SOLUTION: how many liters of 20 percent alcohol solution and how many liters of pure alcohol should be mixed to obtain 8 liters of a 70% alcohol solution ?

We are mixing x liters of a weak alcohol solution with y liters of pure
alcohol and we are ending up with 8 liters of liquid of medium strength.

So looking only at the number of liters of liquid we have:

equation (1)  x + y = 8

Now let's get the pure alcohol out of each of the three terms of equation (1)

First term x [on the left side of equation (1)]
Out of the x liters of weaker solution, only 20% of it is pure, the rest water.
So out of the first term, x,  in equation (1) we get 20%x or 0.20x or 0.2x liters of pure alcohol.   

Second term y [on the left side of equation (1)]
Out of the y liters of pure alcohol, all 100% of it is pure, and no water.
So out of the second term, y,  in equation (1) we get 100%y or 1y or y liters of pure alcohol.

Third term [on the right side of equation (1)]
Out of the 8 liters of final mixture, only 70% of it is pure, the rest water.
So out of the third term, 8,  in equation (1) we get 70%*8 or 0.70*8 or 5.6 
liters of pur alcohol in the final mixture.

So the other equation is just equation (1) with each term multiplied by the
percentage of pure alcohol it contaions.

So we have 

equation(2)  0.2x + y = 5.6

So solve the system of equations (1) and (2):



The first equation is the "liquid" equation and the second equation is the
"pure alcohol" equation.  The first equation is easy x+y=8.  Then the second 
equation is just the first equation with each term multiplied by the percent
of alcohol in each term of the first equation.  

Solve that by substitution or elimination and get x=3 liters and y=5 liters.

That answer is reasonable because 70% is a lot stronger than 20%, so it'll
take more of the pure stuff and less of the weaker to bring the 20% up to 70%,
so it makes sense that we would 5 liters of pure and only 3 liters of the weak
solution.
Edwin