SOLUTION: A certain quantity was taken from a ten liter 30% alcohol solution and replaced by pure alcohol so as to produce 51% alcohol solution. How many liters were taken and replaced?

Let the number of liters of pure added be x
Let the number of liters of 30%alcohol left
    in the jug after the x liters were 
    taken out and replaced be 
    y liters of pure alcohol

                         Percent      Liters
                         alcohol     of pure
  Type        Liters       in        alcohol
   of          of         EACH      contained
liquid       liquid    as decimal    in each
---------------------------------------------
pure alcohol    x         1.00       1.00x
30% alcohol     y         0.30       0.30y
----------------------------------------------------
Final mixture  10         0.51      (0.51)(10) = 5.1

 The first equation comes from the second column.

  

                 x + y = 10

 The second equation comes from the last column.
  

           1.00x + 0.30y = 5.1

Get rid of decimals by multiplying every term by 10:

          10x + 3y = 51

 So we have the system of equations:
           .

We solve by substitution.  Solve the first equation for y:

           x + y = 10
               y = 10 - x

Substitute (10 - x) for y in 10x + 3y = 51

     10x + 3(10 - x) = 51
       10x + 30 - 3x = 51
             7x + 30 = 51
                  7x = 21
answer:            x = 3 liters drained and 
                         replaced with pure alcohol

Edwin

Let "x" be the volume of the 30% alcohol solution taken from and replaced.

After replacement, we have the same volume of 10 liters of the (new) solution.
The alcohol content in the new solution is x + 0.3*(10-x).

Since it is 10 liter of the 51% of alcohol solution, you have this equation

x + 0.3*(10-x) = 0.51*10.

Simplify and solve for x:

x + 3 - 0.3x = 5.1,

0.7x = 5.1 - 3,

0.7x = 2.1,

x =  = 3.

Answer.  3 liters of the original solution were taken off and replaced by the pure alcohol.