Apharmacist wants to mix a 16% saline solution with a 26% saline solution to get 15 l of a 20% saline solution. how much of each solution should she use?

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

i wish i could but this one is weird give me abc and i can solve it.