What was the percentage (by mass) of antimony in the alloy?

Antimony is alloyed with lead to increase the rigidity of components used in the construction of lead storage batteries.
1.340 g of a particular metallic alloy, compounded of only Pb and Sb, can be quantitatively converted into a 1.558-g mixture of the oxides PbO2 and Sb2O4.
What was the percentage (by mass) of antimony in the alloy?
Okay so I'm having a bit of difficulty with the above question so I googled and found

Atomic weights: -

Pb = 207,2
Sb = 121,76
O = 16

Let the mass of Sb be x grams
Then mass of Pb is (1,34 - x ) grams.

In Sb2O4 the ratio of O to Sb is the same as in SbO2

Therefore 121,76 g Sb combines with 32 g O2 and
207,2 grams of Pb combines with 32 grams of O2

So oxygen that reacts with Sb = 32x / 121,76
and O2 that combines with Pb = 32/207,2 X (1,34 - x)

Total combined O2 = 1,558 -1,340 = 0,218 g
Therefore: -
32x/121,76 + (1,34 -x) X 32/207,2 = 0,218
Simplify 32X207,2x + 5221,0688 - 3896,32x = 5499,85
6630,4x + 5221,0688 - 3896,32x = 5499,85

2734,08 x = 278,781
x = 0,10197 grams
Sb = 0,102 g
Pb = 1,238 g

Therefore Sb = 7,61%
I don't really understand this part:

So oxygen that reacts with Sb = 32x / 121,76
and O2 that combines with Pb = 32/207,2 X (1,34 - x)
Could someone please explain how to do this question?