We know that the opposite angles of a parallelogram are equal.
So,
$(3x-10{)}^{\circ }=(x+80{)}^{\circ }\Rightarrow 3x-x={80}^{\circ }+{10}^{\circ }\Rightarrow 2x={90}^{\circ }\Rightarrow x={45}^{\circ }$
Also, the consecutive angles of a parallelogram are supplementary.
So,
$\angle B+\angle C={180}^{\circ }(3x-10{)}^{\circ }+(y+25{)}^{\circ }={180}^{\circ }(3\times {45}^{\circ })-{10}^{\circ }+y+{25}^{\circ }={180}^{\circ }{135}^{\circ }-{10}^{\circ }+y+{25}^{\circ }={180}^{\circ }\Rightarrow {125}^{\circ }+y+{25}^{\circ }={180}^{\circ }\Rightarrow y={180}^{\circ }–{125}^{\circ }-{25}^{\circ }\Rightarrow y={30}^{\circ }$