How many gallons of 35% alcohol solution and 50% alcohol solution must be mixed to get 9 gallons of 40% alcohol solution?

There needs to be
gallons of 35 % alcohol solution and
gallons of 50 % solution.

There needs to be   6     gallons of 35 % alcohol solution

and    3     gallons of 50 % solution.

Explanation:

x = number of gallons of the 35% solution

y = number of gallons of the 50% solution

The two amounts must add to 9 gallons, so x+y = 9

This solves to y = 9-x

Since we want 9 gallons of 40% alcohol solution, this means we want 0.40*9 = 3.6 gallons of pure alcohol.

0.35x = amount of pure alcohol from the first batch

0.50y = amount of pure alcohol from the second batch

0.35x+0.50y = total amount of pure alcohol

Set that expression equal to the 3.6 figure calculated earlier (the total amount of pure alcohol we're after) and we get this equation

0.35x+0.50y = 3.6

Next, we replace y with 9-x and then solve for x.

0.35x+0.50y = 3.6

0.35x+0.50(9-x) = 3.6

0.35x+4.5-0.50x = 3.6

-0.15x+4.5 = 3.6

-0.15x = 3.6-4.5

-0.15x = -0.9

x = -0.9/(-0.15)

x = 6

We'll need 6 gallons of the 35% solution

Use this x value to find y

y = 9-x

y = 9-6

y = 3

We'll also need 3 gallons of the 50% solution

b.-4 units

all you   do is count how many times you move