6x + 3y = 33 4x + y = 15 A. x = 2, y = 7 B. x = - 13, y = 7 C. x = - 2 3, y = 12 2 3 D. x = 5, y = 1

6x+3y=33 ... (1)

4x+y=15

y=15-4x ... (2)

substitute value of y from (2) in (1)

6x+3 (15-4x) = 33

distribute 3 over the bracket

6x+45-12x = 33

-6x=33-45

-6x=-12

x=2

plugging this value of x in equation 2 we get

15-4x=y

15-4 (2) = y

y=15-8

y=7

Answer is x=2 and y=7