Let
denote the amounts (in liters) of the 20%, 30%, and 60% solutions used in the mixture, respectively.
The chemist wants to end up with 72 L of solution, so

while using twice as much of the 60% solution as the 30% solution, so

The mixture needs to have a concentration of 35%, so that it contains 0.35•75 = 26.25 L of pure acid. For each liter of acid solution with concentration
, there is a contribution of
liters of pure acid. This means

Substitute
into the total volume and acid volume equations.

Solve for
and
. Multiply both sides of the second equation by 5 to get

By elimination,

so that

and
