How to solve for concentration of two equilibrium reactions with finite equilibrium constants?
Completely accurate answer :

The main reactions which take place are 
$$
\begin{array}{ll}
\ce{ H2O &lt;=&gt; H+ + OH-} &amp;\quad\left(K_\mathrm{w}= 1\times 10^{-14}\right) \\
\ce{HNO2 &lt;=&gt; H+ + NO2-} &amp;\quad\left(K_\mathrm{a}= 4\times10^{-4}\right) \\
\end{array}
$$
$\ce{HNO2}$ acid is a weak acid, which does not dissociate completely in water. This means it does not give all its hydrogen ions into the water, the resulting solution contains four species ${\ce{H+}, \ce{OH−},\ce{NO2−}}$, and undissociated acid $\ce{HNO2}$. In order to specify the concentrations of the four species
present in an aqueous solution of $\ce{HNO2}$, we need four independent equations
 between them. These are :

1- Equilibria. We have two equilibrium equations:

The dissociation equilibrium of water
$$\ce{ K_\mathrm{w}=[H+][OH−]}\tag{1}$$ 
The dissociation equilibrium of acid
$$\mathrm{ K_\mathrm{a}}=\frac{[H^+]_e[NO_2^−]_e}{ [HNO_2]_e}\tag{2}$$
2- Mass balance. The mass balance equation that relates the concentrations of the various dissociation products of the substance to its initial concentration, which we designate here as $\ce[HNO_2]_0$. For the solution of $\ce{HNO_2}$, this equation would be
$$\ce{[HNO2]_0  = [HNO2]_e + [NO2−]_e}\tag{3}$$
3-Charge balance. In any ionic solution, the sum of the positive and negative electric charges must be zero; in other words, all solutions are electrically neutral. This is known as the electroneutrality principle.
$$\ce{[H+]_e = [NO2−]_e + OH−}_e\tag{4}$$
We can use ${\text{Eq}\mathrm{3}}$ to get an expression for $\ce{[HNO2]_e}$, and this can be solved for$\ce{ [NO2−]_e}$; these are substituted into $\text{ Eq}\mathrm{2}$  to yield
$$ K_\mathrm{a}=\frac{\ce{ [H+]_e} {([H^+]_e - [OH^-]_e)} }{[HNO_2]_0 - [H^+]_e + [OH^-]_e}  \tag{5}$$
This equation is cubic in $\ce{[H+]_e}$ when $\ce{[OH−]}$ is expressed as $\frac{K_\mathrm{w}}{\ce{[H+]_e}}$; it becomes as

$$ K_\mathrm{a}=\frac{\ce{[H+]_e} {([H^+]_e – K_\mathrm{w}/{[H^+]_e})} }{[HNO_2]_0 - [H^+]_e + K_\mathrm {w}/{[H^+]_e}}  \tag{6a}$$
Or 
$$\
 [\ce{H+}]_e^3 
  + K_\mathrm{a} [\ce{H+}]_e^2 
  -(K_\mathrm{w} + [\ce{HNO2}]_0K_\mathrm{a})[\ce{H+}]_e
   - K_\mathrm{w} K_\mathrm{a}
  \tag{6b}
$$
 For most practical applications, we can make approximations that eliminate the need to solve a cubic equation.

a- Unless the acid is extremely weak or the solution is very dilute, the concentration of $\ce{OH−}$ can be neglected in comparison to that of $\ce {[H+]_e}$. If we assume that $\ce{[OH−]_e&lt;&lt;[H+]_e}$, then $\text{Eq}\mathrm{5}$ can be simplified to
$$K_\mathrm{a }=\frac{[H^+]_e^2}{[HNO_2]_0 –[H^+]_e}\tag{7}$$
which is a quadratic equation:
$$\ce{[H^+]_e^2 + K_\mathrm{a}[H+]_e – K_\mathrm{a}[HNO_2]_0 = 0}\tag{8}$$
b-If the acid is fairly concentrated (usually more than $\mathrm{10^{−3}}\text{M}$), a further simplification can frequently be achieved by making the assumption that$\ce{ [H^+]_e &lt;&lt; [HNO_2]_0}$. This is justified when most of the acid remains in its protonated form$\ce{ [HNO_2]_e}$, so that relatively little $\ce{H+}$ is produced. In this event,$\text{Eq}\mathrm{7}$ reduces to
$$K_\mathrm{a}=\frac{[H^+]_e^2}{[HNO_2]_0}\tag{9}$$
Or
$$\ce{[H^+]}  \approx  \sqrt{K_\mathrm{a}\ce{[HNO_2]_0}}\tag{10}$$
-   Comparing of the$\ce{ [H+]_e }$values  resulted from solution the three
 derived equations when the initial acid concentration equal $\mathrm{0.1}\text{M}$

1-  The completely accurate answer solution using cubic equation $\mathrm{(6b)}$by wolframalpha

$\ce{[H+]_e} = 0.00612772\text{M}$

2- The approximate  answer solution using quadratic equation $\mathrm{(8)}$

$\ce{[H+]_e} = 0.00612771680\text{M}$

3- The further approximate  answer solution using the reduced equation $\mathrm{(10)}$ 
$\ce{[H+]_e} = 0.00632455532\text{M}$