X+3
Y= -1 or 4
(-4,-1) and (1,4) are the points of intersection.

Hoping this helps!