Answer:
a. x=57.6, y=44.8
Step-by-step explanation:
Named the angles, check the picture for that
<ABD+<ACD=180 (Opposite angles of a quadrilateral is supplementary (180 degrees)
<CAB+<BDC=180 (Opposite angles of a quadrilateral is supplementary (180 degrees)
therefore:
a. 3y+x-12=180.............. equ 1
2x+y+20=180.............equ 2
From equ 1
x+3y=180+12
x+3y=192...............equ 3
From equ 2
2x+y=180-20
2x+y=160.............equ 4
Equate equ 3 and equ 4
x+3y=192...............equ 3
2x+y=160.............equ 4
Using elimation method
x+3y=192...............equ 3
2x+y=160.............equ 4
Multiple equ 3 by 1 and equ 4 by 3
x+3y=192...............equ 3 ×1
2x+y=160.............equ 4 ×3
x+3y=192
6x+3y=480
Subtract
-5x+0=-288
-5x=-288
x=57.6................... Equ 5
Replace 57.6 for x in equ 3
x+3y=192...............equ 3
57.6+3y=192
3y=192-57.6
3y=134.4
y=44.8
B. <CAB= 2x=2×57.6=115.2
<ABD= 3y=3×44.8= 134.4
<ACD=x-12=57.6-12=45.6
<BDC=y+20=44.8+20=64.8
