let

x = liters of a mixture containing 70% alcohol

y = liters of a mixture containing 20% alcohol

since the new mixture is 16 L we can write

x + y = 16

70% of x alcohol + 20% of y alcohol = 50% of 16 alhcohol

(70/100)*x + (20/100)*y = (50/100)*16

0.7*x + 0.2*y = 0.5*16

0.7*x + 0.2*y = 8

by solving the system of equations

x + y = 16

0.7*x + 0.2*y = 8

we find

x = 9.6 L

y = 6.4 L

We need to mix 9.6 L of the 70% alcohol mixture and 6.4 L of the 20% alcohol to obtain 16 L of a mixture containing 50% alcohol.