Asparkling-water distributor wants to make up 200 gallons of sparkling water to sell for $5.00 per gallon. she wishes to mix three grades of water selling for $8.00, $3.00, and $4.50 per gallon, respectively. she must use twice as much of the $4.50 water as the $3.00, water. how many gallons of each should she use?

Assuming she make no profit and no loss from the business.

Let the number of gallons of the $8 grade water used be x, that of the $3 grade water, y, and that of the $4.50 grade water be z, then:

x + y + z = 200 . . . (1)
8x + 3y + 4.5z = 200(5) = 1,000 . . . (2)
z = 2y . . . (3)

Putting equation (3) into equations (1) and (2), we have:

x + y + 2y = 200
or x + 3y = 200 . . . (4)
and
8x + 3y + 4.5(2y) = 1000
or 8x + 3y + 9y = 1000
or 8x + 12y = 1000 . . . (5)

Multiplying equation (4) by 4, we have:
4x + 12y = 800 . . . (6)

Subtracting equation (6) from equation (5), we have:
4x = 200
or x = 200 / 4 = 50

Substituting for x into equation (4), we have:
50 + 3y = 200
or 3y = 200 - 50 = 150
or y = 150 / 3 = 50

Substituting for z into equation (3) gives:

z = 2(50) = 100

Therefore, 50 gallons each of the $8 grade water and the $3 grade water should be used and 100 gallons of the $4.50 grade water.

Is there any commas or anything

Answers: the graph is translated down 1 unit and the domain and range of g(x) is the same as the domain and range of f(x).