Finding mass percentage of components of a solid mixture
We first start by writing the system of equations in this reaction, assuming that we don't know the ratios that the elements combine in.

Because we know the mass of the unknown mixture, we'll use $x$ to represent the mass of $\ce{K2CO3}$ and $y$ to represent the mass of $\ce{KHCO3}$, in grams. This is our first equation:
$$x + y = 2.184$$

Next, we know the final amount of moles of $\ce{HCl}$ that reacted. It was
$$\mathrm{0.03129\; \pu L \times \frac{0.742\; mol\; \ce{HCl}}{\pu L} = 0.0232\; mol\; \ce{HCl}}$$

Now, let's reconsider the balanced equations for the reaction. We have
$$\ce{2HCl_{(aq)} + K2CO3_{(aq)} -&gt; 2KCl_{(aq)} + CO2_{(g)} + H2O_{(l)}}$$
$$\ce{HCl_{(aq)} + KHCO_{3(aq)} -&gt; KCl_{(aq)} + CO2_{(g)} + H2O_{(l)}}$$

We can see that it takes $\ce{2 HCl}$ to neutralize one $\ce{CO3}$, but it only takes $\ce{1 HCl}$ to neutralize one $\ce{HCO3}$.

We also know the formula weights of the individual components. Now, we can write the second equation as an expression of the sum of the moles of the reactants. Our coefficients for $x$ and $y$ become:
$$
\ce{\frac{2\; mol\; HCl}{1\; mol\; K2CO3} \times \frac{1\; mol\; K2CO3}{138.2058\; g}} \times x
$$

$$
\ce{\frac{1\; mol\; HCl}{1\; mol KHCO3} \times \frac{1\; mol\; KHCO3}{100.1154\; g}} \times y
$$

Now we can cancel out units and write out the second equation:
$$\ce{\frac{2\; mol\; HCl}{138.2058\; g}}x + \ce{\frac{1\; mol\; HCl}{100.1154\; g}}y = \ce{0.0232\; mol\; HCl}$$

After that, this is straightforward linear algebra. This system of equations is simple enough to be solved by basic substitution and elimination. However, you can choose to use a matrix. In that case, we have

$$
\mathrm{
\begin{bmatrix}
    1 &amp; 1 \\
    \frac{2}{138.2058} &amp; \frac{1}{100.1154}
\end{bmatrix}
\begin{bmatrix}
x \\
y
\end{bmatrix}
}
=
\begin{bmatrix}
2.184 \\
0.0232
\end{bmatrix}
$$

Rearranging gives
$$
\mathrm{
\begin{bmatrix}
    1 &amp; 1 \\
    \frac{2}{138.2058} &amp; \frac{1}{100.1154}
\end{bmatrix}^{-1}
\begin{bmatrix}
2.184 \\
0.0232
\end{bmatrix}
=
\begin{bmatrix}
x \\
y
\end{bmatrix}
}
$$

Using a calculator we find that
$$\mathrm{
\begin{bmatrix}
x \\
y
\end{bmatrix}
=
\begin{bmatrix}
0.309\\
1.875
\end{bmatrix}
}
$$
$$x = \ce{0.309\; g\; K2CO3}$$
$$y = \ce{1.875\; g\; KHCO3}$$

Finally, the mass percentages are determined to be:
$$\ce{\frac{0.309\; g\; K2CO3}{2.184\; g\; total} \times 100{\%} = 14.1{\%} K2CO3}$$
$$\ce{\frac{1.875\; g\; KHCO3}{2.184\; g\; total} \times 100{\%} = 85.9{\%} KHCO3}
$$