# Solving for an unknown variable

• Sep 19th 2011, 12:28 PM
KayPee
Solving for an unknown variable
Hi

How best can I approach this question?

A solution containing 45% water is mixed with another solution containing 10% bromine. If the chemist needs 30 liters of the mixture, how many liters of each solution are needed to make a 34.5% water/bromine solution?

Let x= water and y= the salt solution

"If the chemist needs 30 liters of the mixture"
x + y = 30 ...is that right?
• Sep 19th 2011, 03:40 PM
Wilmer
Re: Solving for an unknown variable
Quote:

Originally Posted by KayPee
A solution containing 45% water is mixed with another solution containing 10% bromine. If the chemist needs 30 liters of the mixture, how many liters of each solution are needed to make a 34.5% water/bromine solution?
Let x= water and y= the salt solution
"If the chemist needs 30 liters of the mixture"
x + y = 30 ...is that right?

Easier this way: let w = water ; then bromine = 30 - w
Code:

```  w  @  .45 30-w  @  .10 ====      ====   30  @  .345```
Can you turn that into an equation, then solve for w?