An automobile radiator contains 8 liters of a solution that is 10 percent antifreeze and 90 percent water. How much of the solution should be drained and replaced with pure antifreeze to obtain a solution that is 25 percent antifreeze? Please help with complete solutions thank you :)

Question

zpupster · Answer

x= 10% of the remaining liquid
y= the 100% anti freeze we add

We know the radiator contains 8L

 so x + y = 8 

but we want  to  that to equal a 25 % of 8L

system of equations is

.1x+1y= .25(8)
  x + y = 8

can you do it from here?

anonymous · Answer

I'll try thank you so much @zpupster :)

zpupster · Answer

you're welcome

anonymous · Answer

@zpupster I got stuck with the .1x + 1y = .25(8)

zpupster · Answer

ok
i was away 

.1x+1y= .25(8)
  x + y = 8


.1x+1y= .25(8)    .25 is also 1/4  so 8*(1/4) is 2
  and y = 1y

we have now
.1x+y =2

x+y=8  or    8-x=y   substitute that in for y in

.1x + (8-x) =2  then
-1x+.1x + 8 = 2  combine like terms and subtract 8 both sides
-.9x=-6   divide both sides by .9

what did you get

zpupster · Answer

hope i helped i have to be out for awile.

anonymous · Answer

Thank you uhh I got 6.66667

zpupster · Answer

yes

 (or 6 and 2/3) 

Since the quantities  add to 8L, y = 4/3. 

So drain 4/3 L  add 4/3 L pure antifreeze to get 25% anti freeze

anonymous · Answer

Ohhh! Thank you, thank you very much :D