Question

I'm not sure how to do this. Please include steps. WILL GIVE BRAINLY IF RIGHT!!!

Answer 1

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