A chemist must mix 6 liters of a 30% acid solution with some of an 80% acid solution to get a 50% acid solution. How many liters of the 80% acid solution should be used?

Quick Answer: 4 Liters of 80% Acid should be used.
Comment
One: 50% of the final volume must be acid.
Two: We don't know the volume of 80% acid needed.
Three: We don't know the volume of the final mixture.

What we do know.
One: We know how much of the 30% acid we have. that's really all we do know.

Volume of the 30% acid = 6 L
Volume of the acid = (30/100 * 6) = 180/100 = 1.8 liters of the 6 liters is acid.

What we can set up.
Let the volume of the 80% acid = x Liters.
Let the acid content = (80/100) * x and this will be in liters
Let the acid content = 0.8*x 

Let the total volume = 6 + x (of both acid and water).
Let the acid content of this volume = 50% of the total volume
So the acid content of the result = 50% (x + 6) = 50/100(x + 6) = 0.5(x + 6)
That's the right side of the equation.

The left side is what we know about the acid itself
We know that the 30% content will contribute 6* 30/100 = 1.8 L of acid.
The 80% acid will contribute 0.8*x 
Total acid content = 0.8x + 1.8 This is the left hand side.

Equation
0.8x + 1.8 = 0.5(x + 6)

Solve
0.8x + 1.8 = 0.5x +3 Subtract 0.5x from both sides.
0.8x - 0.5x + 1.8 = 3 Combine like terms.
0.3x + 1.8 = 3           Subtract 1.8 from both sides
0.3x  = 1.2                Divide by 0.3

x = 4  <<<< answer 4 liters of 80% acid should be used.
 
Check
Amount of acid contributed by each acid
1.8 L of the 30% acid
3.2 L of the 80% acid
Total volume of acid in the mixture = 5 L

The total volume of 80% + 30% acid = 6 + 4 = 10 L
50% of that amount is 5 Liters. We have solved the problem correctly

Note
You may wonder where I got that 3.2 L from for the 80% acid.
You have 4L total. 80% of it is acid. 80/100 * 4 = 3.2 Liters are acid.