x2 y2 22x 13 0 and 2x y 4Solutionx2 y2 22x 13 0 2x y

x^2 - y^2 - 22x + 13 = 0 and 2x - y = 4

x^2 - y^2 -22x + 13 =0

2x -y =4

point ofintersection of two equations:

substitute y = 2x -4 in 1st equ.

x^2 -( 2x -4)^2 -22x +13 =0

x^2 -(4x^2 +16 -16x) -22x +13 =0

-3x^2 - 6x -3 =0

solve the quadratic: x = -1

So, y =2x -4 = 2*(-1) -4 = -6

Solution ( -1 , -6)