How to derive Hicksian demand?
Expenditure minimization problem in the question is as follows :
\begin{eqnarray*} \min_{x\geq 0, y\geq 0} &amp; \ \ p_Xx + p_Yy \\ \text{s.t.}&amp; \ \ x + 3y \geq U \end{eqnarray*}
where $p_X &gt; 0$, $p_Y &gt; 0$ and $U \geq 0$ are given.
Since prices are positive, the cost minimizing choice will satisfy the condition that $x + 3y = U$. So, we can rewrite the above problem as :
\begin{eqnarray*} \min_{x\geq 0, y\geq 0} &amp; \ \ p_Xx + p_Yy \\ \text{s.t.}&amp; \ \ x + 3y = U \end{eqnarray*}
We can now substitute the value of $y = \frac{U-x}{3}$ in the objective and rewrite the problem as :
\begin{eqnarray*} \min_{0 \leq x \leq U} &amp; \ \ p_Xx + p_Y\left(\dfrac{U-x}{3}\right)  \end{eqnarray*}

Since the objective ${\color{red} {p_Xx + p_Y\left(\frac{U-x}{3}\right) =\frac{p_YU}{3} + \left(p_X- \frac{p_Y}{3}\right)x }}$  is linear in $x$, solution is straight forward :
\begin{eqnarray*} x^h(p_X, p_Y, U)  \begin{cases} = U &amp; \text{if } p_X &lt; \frac{p_Y}{3} \\ \in [0, U] &amp; \text{if } p_X = \frac{p_Y}{3} \\ = 0 &amp; \text{if } p_X &gt; \frac{p_Y}{3}\end{cases} \end{eqnarray*}