Find the intersection points, if any, for each system of equations

x^2/4 + y^2 = 1
y=x+1



x^2+y^2-16y+39=0
y^2-x^2-9=0

1) Use y = x+1 in the first equation then,

x^2/4 +(x+1)^2 = 1 ---->

x^2/4 + x^2 + 2*x +1 = 1 ---->

5/4 *x^2 + 2x = 0 ---> x*(5/4*x + 2) = 0, then,

x= 0 or -8/5 and the corresponding y = 1 or -3/5

Thus there are two intersection points one at (0,1) and the second at (-8/5,-3/5).

2) from the second equation, x^2 = y^2-9. Use it in the first one, then,

y^2 -9 +y^2 -16*y +39 = 0 ---->
2*y^2 -16*y +30= 0 ---->(divide all by 2)
y^2 -8*y + 15 = 0 ------>
(y-5)*(y-3) = 0 , then,
y= 5 or 3.
When y = 5, then x^2 = 25-9 = 16 ---> x= 4 or -4.
When y = 3, then x^2 = 9-9 = 0 ----> x =0.
Thus, there are three intersection points, one at (4,5), second at (-4,5) and the third at (0,3).

Best Regards.