MATH SOLVE

3 months ago

Q:
# A chemical company makes two brands of antifreeze. The first brand is 65% pure antifreeze, and the second brand is 90% pure antifreeze. In order to obtain 50gallons of a mixture that contains 70% pure antifreeze, how many gallons of each brand of antifreeze must be used?

Accepted Solution

A:

Let x = amount of 65% antifreeze

Let y = amount of 90% antifreeze

EQUATION 1: x + y = 50 (total of 50 gallons mixed)

EQUATION 2: .65x + .90y = .70(x + y)

Simplify and solve the system of equations

Multiply second equation by 100 on both sides to remove the decimals

65x + 90y = 70(x + y)

Combine like terms

65x + 90y = 70x + 70y

65x -70x+ 90y-70y = 0

-5x + 20y = 0

Now we have the following system of equations:

x + y = 50

-5x + 20y = 0

Multiply the first equation by 5 to get opposite coefficients for x;

add the equations to eliminate x

5x + 5y = 250

-5x + 20y = 0

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

25y = 250

Solve for y

y = 10

Since the total mixed gallons is 50,

x = 50 - 10 = 40

So we need 40 gallons of the 65% antifreeze and 10 gallons of the 90% antifreeze

Let y = amount of 90% antifreeze

EQUATION 1: x + y = 50 (total of 50 gallons mixed)

EQUATION 2: .65x + .90y = .70(x + y)

Simplify and solve the system of equations

Multiply second equation by 100 on both sides to remove the decimals

65x + 90y = 70(x + y)

Combine like terms

65x + 90y = 70x + 70y

65x -70x+ 90y-70y = 0

-5x + 20y = 0

Now we have the following system of equations:

x + y = 50

-5x + 20y = 0

Multiply the first equation by 5 to get opposite coefficients for x;

add the equations to eliminate x

5x + 5y = 250

-5x + 20y = 0

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

25y = 250

Solve for y

y = 10

Since the total mixed gallons is 50,

x = 50 - 10 = 40

So we need 40 gallons of the 65% antifreeze and 10 gallons of the 90% antifreeze