SOLUTION: A 60% antifreeze solution is to be mixed with a 10% antifreeze solution.In what ratio (x/y)must they be mixed to obtaiin a 40% antifreeze solution?(hint:Let N be the number of litr

Algebra ->  Linear-equations -> SOLUTION: A 60% antifreeze solution is to be mixed with a 10% antifreeze solution.In what ratio (x/y)must they be mixed to obtaiin a 40% antifreeze solution?(hint:Let N be the number of litr      Log On


   



Question 179738: A 60% antifreeze solution is to be mixed with a 10% antifreeze solution.In what ratio (x/y)must they be mixed to obtaiin a 40% antifreeze solution?(hint:Let N be the number of litres of the 40% solution)
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let n= liters of 40% solution
Let x= liters of 60% solution
Let y= liters of 10% solution
How much antifreeze is in the 40% solution? .4n
(1) .6x+%2B+.1y+=+.4n
(2) x+%2B+y+=+n
Multiply (2) by .4 and subtract from (1)
(1) .6x+%2B+.1y+=+.4n
(2) -.4x+-+.4y+=+-.4n
.2x+-+.3y+=+0
.2x+=+.3y
x%2Fy+=+3%2F2 answer
check:
Suppose n+=+10 liters
(1) .6x+%2B+.1y+=+4
(2) x+%2B+y+=+10
Multiply both sides of (1) by .1 and
subtract (2) from (1)
(1) .6x+%2B+.1y+=+4
(2) -.1x+-+.1y+=+-1
.5x+=+3
x+=+6
(2) x+%2B+y+=+10
y+=+4
6%2F4+=+3%2F2
OK