Question 1031015
Yes, from 
{{{x^2 + y^2 = 4}}} and {{{y - x = 1}}}
we solve the second one and substitute into the first one...we get
y = x + 1 and then
x^2 + (x+1)^2 = 4
x^2 + x^2 + 2x + 1 = 4
2x^2 + 2x - 3 = 0
This cannot be factored so we use the quadratic formula...
x = (-2 + sqrt(4 + 24))/4 = (-2 + sqrt(28))/4 = (-1 + sqrt(7))/2
and
x = (-2 - sqrt(4 + 24))/4 = (-2 - sqrt(28))/4 = (-1 - sqrt(7))/2
Plug in and solve for y, as in y = x + 1...