### 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 the equation of a circle, except the right-hand side is not a constant.
Maybe I can complete the square on both sides to get a more standard form.
To do that, I need to subtract $x$ and $y$ from both sides, and then add $\frac{1}{4}$ and $\frac{1}{4}$ to both sides.
This gives me
\[(x - \frac{1}{2})^2 + (y - \frac{1}{2})^2 = \frac{1}{2}\]
Now I can see that 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 $x$ that satisfies this equation, I need to find the rightmost point on the circle.
That happens when $y = \frac{1}{2}$, so I can plug that into the equation and solve for $x$.
I get
\[(x - \frac{1}{2})^2 + 0 = \frac{1}{2}\]
Taking the square root of both sides, I get two possible values for $x - \frac{1}{2}$: $\pm \frac{1}{\sqrt{2}}$.
Adding $\frac{1}{2}$ to both sides, I get two possible values for $x$: $\frac{1}{2} \pm \frac{1}{\sqrt{2}}$.
The largest one is $\frac{1}{2} + \frac{1}{\sqrt{2}}$, so that is my final answer.