To use a simple example, assume that consumer $i$ maximizes
$$U(x_i,y_i,I_i) = \alpha\ln x_i +(1-\alpha)\ln y_i + (\bar I_i-p_xx_i -p_yy_i)\\
s.t. \bar I_i \geq p_xx_i +p_yy_i$$
In other words, it may spend all his income on the two goods (but not more), or he may keep some for other goods, not modeled here. Due to the inequality in the constraint we need to use a Karush-Kuhn-Tucker non-negative multiplier, rather than the usual Lagrange multiplier.
Assuming a fixed income, the lagrangean function is
$$\Lambda = \alpha\ln x_i +(1-\alpha)\ln y_i + (\bar I_i-p_xx_i -p_yy_i) +\lambda_i(\bar I_i-p_xx_i -p_yy_i)$$
Assuming income is fixed, the first order conditions are
$$\frac {\alpha}{x_i} - (1+\lambda_i)p_x \leq 0,\;\; \frac {1-\alpha}{y_i} - (1+\lambda_i)p_y \leq 0$$
$$\lambda(\bar I_i-p_xx_i -p_yy_i) = 0 ,\; x_i\cdot \left(\frac {\alpha}{x_i} - (1+\lambda_i)p_x\right)=0, \; y_i\cdot \left(\frac {1-\alpha}{y_i} - (1+\lambda_i)p_y\right) =0$$
it follows that at the optimum both goods will be demanded in positive quantities, which in turn requires that the first-derivatives be set equal to zero. This gives us
$$\frac {\alpha}{p_xx_i} = \frac {1-\alpha}{p_yy_i} \implies x_i^D=\frac {\alpha}{1-\alpha}\frac {p_y}{p_x}y_i^D \tag{1}$$
from which we can obtain the optimal relation for the Expenditure $E_i$ on the two goods
$$E_i^* = \frac {1}{1+\lambda_i^*} , \;\; E_i^* \leq \bar I_i$$
Since $\lambda_i \geq 0 \implies \max E_i^* \leq 1$ so if it so happens that $\bar I_i >1$ then not all income will be spent on the two goods, the constraint won't be binding and so $\lambda_i^* =0 $. (Is $\max E_i^* \leq 1$ a strange result?) On the other hand if $\bar I_i \leq 1$ then we will necessarily have $E_i^* = \bar I_i$, and $\lambda_i^* = (1-\bar I_i)/\bar I_i$.
In either case, optimal relation $(1)$ remains valid. So, assuming identical consumers with respect to preferences (not necessarily with respect to Income), aggregating $(1)$ we obtain the market-level relation
$$ X^D=\frac {\alpha}{1-\alpha}\frac {p_y}{p_x}Y^D \tag {2}$$
At equilibrium, we have $X^D = X^S,\; Y^D = Y^S \tag{3}$.
Equations $(2)$ and $(3)$ hold for any supply.
We now change the supply of $X$ but leave the supply of $Y$ unchanged. Indexing the initial situation by $0$ and the second situation by $1$, they are described by
$$ X^D_0=\frac {\alpha}{1-\alpha}\frac {p_{y0}}{p_{x0}}Y^D_0,\;\;X^D_0 = X^S_0,\;\; Y^D_0 = Y^S_0 \tag {4}$$
$$ X^D_1=\frac {\alpha}{1-\alpha}\frac {p_{y1}}{p_{x1}}Y^D_1,\;\;X^D_1 = X^S_1>X^S_0,\;\; Y^D_1 = Y^S_1= Y^S_0\tag {5}$$
Then we have
$$X^D_1 - X^D_0 = \frac {\alpha}{1-\alpha}\frac {p_{y1}}{p_{x1}}Y^D_1 - \frac {\alpha}{1-\alpha}\frac {p_{y0}}{p_{x0}}Y^D_0 $$
and using the various relations
$$\implies X^S_1 - X^S_0 = \frac {\alpha}{1-\alpha}\frac {p_{y1}}{p_{x1}}Y^S_1 - \frac {\alpha}{1-\alpha}\frac {p_{y0}}{p_{x0}}Y^S_0 >0 $$
$$\implies \frac {\alpha}{1-\alpha}Y^S_0\cdot \left[\frac {p_{y1}}{p_{x1}}-\frac {p_{y0}}{p_{x0}}\right] >0$$
$$\implies \frac {p_{y1}}{p_{x1}}-\frac {p_{y0}}{p_{x0}} >0 \implies \frac {p_{y1}}{p_{y0}}>\frac {p_{x1}}{p_{x0}} \tag{6}$$
$(6)$ tells us that the price of $Y$, if it falls, it will fall certainly less than the price of $X$, but in proportional terms.
It appears that in this benchmark example, we cannot say anything about the price changes in terms of levels, as the OP asks about.
If the levels of the two prices are very different we can easily have a smaller proportional fall for $p_y$, while at the same time a higher fall in monetary units. Set for example $p_{y0} = 100, p_{x0} = 1$, and assume that the price of $Y$ falls only by $10$% while the price of $X$ falls by $20$%.
In turn the levels of prices depends on the magnitude of the supply of the two goods also, which are treated as exogenous here.
