The answer is 'D'. Method: substitute y with x-5 in each of the equations for solving simultaneously. We get a quadratic equation of the form Ax2 +Bx+ C = 0. It has only one real solution if B2-4AC = 0. This holds true only with the last option where after substitution of y we get x2+6x+9 = 0. B2-4AC =36-4*9=0. So the parabola y= x2 +7x +4 has one real solution with the line y = x- 5. The solution point is (-3,-8).