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?

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?