SOLUTION: How many pints of a 5% antifreeze solution and how many pints of a 20% antifreeze solution must be mixed to obtain 12 pints of a 15% solution? Write a system of two equations in tw

Algebra ->  Percentage-and-ratio-word-problems -> SOLUTION: How many pints of a 5% antifreeze solution and how many pints of a 20% antifreeze solution must be mixed to obtain 12 pints of a 15% solution? Write a system of two equations in tw      Log On


   

Question 839247: How many pints of a 5% antifreeze solution and how many pints of a 20% antifreeze solution must be mixed to obtain 12 pints of a 15% solution? Write a system of two equations in two variables to solve each problem.
Answer by alicealc(293) About Me  (Show Source):
You can put this solution on YOUR website!
5% antifreeze solution = x pints
20% antifreeze solution = y pints
15% antifreeze solution = 12 pints
x + y = 12 -> x = 12 - y
5%2F100%2Ax%2B20%2F100%2Ay=15%2F100%2A12
multiply all by 100:
5x + 20y = 15 * 12
5(12 - y) + 20y = 180
60 - 5y + 20y = 180
60 + 15y = 180
15y = 180 - 60
15y = 120
y = 120/15
y = 8
x = 12 - y = 12 - 8 = 4
so, to make 12 pints of 15% antifreeze solution, we need 4 pints of 5% antifreeze solution and 8 pints of 20% antifreeze solution.