Solve algebratically the simultaneous equations

x^2 + y^2 =  25    ⇒ y^2  =  25 - x^2     (1)

5x - y = 11     ⇒  5x - 11  = y     (2)

Square  both sides of (2)  and we have that

(5x - 11)^2  = y^2

25x^2 - 110x + 121  = y^2          set this equal to (1)

25x^2 - 110x + 121  =  25 - x^2     add x^2 to both sides, subtract 25 from both sides

26x^2 -110x + 96 = 0      divide through by 2

13x^2 - 55x +  48  =  0      factor as

(13x  - 16) (x - 3)  = 0

Set both factors to 0, solve for x    and we have that  x = 16/13 and x  = 3

So using   5x - 11  = y 

5(16/13) - 11 =   80/13 - 11  =  [ 80 - 143] / 13 =  -63/13

5 (3) - 11 =  15 - 11  =  4

So.......the solutions are  (16/13, -63/13) and ( 3, 4)