Perfect Complements - Walrasian Equilibrium For a homework , I struggled to solve the following question but couldn't go further:

endowment of person 1 = (30,0)

endowment of person 2 = (0,20)


utility functions are such that:
\begin{eqnarray*}
    U (a_1,b_1) &amp; = &amp; \min(a_1,b_1) \\
\\
    U (a_2,b_2) &amp; = &amp; \min(4a_2,b_2).
\end{eqnarray*}
What I am doing is setting a1 equal to b1 and 4a2 equal to b2. After that I'm writing these;
$$
p_1 a_1 + p_2 b_1 = 30p_1 \mbox{ and } p_1a_2 + p_24a_2 = 20p_2
$$
Finally, by looking at feasibility condition, I'm writing down:
\begin{eqnarray*}
    \frac{30p_1}{p_1+p_2} + \frac{20p_2}{p_1+4p_2} &amp; = &amp; 30 \\
\\
    \frac{30p_1}{p_1+p_2} + \frac{4 \cdot 20p_2}{p_1+4p_2} &amp; = &amp; 20.
\end{eqnarray*}

Here, if I do some calculations, they result in $p_1 = p_2$ and accordingly $a_1=b_1=15, a_2=4$ and $b_2=16.$ But then the problem is that there is excess demand for good2 and there isn't a WE.

Alternatively, I am considering to have $p_2 = 0$ so that there exist a Walrasian Equilibrium.

BUT, I'm stuck at this point and lack the correct intuition for solving next steps. Or, I might be in a completely incorrect way. Please, explain me what will be done with this excess demand?

Thank you in advance.
If the prices $p_1$ and $p_2$ are positive than as you pointed out the equations
\begin{eqnarray*}
    \frac{30p_1}{p_1+p_2} + \frac{20p_2}{p_1+4p_2} &amp; = &amp; 30 \\
\\
    \frac{30p_1}{p_1+p_2} + \frac{4 \cdot 20p_2}{p_1+4p_2} &amp; = &amp; 20.
\end{eqnarray*}
hold. This is troublesome, because subtracting the first equation from the second yields
\begin{eqnarray*}
\frac{3 \cdot 20p_2}{p_1+4p_2} &amp; = &amp; -10
\end{eqnarray*}
which has no solution for positive prices. However if a good has zero price, than the demand equations are different as no one minds buying superflous goods as long as they are free. For $p_2 = 0$ you would have
\begin{eqnarray*}
    \frac{30p_1}{p_1+p_2} + \frac{20p_2}{p_1+4p_2} &amp; = &amp; 30 \\
\\
    \frac{30p_1}{p_1+p_2} + \frac{4 \cdot 20p_2}{p_1+4p_2} &amp; \leq &amp; 20.
\end{eqnarray*}
This is because $\frac{30p_1}{p_1+p_2} + \frac{4 \cdot 20p_2}{p_1+4p_2}$ is a minimum that the buyers need but they do not mind having more as good 2 is free. There is also a formulation of equilibrium, the 'value-equilibrium' which is that for good $i$
\begin{eqnarray*}
p_i \cdot D_i(p) = p_i \cdot S_i(p) \mbox{ and } D_i(p) \leq S_i(p)
\end{eqnarray*}
meaning that there can be oversupply of zero price goods. 
(Think about air. Would you buy air from someone?)

Unfortunately this equation system would still imply
\begin{eqnarray*}
\frac{3 \cdot 20p_2}{p_1+4p_2} &amp; \leq &amp; -10
\end{eqnarray*}
which has no solution in non-negative prices. On the other hand, for $p_1 = 0$ you would have
\begin{eqnarray*}
    \frac{30p_1}{p_1+p_2} + \frac{20p_2}{p_1+4p_2} &amp; \leq &amp; 30 \\
\\
    \frac{30p_1}{p_1+p_2} + \frac{4 \cdot 20p_2}{p_1+4p_2} &amp; = &amp; 20.
\end{eqnarray*}
which implies
\begin{eqnarray*}
\frac{3 \cdot 20p_2}{p_1+4p_2} &amp; \geq &amp; -10.
\end{eqnarray*}
As $p_1 = 0$ any positive $p_2$ is a solution for this.