a graduate student in chemistry needs a 40% saline solution. How much of a 20% saline solution must be mixed?

Let
x = amount of 20% solution needed

.2x + .5*10 = .4(x + 10)
.2x + 5 = .4x + 4
1 = .2x
5 = x

You need to add 5 liters of 20% solution to 10 liters of 50% solution to get 40% solution.

The answer is 5.  Here's the math.
x is the number of liters of solution that will be 40% saline.

x(.4) = 10(.5)+(x-10)(.2)
.4x = 5 + .2x - 2
.4x = .2x +3
.2x = 3
x = 15

But remember that 15 is the number of liters in the final solution.  Subtract 10 to find that it's 5 liters of 20% solution that needs to be added.

1) Let X be the answer; that is, X is the number of liters of a 20% solution that needs to be added to get the desired result.

2) Therefore, (X+10) * 0.4 must equal 10 * 0.5 + X * 0.2.

3) Just solve the equation for X; the answer is 5 liters

In answering this problem, I assume that the concentrations given are by volume.  The first equation would be x/(x+y) = 0.5.  We know that y is 10L or 10000g, since the density of water is 1g/mL.  Then we solve for x and find that x = 10000g of salt.

Next we consider the mass that is added from the 20%.  The equation is (10000g + x) / (10000 + 10000 + x + y) = 0.4.  We also know that it is a 20% solution.  So another equation is x / (x + y) = 0.2.  

Solving these equations simultaneously, you solve for y.  You will find that y = 8000mL.  So 8L of the 20% solution is required.