Question 86·Medium·Nonlinear Equations in One Variable; Systems in Two Variables

A solution to a system is a point on both graphs. Type $x^2+y=11$ and $y-x=1$ on separate Desmos lines, then click their intersections. Desmos shows about $(-3.702,-2.702)$ and $(2.702,3.702)$.

In $(x,y)$, $x$ is the first coordinate. So the point $(2.702,3.702)$ has the requested positive $x$, about $2.702$. Use the condition to choose the point, then read the coordinate the question asks for. The $3.702$ is its $y$-coordinate.

Type the two positive expressions from the choices. Desmos gives about $3.702$ for $\frac{1+\sqrt{41}}{2}$ and $2.702$ for $\frac{-1+\sqrt{41}}{2}$. The matching expression is the positive value of $x$ that satisfies the system: $\frac{-1+\sqrt{41}}{2}$. Choice D.

The second equation lets you get $y$ by itself. Add $x$ to both sides: $y=x+1$. You can use substitution, replacing $y$ in the other equation.

Both equations describe the same $y$ at a solution. Replace $y$ in $x^2+y=11$ with $x+1$: $x^2+(x+1)=11$.

Remove the parentheses: $x^2+x+1=11$.

Subtract $11$ from both sides: $x^2+x-10=0$. This is a quadratic equation, with an $x^2$ term.

For $ax^2+bx+c=0$, the quadratic formula gives $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$. Here $a=1$, $b=1$, and $c=-10$; keep the negative sign on $c$ when you substitute: $x=\frac{-1\pm\sqrt{1^2-4(1)(-10)}}{2(1)}$.

Multiply inside the square root and in the denominator: $x=\frac{-1\pm\sqrt{1+40}}{2}$.

Add inside the square root: $x=\frac{-1\pm\sqrt{41}}{2}$. The $\pm$ keeps both possible solutions until you apply the question's condition.

Because $\sqrt{41}>1$, the numerator $-1+\sqrt{41}$ is positive, while $-1-\sqrt{41}$ is negative. The denominator $2$ is positive, so the positive value of $x$ is $\frac{-1+\sqrt{41}}{2}$. Choice D.