Answer:
The sum of the internal ángles = 360°
(3y+40)° and (3x-70°) are suplementary angles = 180°
then:
(3x-70) + (3y+40) + 120 + x = 360 ⇒ first eq.
(3y+40) + (3x-70) = 180 ⇒ second eq
development:
from the first eq.
3x + x + 3y = 360 + 70 - 40 - 120
4x + 3y = 430 - 160
4x + 3y = 270 ⇒ third eq.
3y = 270 - 4x
y = (270 - 4x) / 3 ⇒ fourth eq.
from the secon eq.:
3y + 3x = 180 + 70 - 40
3y + 3x = 250 - 40
3y + 3x = 210 ⇒ fifth eq.
multiply by -1 the fifth eq and sum with the third eq.
-3y - 3x = -210 ⇒ (fifth eq. *-1)
3y + 4x = 270
⇒ 0 + x = 60
x = 60°
from the fourth eq.
y = (270-4x)/3
y = (270-(4*60)) / 3
y = (270 - 240) / 3
y = 30/3
y = 10°
Probe:
from the first eq.
(3x-70) + (3y+40) + 120 + x = 360
3*60 - 70 + 3*10 + 40 + 120 + 60 = 360
180 - 70 + 30 + 40 + 120 + 60 = 360
180 + 30 + 40 + 120 + 60 - 70 = 360
430 - 70 = 360
Answer:
y = 10