Question

A chemical company makes two brands of antifreeze. The first brand is 20% pure antifreeze, and the second brand is 70% pure antifreeze. In order to obtain 30 gallons of a mixture that contains 35% pure antifreeze, how many gallons of each brand of antifreeze must be used?

Answer 1

Answer:

First brand of antifreeze: 21 gallons

Second brand of antifreeze: 9 gallons

Step-by-step explanation:

Let's call A the amount of  first brand of antifreeze. 20% pure antifreeze

Let's call B the amount of second brand of antifreeze. 70% pure antifreeze

The resulting mixture should have 35% pure antifreeze, and 30 gallons.

Then we know that the total amount of mixture will be:

$A + B = 30$

Then the total amount of pure antifreeze in the mixture will be:

$0.2A + 0.7B = 0.35 * 30$

$0.2A + 0.7B = 10.5$

Then we have two equations and two unknowns so we solve the system of equations. Multiply the first equation by -0.7 and add it to the second equation:

$-0.7A -0.7B = -0.7*30$

$-0.7A -0.7B = -21$

$-0.7A -0.7B = -21$

               +

$0.2A + 0.7B = 10.5$

--------------------------------------

$-0.5A = -10.5$

$A = \frac{-10.5}{-0.5}$

$A = 21\ gallons$

We substitute the value of A into one of the two equations and solve for B.

$21 + B = 30$

$B = 9\ gallons$