An infinite set of equations

Question

Let $x,y,p\in[0,1]$ and consider the following equation

$$px+(1-p)y = \frac{1}{2}$$

Question: I'm trying to find $x$ and $y$ such that the above is satisfied for all $p$.

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What I've tried:

If $p=0$, then $y=1/2$, $x\in[0,1]$

If $p=1$, then $x=1/2$, $y\in[0,1]$

If $p=1/2$, then $x+y = 1$

If $p=1/3$, then $x+2y = 3/2$

Answer 1

Yes, if $p=0$, you see that $y=\frac12.$ And if $p=1$, you see that $x=\frac12.$ at that point, you know that if $x,y$ exist that satisfy the above equation for all $p$, then $x$ must be $\frac12$, and $y$ must be $\frac12$. All you need to do now is to show that indeed, if $x=y=\frac12$, then the equation is satisfied for all $p$. Should be easy to do.

Answer 2

Treat the LHS as a polynomial in $p$, then $$(x-y)p+y=\frac12$$ Comparing coefficients (first constant term, then linear term, both in $p$, since the equation is true for all $p$) yields the only solution $x=y=\frac12$.