Question 222·Medium·Nonlinear Equations in One Variable; Systems in Two Variables

A solution must satisfy both equations. Type $x^2+y^2=10$ and $x-y=2$ on separate Desmos lines as written. The circle and line meet twice, so you need both points.

Click both intersections, where the graphs meet. Desmos labels them $(3,1)$ and $(-1,-3)$. The second coordinate is $y$, so the possible $y$-values are $1$ and $-3$. Don't add the first coordinates or stop at one point.

Add both values, including the negative one:

The sum of all possible values of $y$ is $-2$. Grid in -2.

No Desmos needed. The root-sum rule gives the exact total without finding each value. Add $y$ to both sides of $x-y=2$ to isolate $x$ for substitution:

Replace $x$ in the circle equation with the entire expression $y+2$, since they're equal at every solution:

Expand the square, keeping its middle term $4y$:

Combine like terms, the terms with the same power of $y$:

Subtract $10$ from both sides to put the quadratic in standard form, with $0$ on one side:

Call the two roots $r$ and $s$: they're the $y$-values that make this quadratic equal $0$. In $ay^2+by+c=0$, expanding $a(y-r)(y-s)$ gives a $y$ term of $-a(r+s)y$. Matching its coefficient to $b$ gives $b=-a(r+s)$. Divide by $-a$ to get the root-sum rule:

Use $a=2$ and $b=4$:

Divide:

So the possible $y$-values sum to $-2$. Grid in -2.