SOLUTION: Kathy has 5 liters of a 32% acid solution and she4 also has a large amount of a 26% acid solution. How many liters of the 26% solution can Kathy mix with the 5 liters of 32% solut

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Question 168604: Kathy has 5 liters of a 32% acid solution and she4 also has a large amount of a 26% acid solution. How many liters of the 26% solution can Kathy mix with the 5 liters of 32% solution in order to produce a 50% acid solution?
the answer isn't possible in the end but i need to know the values of x and y and write out a system of equations from the information given and then solve hte system of equations using substitution or elimination
Found 2 solutions by josmiceli, ankor@dixie-net.com:
Answer by josmiceli(19441) (Show Source):
You can put this solution on YOUR website!
I believe it's impossible to get a stronger concentration mixture
from 2 weaker concentrations
To prove that, I'll go ahead and try to solve it
Let = the liters of 26% solution she needs

Multiply both sides by

Clearly, the answer is impossible
check:

OK

Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website!
Kathy has 5 liters of a 32% acid solution and she also has a large amount of a
26% acid solution. How many liters of the 26% solution can Kathy mix with
the 5 liters of 32% solution in order to produce a 50% acid solution?
:
I think you would just use a typical mixture equation, if it's impossible, it
will let you know, by giving a negative solution.
:
Let x = amt of 26% solution (why would you need y?)
:
.32(5) + .26x = .50(x+5)
:
1.6 + .26x = .50x = 2.5
:
.26x - .50x = 2.5 - 1.6
-.24x = .9
x =
x = -3.75; clearly, not of the real world
:
:
Substitution will show the equation is satisfied, but negative substances don't exist