### 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 this equation looks like a circle, since it has $x^2$ and $y^2$ terms with the same coefficient.
I wonder if I can rewrite it in the standard form of a circle, $(x-h)^2 + (y-k)^2 = r^2$, where $(h,k)$ is the center and $r$ is the radius.
To do that, I need to complete the square for both $x$ and $y$ terms.
I can start by moving the linear terms to the right side of the equation:
\[x^2 - x + y^2 - y = 0\]
Then I can add the missing terms to make perfect squares on both sides.
I need to add $(1/2)^2 = 1/4$ to both sides for the $x$ terms, and $(1/2)^2 = 1/4$ to both sides for the $y$ terms:
\[x^2 - x + \frac{1}{4} + y^2 - y + \frac{1}{4} = \frac{1}{4} + \frac{1}{4}\]
Now I can factor the left side and simplify the right side:
\[\left(x - \frac{1}{2}\right)^2 + \left(y - \frac{1}{2}\right)^2 = \frac{1}{2}\]
This is the equation of a circle with center $(1/2, 1/2)$ and radius $\sqrt{1/2}$.
To find the largest value of $x$ that satisfies this equation, I need to find the point on the circle that is farthest to the right.
This will happen when $y = 1/2$, since the center is on the horizontal line $y = 1/2$.
Plugging in $y = 1/2$ into the equation, I get:
\[\left(x - \frac{1}{2}\right)^2 + \left(\frac{1}{2} - \frac{1}{2}\right)^2 = \frac{1}{2}\]
Simplifying, I get:
\[\left(x - \frac{1}{2}\right)^2 = \frac{1}{2}\]
Taking the square root of both sides, I get two possible values for $x$:
\[x - \frac{1}{2} = \pm \sqrt{\frac{1}{2}}\]
Adding $1/2$ to both sides, I get:
\[x = \frac{1}{2} \pm \sqrt{\frac{1}{2}}\]