The correct answer is $StartFraction 35 Over 2 EndFraction$ . Subtracting the second equation from the first equation yields $(y + k) - (y - k) = x + 26 - (x^{2} - 5 x)$ , or $2 k = - x^{2} + 6 x + 26$ . This is equivalent to $x^{2} - 6 x + (2 k - 26) = 0$ . It's given that the system has exactly one distinct real solution; therefore, this equation has exactly one distinct real solution. An equation of the form $a x^{2} + b x + c = 0$ , where $a$ , $b$ , and $c$ are constants, has exactly one distinct real solution when the discriminant, $b squared minus 4 a c$ , is equal to $0$ . The equation $x^{2} - 6 x + (2 k - 26) = 0$ is of this form, where $a equals 1$ , $b = - 6$ , and $c = 2 k - 26$ . Substituting these values into the discriminant, $b squared minus 4 a c$ , yields ${(- 6)}^{2} - 4 (1) (2 k - 26)$ . Setting the discriminant equal to $0$ yields ${(- 6)}^{2} - 4 (1) (2 k - 26) = 0$ , or $- 8 k + 140 = 0$ . Subtracting $140$ from both sides of this equation yields $- 8 k = - 140$ . Dividing both sides of this equation by $negative 8$ yields $k = \frac{35}{2}$ . Note that 35/2 and 17.5 are examples of ways to enter a correct answer.