I think this is algebra, help

A cars radiator contains 12 L of antifreeze at 25% concentration. How many liters must be drained and then replaced by pure antifreeze to bring the concentration to 50%? please show work/explain your thinking. Preferable simply because I'm only 12.

This is a simultaneous equation because you have two variables. You can solve it this way:

Let A = the number of liters needed of the 0.25 antifreeze
Let B = the number of liters needed of the 1.00 antifreeze

Now, set up the first equation:

0.25A + 1.00B = 12 (0.50)

You are saying "A liters of the .25 solution + B liters of the 1.00 solution equals 12 liters of the 0.50 solution".

Now you need a second equation. You know that the total of A and B has to equal 12:

A + B = 12

Put the two equations together:

0.25A + 1.00B = 12(0.50) = 6.00
A + B = 12

Now subtract the bottom from the top because the 'B's cancel out. You get:

-0.75A = -6

Divide both sides by -0.75A. You get:

A = 8

Plug that in to one of the equations and solve for B. You get B = 4
So your answer is 8 liters of the 25% antifreeze and 4 liters of the 100% antifreeze = 12 liters of 50% antifreeze.