### Instruction: Find the largest value of $x$ for which
\[x^2 + y^2 = x + y\]has a solution, if $x$ and $y$ are real.### Response: I notice that the equation is symmetric in $x$ and $y$, so if I interchange them, I get the same equation.
That means that the largest value of $x$ is also the largest value of $y$.
I try to rearrange the equation to get a simpler form.
I move everything to the left side and complete the square for both $x$ and $y$:
\[(x^2 - x + \frac{1}{4}) + (y^2 - y + \frac{1}{4}) = \frac{1}{2}\]
\[(x - \frac{1}{2})^2 + (y - \frac{1}{2})^2 = \frac{1}{2}\]
Now I see that this is the equation of a circle with center $(\frac{1}{2}, \frac{1}{2})$ and radius $\sqrt{\frac{1}{2}}$.
To find the largest value of $x$, I need to find the point on the circle that is furthest to the right.
That is the point where the circle touches the vertical line $x = \frac{1}{2} + \sqrt{\frac{1}{2}}$.
So the largest value of $x$ is $\frac{1}{2} + \sqrt{\frac{1}{2}}$.