Solving for $x_i$ and $y_i$ in this system: $(\mu-x_1)y_1+(x_2-\mu)y_2=d$, $x_1y_1+x_2y_2=\mu$, $y_1+y_2=1$ 
Suppose we have a system of equations:
  \begin{align}
(\mu-x_1)y_1+(x_2-\mu)y_2&=d \\
x_1y_1+x_2y_2&=\mu\\
y_1 + y_2 &=1\\
\end{align}
  The decision variables are $x_1$, $x_2$, $y_1$, $y_2$, while $\mu$ and $d$ are given parameters.

I think it's a non-convex QCQP but I don't know how to solve it, because it doesn't have a positive semidefinite matrix.
Can anyone simplify it to an LP or reformulate it as an SOCP or an SDP? (I know it can be formulated as an SDP. But LP and SOCP is better since SDP is difficult to solve)
 A: Solving the last two equations gives
$$y_1=\frac{m-x_2}{x_1-x_2} \qquad y_2=\frac{x_1-m}{x_1-x_2} \tag{1}$$ 
(which require $x_1\neq x_2$). Substituting into the first equation, and clearing fractions, gives
$$2(x_1-m)(x_2-m)=d(x_1-x_2) \tag{2}$$
Solving for, say, $x_2$ ...
$$x_2 = \frac{d x_1 + 2 m (x_1-m)}{d + 2 (x_1-m)} \tag{3}$$
Substituting this into $(1)$ ...
$$y_1 =-\frac{d}{2(x_1-m)} \qquad y_2 = 1-y_1 \tag{4}$$
which requires $x_1\neq m$ but leaves $x_1$ otherwise unconstrained. $\square$ 

Note. An edit to the question introduced the conditions $y_1\geq 0$ and $y_2\geq 0$; they have since been edited back out, but I'll include the discussion of them anyway. Those conditions, along with $y_1+y_2=1$, tell us that $y_1$ is constrained between $0$ and $1$ (inclusive). Thus, we must have
$$0\leq-\frac{d}{2(x_1-m)}\leq 1$$
Leaving the case of $d=0$ to the reader, we see otherwise that $x_1-m$ and $d$ must have opposite signs, and that
$$\begin{cases}
x_1 \geq m-\frac12d, &\text{for}\; d < 0 \\[6pt]
x_1 \leq m-\frac12d, &\text{for}\; d > 0
\end{cases}$$
Neither of these restricts $x_1$ to a particular value. $\square$
