Equilibrium partial pressure

[605938236]

For this question from a worksheet I dowloaded on Chemistry community, can someone explain why the answer is 0.152 atm?

If one starts with pure NO2(g) at a pressure of 0.500 atm, the total pressure inside the reaction vessel when 2 NO2 (g) ⇋ 2 NO (g) + O2 (g) reaches equilibrium is 0.674 atm. Calculate the equilibrium partial pressure of NO2.

[Harry Forbeck 2A]

The way I looked at this question was by first setting up ICE table:

2 NO2 (g) ⇋ 2 NO (g) + O2 (g)
0.5 0 0
-2x +2x +x
0.5-2x. 2x x

and you know the sum of all equilibrium pressures is 0.674 atm. So, you can make the equation:

0.5-2x+2x+x = 0.674
0.5+x = 0.674
x = 0.174

then it is asking for equilibrium partial pressure of NO2, which is 0.5 - 2x, which gives you 0.5 - 2(0.174) = 0.152 atm