Figure ABCD is a trapezoid. What are the values of x and y

360= 2x+3y+(x-12)+(y+20)
360= 3x+4y+8
352=3x+4y
352-3x=4y
88-3x/4=y
352-4y=3x
117 1/3 -4y/3=x
I'm proabably wrong, but I tried.
Answer : The value of x and y are, 64° and 50° respectively.
Step-by-step explanation :
As we know that, ABCD is a trapezium and AB || CD. So, the sum of opposite angles are 180°
[tex]\angle A+\angle D=180^o[/tex]
Given:
[tex]\angle A=2x^o\\\\\angle D=(x-12)^o[/tex]
[tex]\angle A+\angle D=180^o[/tex]
[tex]2x^o+(x-12)^o=180^o[/tex]
[tex]3x^o-12^o=180^o[/tex]
[tex]3x^o=192^o[/tex]
[tex]x=64^o[/tex]
and,
Given:
[tex]\angle B=3y^o\\\\\angle C=(y+20)^o[/tex]
[tex]\angle A+\angle D=180^o[/tex]
[tex]3y^o+(y+20)^o=180^o[/tex]
[tex]4y^o+20^o=180^o[/tex]
[tex]4y^o=200^o[/tex]
[tex]y=50^o[/tex]
Thus, the value of x and y are, 64° and 50° respectively.