A chemical company makes two brands of antifreeze. The first brand is 60% pure antifreeze, and the second brand is 90% pure antifreeze. In order to obtain 180 gallons of a mixture that contains 85% pure antifreeze, how many gallons of each brand of antifreeze must be used?

We have two unknown quantities, [tex]V_1[/tex] and [tex]V_2[/tex], so it is necessary to write two equations that relate that variables. First, we are going to consider that the volume is an additive property, which means that if you mix [tex]V_1[/tex] and [tex]V_2[/tex] gallons the final volume V will be:

[tex]V=V_1+V_2[/tex]

The other equation we will write is based on a mass balance of the antifreeze. We know that the content of antifreeze in a given volume is the multiplication of that volume by the concentration, so the mass balance would be:

[tex]V_M*x_M=V_1*x_1+V_2*x_2[/tex]

It means that the total of the antifreeze in the final mix (which is [tex]V_M*x_M[/tex]) is equal to the sum of the antifreeze that was on the first brand volume (V_1*x_1) and the antifreeze that was on the second brand volume (V_2*x_2).

Now we have a two equation two unknown system, so let's solve it:

[tex]180=V_1+V_2[/tex]

[tex]0.85*180=0.60*{V_1}+0.90*{V_2}\\153=0.60*{V_1}+0.90*{V_2}[/tex]

From first equation:

[tex]V_1=180-{V_2}[/tex],

Replace [tex]V_1[/tex] on the second one.

[tex]153=0.60*(180-{V_2})+0.90*{V_2}\\153=0.60*180-0.60*{V_2}+0.90*{V_2}\\153=108+0.30*{V_2}\\{153-108}=0.30*{V_2}\\\\0.30*{V_2}=45\\{V_2}={\frac{45}{0.30} }= 150[/tex]

Finally, from [tex]V_1=180-{V_2}[/tex], [tex]V_1=30[/tex]