SAT Question 2cb17792: Nonlinear Equations In One Variable And Systems Of Equations In Two Variables answer and explanation with Desmos solution

Subtracting the second equation from the first equation yields $\left(y+k\right)-\left(y-k\right)=x+26-\left(x^2-5x\right)$, or $2k=-x^2+6x+26$. This is equivalent to $x^2-6x+\left(2k-26\right)=0$. 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 $a{x}^2+bx+c=0$, where $a$, $b$, and $c$ are constants, has exactly one distinct real solution when the discriminant, $b^2-4ac$, is equal to $0$. The equation $x^2-6x+(2k-26)=0$ is of this form, where $a=1$, $b=-6$, and $c=2k-26$. Substituting these values into the discriminant, $b^2-4ac$, yields ${\left(-6\right)}^2-4\left(1\right)\left(2k-26\right)$. Setting the discriminant equal to $0$ yields ${\left(-6\right)}^2-4\left(1\right)\left(2k-26\right)=0$, or $-8k+140=0$. Subtracting $140$ from both sides of this equation yields $-8k=-140$. Dividing both sides of this equation by $-8$ yields $k=\frac{35}{2}$. Note that 35/2, 17.5, and 17.50 are examples of ways to enter a correct answer.