How to solve for these simultaneous equations I have the following set of equations
$$\pi_1 = \pi_3 + [1 - \alpha(1 - p)]\pi_4$$
$$\pi_2 = \alpha(1 - p)\pi_4$$
$$\pi_3 = \alpha(1 - p)]\pi_1$$
$$\pi_4 = [1 - \alpha(1 - p)]\pi_1 + \pi_2$$
$$\pi_1 + \pi_2 + \pi_3 + \pi_4 = 1$$
Using this, I'm supposed to get the answer
$$\pi_1 = \pi_4 = \frac{1}{2 + 2\alpha(1 - p)} \hspace{1.5cm} \pi_2 = \pi_3 = \frac{\alpha(1 - p)}{2 + 2\alpha(1 - p)} $$
But I can't seem to do this. I always just end up with all the $\pi$'s equaling each other.
How do I do this question
 A: In all honesty, "Just do it."
For example, if you substitute the third equation into the first, you get that
$$ [1-\alpha(1-p)](\pi_1-\pi_4)=0$$
For now, I'm going to assume that the coefficient is non-zero, which gives $\pi_1 = \pi_4$. The second and third equation then give that $\pi_2=\pi_3$. Substituting this into the last equation, we get that $\pi_1 = \frac {1}{ 2 +2\alpha(1-p)}$ (which is equal to $\pi_4$). And the value of $\pi_3 = \pi_2$ drops out immediately.
Now, if the coefficient is 0, then $[1-\alpha(1-p)]=0$, which gives $\pi_1=\pi_3$, $\pi_2=\pi_4$, and you have infinitely many solutions subject to $\pi_1+\pi_2 = \frac {1}{2}$.
A: We have to make this easier to read, so please forgive my rewriting of terms.
Let: $s = \pi_1, t = \pi_2, u = \pi_3, v = \pi_4, w = \alpha(1 - p)$.
Rewriting your system yields:
$\tag 1 s = u + (1 - w)v$
$\tag 2 t = wv$
$\tag 3 u = ws$
$\tag 4 v = (1 - w)s + wv$
$\tag 5 s + t + u + v = 1$ 
From $(4)$, we collect like terms on each side of the equation by subtracting $wv$ and get: 
$$\tag 6 v(1 - w) = (1 - w)s, \text{hence} ~v = s$$
Now, just substitute all the values you have into (5), yielding:
$u + (1-w)v + wv + ws + (1-w)s + wv = 1$, and collecting like terms and simplifying, we get:
$u + v + s + wv = 1$, and from $(3)$ we have $u = ws$, so we get:
$ws + s + v + wv = 1$
$s(1+w) + v(1+w) = 1$, or
$(s+v) = \frac{1}{1+w}$, but we know from (6) that $v = s$ and now have:
$2s = \frac{1}{1+w}$, or
$s = \frac{1}{2(1+w)}$
We now can easily substitute into $(2)$, yielding:
$t = w v = w s = w \frac{1}{2(1+w)}$, and
Substituting into $(3)$, yields:
$u = w s = w \frac{1}{2(1+w)}$
Regards
