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the following system is at equilibrium with [N2O4] = 0.55 M and [NO2] = 0.25 M....

the following system is at equilibrium with [N2O4] = 0.55 M and [NO2] = 0.25 M. If an additional 0.10 M of NO2 is added to the reaction mixture with no change in volume, what will be the new equilibrium concentration of N2O4? This is from a practice exam and the correct answer is given as 0.62 M. I have searched the internet high and low and have found only alot of answers to this question but none of them are the correct one. I just need the solution steps to arriving at 0.62M please.

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