PLEASE!!!!, can someone help with this Geometry question. I have a quiz.

Step-by-step explanation:
4y-10 and 3x are supplementary angles
so 4y-10+3x=180° (1)
4y-10 and 9x+12 are alternate exterior angles and because we have two parallel lines then these angles are congruent.
so 4y-10=9x+12 (2)
so we have
(1) 4y+3x=180+10
(2) 4y-9x=12+10
(1) 4y=190-3x
we replace this equation to the second equation
190-3x-9x=22
190-22=9x+3x
168=12x
x=168/12
x=14
4y =190-3*(14)=190-42=148
y=148/4=37
3x=3*14=42°
4y-10=4*37-10=148-10=138°
9x+12=9*14+12=126+12=138°