$$-V D_V - P_1 D_1 - P_2 D_2 = 0 \ldots (i)$$ $$V C_V + P_1 C_1 + P_2 C_2 = 0 \ldots (ii)$$

Where
$P_1, D_1,$ and $C_1$ are value, duration, and convexity of the first bond
$P_2, D_2,$ and $C_2$ are value, duration, and convexity of the second bond
$V, D_V,$ and $C_V$ are value, duration, and convexity of the positioned to be hedged.

Working with USD million, the above equations, in this case, will be as follows:

$$-20 - 6P_1 - 4P_2 = 0 \ldots (i)$$ $$120 + 50P_1 + 20P_2 = 0 \ldots (ii)$$

Solving the above equations simultaneously, we get

$$P_1 = -1 \text{ and } P_2 = -3.5$$

Therefore, for the bank to hedge against its position, it must take a short position of 1 million in bond A and a short position of 3.5 million in bond B.

Note:
We use the elimination method to solve the simultaneous equations:
First, rewrite the equations as follows:

$$-6p_1 - 4p_2 = 20 \ldots \text{eq 1}$$ $$50p_1 + 20p_2 = -120 \ldots \text{eq 2}$$

Now we can choose to eliminate either $p_1$ or $p_2$. Let's go with $p_1$
From eq 1, we can make $p_1$ the subject as follows:

$$-6p_1 = 4p_2 + 20$$

so that $p_1 = \dfrac{[-4p_2 - 20]}{6}$

Let's now eliminate $p_1$ in equation 2 by replacing it with $\dfrac{[-4p_2 - 20]}{6}$:

$$50\left(\dfrac{[-4p_2 - 20]}{6}\right) + 20p_2 = -120$$

Multiply by 6 on both sides:

$$50[-4p_2 - 20] + 120p_2 = -720$$

Simplifying further,

$$\begin{aligned} -200p_2 - 1,000 + 120p_2 &= -720 \\ -80p_2 &= 280 \\ p_2 &= \dfrac{280}{-80} = -3.5 \\ p_1 &= \dfrac{[-4(-3.5) - 20]}{6} = -1 \end{aligned}$$