Question 104·Medium·Nonlinear Equations in One Variable; Systems in Two Variables

Type $x^2+y^2=65$ and $y=x+1$ on separate Desmos lines. Click both intersections, the points shared by the two graphs. Desmos shows approximately $(-6.179,-5.179)$ and $(5.179,6.179)$. Each shared point satisfies both equations. Since $x$ is the first coordinate, its possible values are about $-6.179$ and $5.179$.

Type $\frac{-1+\sqrt{129}}{2}$ on a new Desmos line. It displays about $5.179$, matching the larger $x$-coordinate. The graph rounds, but this expression keeps the square root exact. So the larger possible value of $x$ is $\frac{-1+\sqrt{129}}{2}$. Choice D.

The line gives $y=x+1$. Substitution means replacing $y$ with that whole expression in the other equation, so a solution must satisfy:

Expand the square: $x^2+x^2+2x+1=65$. Combine the $x^2$ terms: $2x^2+2x+1=65$. Subtract $65$ from both sides: $2x^2+2x-64=0$. Divide every term by $2$:

For a quadratic $ax^2+bx+c=0$, the quadratic formula finds its solutions. Here $a=1$, $b=1$, and $c=-32$. Keep both signs when you substitute: $x=\frac{-1\pm\sqrt{1^2-4(1)(-32)}}{2(1)}$. Simplify inside the square root:

Because $\sqrt{129}$ is positive, adding it gives the larger of the two values. So the larger possible value of $x$ is $\frac{-1+\sqrt{129}}{2}$. Choice D.