How to calculate pH of a 0.1 M solution of ammonium bicarbonate considering simultaneous equilibrium along with hydrolysis?

We need to calculate the $\ce{[H+]}$ of 0.1M $\ce{(NH4+)(HCO3^-)}$ given the $\ce{k\_b(NH4OH), k\_{a\_1}(H2CO3), k\_{a\_2}(HCO3^-)}$   

The hint tells me to directly use the result $\ce{[H+]=\sqrt{k\_{a\_1}(\frac{k\_w}{k\_b}-k\_{a\_2})}}$  

I'm trying to reach this result by systematic equilibrium analysis.  

I considered the following simultaneous equilibrium:   

$\ce{NH3 +H2O<=>NH4+ +OH^-:k\_b = \frac{[NH4+][OH-]}{[NH3]}}$  

$\ce{H2CO3<=>H+ +HCO3-:k\_{a\_1}=\frac{[H+][HCO3-]}{[H2CO3]}}$   

$\ce{HCO3^-<=>H+ +CO3^{2-}:k\_{a\_2}=\frac{[H+][CO3^{2-}]}{[HCO3^-]}}$   

$\ce{H2O(l)<=>H+ +OH-:k\_w=[H+][OH-]}$   

I'm not sure how I shall proceed further. I also tried to write the following charge balance equation:   

$\ce{[NH4+] +[H+]=[HCO3-] +2[CO3^{2-}] +[OH-]}$   

But in this case, wouldn't the $\ce{[NH4+],[HCO3-]}$ cancel out as they both are equal to 0.1M? Upon solving the charge balance I don't reach the given result. Where am I going wrong? How do I derive the formula?


