After you write $x$ in terms of $y$, substitute that expression into $x^2 + y^2 = 10$ so that the equation only has $y$ in it.

Once you substitute, you should get a quadratic equation of the form $ay^2 + by + c = 0$. You can factor it or use the quadratic formula to find all possible $y$-values.

For a quadratic $ay^2 + by + c = 0$, the sum of its solutions for $y$ equals $-\dfrac{b}{a}$. You can use this to get the sum without extra work once you have the quadratic.

In Desmos, type x^2 + y^2 = 10 to graph the circle with radius $\sqrt{10}$ centered at the origin.

On a new line, type x - y = 2 (or y = x - 2) to graph the line.

Click on each point where the line intersects the circle. Desmos will display the coordinates $(x, y)$ of each intersection; note both $y$-values.

Add the two $y$-values shown by Desmos (you can type them into a new Desmos line as y1 + y2 using the actual numbers) to check that this matches the sum you found algebraically.

From the linear equation

solve for $x$ in terms of $y$:

We will substitute this expression for $x$ into the first equation.

Substitute $x = y + 2$ into $x^2 + y^2 = 10$:

Expand and combine like terms:

Subtract 10 from both sides:

Divide the whole equation by 2 to simplify:

Now you have a quadratic equation in $y$.

Factor the quadratic equation:

We look for two numbers that multiply to $-3$ and add to $2$. Those numbers are $3$ and $-1$, so

Set each factor equal to zero:

The question asks for the sum of all possible values of $y$:

So the sum of all possible values of $y$ is $-2$.