### 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: The equation looks like a circle, but it is not in standard form.
To convert it to standard form, I need to complete the square for both $x$ and $y$ terms.
To complete the square for $x$, I need to add $(\frac{1}{2})^2 = \frac{1}{4}$ to both sides.
To complete the square for $y$, I need to add $(\frac{1}{2})^2 = \frac{1}{4}$ to both sides as well.
After doing that, I get
\[(x - \frac{1}{2})^2 + (y - \frac{1}{2})^2 = \frac{1}{2}\]
This is the equation of a circle with center $(\frac{1}{2}, \frac{1}{2})$ and radius $\frac{1}{\sqrt{2}}$.
To find the largest value of $x$ for which the circle has a solution, I need to find the rightmost point on the circle.
The rightmost point is where $y = \frac{1}{2}$, since that is the y-coordinate of the center.
Plugging in $y = \frac{1}{2}$ into the equation, I get
\[(x - \frac{1}{2})^2 + (\frac{1}{2} - \frac{1}{2})^2 = \frac{1}{2}\]
Simplifying, I get
\[(x - \frac{1}{2})^2 = \frac{1}{2}\]
Taking the square root of both sides, I get
\[x - \frac{1}{2} = \pm \frac{1}{\sqrt{2}}\]
Adding $\frac{1}{2}$ to both sides, I get
\[x = \frac{1}{2} \pm \frac{1}{\sqrt{2}}\]