the pharmacist wants to mix 2 solutions to make 15 L of a solution with 20% percentage
so if he mixes a volume of y of 16%, the quantity of solute added from the 16% solution is 16% x y = 0.16y
and he mixes a volume of z of 26%, the quantity of solute added from 26 % solution is 26% x z = 0.26z
the final solution has a solute quantity of 20 % x 15 L = 0.2 x 15 = 3
quantities of solute added from both solutions is equal to the quantity of the final solution
0.16y + 0.26z = 3 ---1)
the 2 volumes added equals 15 L
y + z = 15 ---2)
then we have a set of simultaneos equations
multiply 2nd equation by 0.16
0.16y + 0.16z = 2.4 --3)
subtract 3rd equation from the 1st equation
0.26 z - 0.16z = 0.6
0.10 z = 0.6
z = 0.6/0.10 = 6
since z = 6
then substitute z = 6 in 2nd equation
y + z = 15
y + 6 = 15
y = 9
the volume of 26% solution added is 6 L
volume of 16 % solution added is 9 L