diberikan sistem persamaan (x-2)/3 + (y+1)/6=2 dan (x+3)/4 - (2y-1)/2=1 maka nilai x+y = .....?

(x - 2)/3 + (y + 1)/6 = 2 (x 6)
2(x - 2) + y + 1 = 12
2x - 4 + y + 1 = 12
2x + y = 15 ... (i)

(x + 3)/4 - (2y - 1)/2 = 2 (x 4)
x + 3 - 2(2y - 1) = 8
x + 3 - 4y + 2 = 8
x - 4y = 3 |x 2|
2x - 8y = 6 ...(ii)

Eliminasi i dgn ii
2x + y =15
2x - 8y=6
---------------- (-)
9y = 9
y = 1

x - 4y = 3
x - 4(1) = 3
x = 7

HP : {7 , 1}