Algerbra II Word Problem Help

Please help solve this question.

How much pure antifreeze must be added to 12L of a 40% solution of antifreeze to obtain a 60% solution?

Thank you for all of your answers!

If there are 12 liters of a solution containing 40% antifreeze. Then there are:
0.40 * 12L = 4.8 L of antifreeze in the solution or
12 - 4.8L = 7.2 L of water.

So to obtain a 60% solution, we set up an equality. Liters of antifreeze / liters of solution = % antifreeze. Let X represent the liters of pure antifreeze added to the initial solution:
(4.8L antifreeze + X liters of antifreeze) / (4.8 L (antifreeze) + X liters of antifreeze + 7.2 L (water)) = 0.60

4.8L + X = 0.60 (4.8L + X + 7.2L)
4.8L + X = 7.2L + 0.60*X
0.40X = 2.4 L
X = 6L of antifreeze added.

Hope this helps!

6L duhhhhh. Daggummm 40% to 60% is a 50% increase drrrrr, so increase 12 50%, which is 18, which is a 6L difference. Drrrrhh gorilla math is easy. Hope you're not planning to become an engineer or a scientist.

No. of liters of pure antifreeze to be added (x):
0.4(12) + x = 0.6(12 + x)
2(12) + 5x = 3(12 + x)
24 + 5x = 36 + 3x
2x = 12
x = 6