### 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: This problem involves a quadratic equation and some inequalities, so I will try to manipulate the equation to get a more useful form.
I can start by moving everything to one side and simplifying:
\[x^2 + y^2 - x - y = 0\]
\[(x - \frac{1}{2})^2 + (y - \frac{1}{2})^2 = \frac{1}{2}\]
Now I can see that this equation represents a circle with center $(\frac{1}{2}, \frac{1}{2})$ and radius $\frac{1}{\sqrt{2}}$.
To find the largest value of $x$ that satisfies the equation, I need to find the rightmost point on the circle.
This point will have the same $y$-coordinate as the center, so I can plug in $y = \frac{1}{2}$ and solve for $x$:
\[(x - \frac{1}{2})^2 + (\frac{1}{2} - \frac{1}{2})^2 = \frac{1}{2}\]
\[(x - \frac{1}{2})^2 = \frac{1}{2}\]
\[x - \frac{1}{2} = \pm \frac{1}{\sqrt{2}}\]
\[x = \frac{1}{2} \pm \frac{1}{\sqrt{2}}\]