Answer:
30 gallons of first brand & 20 gallons of second brand
Step-by-step explanation:
Let the amount in gallons of 65% antifreeze needed be x.
Let the amount in gallons of 90% antifreeze needed be y.
You need to make a total of 50 gallons, so that gives us our first equation:
x + y = 50
Now we deal with the amount of pure antifreeze in each amount of antifreeze and in the result.
x gallons of 65% antifreeze contain 0.65x gallons of pure antifreeze.
y gallons of 90% antifreeze contain 0.9x gallons of pure antifreeze.
The result is 50 gallons of 75% pure antifreeze which contains 0.75 * 50 gallons of pure antifreeze. 0.75 * 50 = 37.5
Now we add the two amounts of pure antifreeze and set equal to the result.
0.65x + 0.9y = 37.5
This is the second equation. We have a system of two equations in two unknowns.
x + y = 50 Equation 1
0.65x + 0.9y = 37.5 Equation 2
Multiply the first equation by -0.65
-0.65x - 0.65y = -32.5
0.65x + 0.9y = 37.5 This is Equation 2.
Add the two equations above to eliminate x.
0.25y = 5
y = 20
Substitute 20 for y in the original equation and solve for x.
x + y = 50
x + 20 = 50
x = 30
Answer: 30 gallons of first brand & 20 gallons of second brand
