The angles in the question are vertically opposite angles and Vertically opposite angles are equal.
Therefore,
[tex]\begin{gathered} (12x-37)^0=(9x+5)^0 \\ by\text{ collecting like terms we will have} \\ 12x^0-9x^0=5^0+37^0 \\ 3x^0=42^0 \\ \text{divide both sides by 3} \\ \frac{3x}{3}=\frac{42}{3} \\ x=14^0 \end{gathered}[/tex][tex]\begin{gathered} to\text{ measure }\angle y \\ 12x-37+y=180^0 \\ \text{substitute x=}14^0\text{ in the above equation to get y} \\ 12(14)-37^0+y^0=180^0 \\ 168^0-37^0+y=180^0 \\ 131^0+y=180^0 \\ y=180^0-131^0 \\ y=49^0 \end{gathered}[/tex]
Therefore,
[tex]\angle y=49^0[/tex]
