I. 2x + 4y = 2II. x - 3y = 16
From equation II, let's express x in terms of y:x = 16 + 3y
Now substitute this into equation I:2(16 + 3y) + 4y = 232 + 6y + 4y = 210y = -30y = -3
Now substitute y back into x = 16 + 3y:x = 16 + 3(-3)x = 16 - 9x = 7
Therefore, the solution to the system of equations is x = 7, y = -3.
I. 2x + 4y = 2
II. x - 3y = 16
From equation II, let's express x in terms of y:
x = 16 + 3y
Now substitute this into equation I:
2(16 + 3y) + 4y = 2
32 + 6y + 4y = 2
10y = -30
y = -3
Now substitute y back into x = 16 + 3y:
x = 16 + 3(-3)
x = 16 - 9
x = 7
Therefore, the solution to the system of equations is x = 7, y = -3.