Question

A chemical company makes two brands of antifreeze. The first brand is 45% pure antifreeze, and the second brand is 95% pure antifreeze. In order to obtain 90 gallons of a mixture that contains 60% pure antifreeze, how many gallons of each brand of antifreeze must be used?

Answer 1

Let x = amount of 45% antifreezeLet y = amount of 70% antifreeze EQUATION 1: x + y = 150 (total of 150 gallons mixed) EQUATION 2: .45x + .75y = .55(x + y) Simplify and solve the system of equations Multiply second equation by 100 on both sides to remove the decimals 45x + 75y = 55(x + y) Combine like terms 45x + 75y = 55x + 55y 45x - 55x + 75y - 55y = 0 -10x + 20y = 0 Now we have the following system of equations: x + y = 150 -10x + 20y = 0 Multiply the first equation by -10 to get opposite coefficients for x; add the equations to eliminate x 10x + 10y = 1500 -10x + 20y = 0 ------------------------------ 30y = 1500 Solve for y 30y = 1500 y = 50 Since the total mixed gallons is 150, x = 150 - 50 = 100 So we need 100 gallons of the 45% antifreeze and 50 gallons of the 70% antifreeze