Question 1003758
Hi there,
x^2 + y^2 = 13
y = x + 1
Substitute (x + 1) for 'y'
in x^2 + y^2 = 13
x^2 +(x + 1)^2 = 13
Multiply out bracket
x^2 +(x^2 + 2x + 1)= 13
Remove brackets
x^2 + x^2 + 2x + 1 = 13
Collect like terms
2x^2 + 2x + 1 - 13 = 0
2x^2 + 2x - 12 = 0
Divide throughout by 2
2(x^2 + x - 6)= 0
2(x + 3)(x - 2) = 0
x + 3 = 0
x = -3
x - 2 = 0
x = 2
Substitute x = -3 and 2 
into the equation y = x + 1
to get the y coordinates.
x = -3
y = x + 1
y = -3 + 1
y = -2
x = 2
y = x + 1
y = 2 + 1
y = 3
Points of intersection are:
(-3,-2) and (2,3)
Hope this helps :-)