check the pictured below.
now, notice that APB is just a flat-line and therefore those two angles there add up to 180°.
also notice that since ∡ACB is 90°, then those two angles add up to that much.
[tex]\bf \stackrel{\measuredangle ACP}{(3x+2y)}+\stackrel{\measuredangle BCP}{(3x+4y)}=90\implies 6x+6y=90\implies \stackrel{\textit{common factor}}{6(x+y)}=90
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x+y=\cfrac{90}{6}\implies x+y=15\implies \boxed{x=15-y}
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~\dotfill[/tex]
[tex]\bf \stackrel{\measuredangle APC}{(7x+3)}+\stackrel{\measuredangle BPC}{(16y)}=180\implies 7x+16y=177\implies 7\left( \boxed{15-y} \right)+16y=177
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105-7y+16y=177\implies 9y=72\implies y=\cfrac{72}{9}\implies \blacktriangleright y=8 \blacktriangleleft
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\measuredangle BPC=16(8)\implies \measuredangle BPC=128[/tex]
