Question 203·Medium·Nonlinear Equations in One Variable; Systems in Two Variables

Type $x+y=10$ and $x^2+y^2=58$ on separate lines, then click both intersections. Desmos shows $(3,7)$ and $(7,3)$. Each point is a solution: its coordinates satisfy both equations. Swapping the coordinates changes neither the sum nor the sum of squares, so keep both points.

Use $(7,3)$ and type $|7-3|$ on a new Desmos line. It prints $4$. At $(3,7)$, the difference $x-y$ is negative, but absolute value makes it the same positive distance. Swapping $x$ and $y$ changes the sign of their difference, not its absolute value. So $|x-y|$ is $4$. Grid in 4.

The target is $|x-y|$, so work toward $(x-y)^2$. Expand each square; the $2xy$ terms have opposite signs:

Subtract the squared sum from twice the sum of squares to get $(x-y)^2$:

This works because both quantities on the left are given.

Replace $x^2+y^2$ with $58$ and $x+y$ with $10$. Type $2(58)-10^2$ in Desmos; it prints $16$, so $(x-y)^2=16$. Don't stop at $16$: that's the gap squared, not the gap.

Because absolute value is nonnegative, $|x-y|=\sqrt{(x-y)^2}$. Type $\sqrt{16}$ on the next Desmos line; it prints $4$. So the distance between $x$ and $y$ is $4$. Grid in 4.