Question

An ounce of milk contains 20 milligrams of calcium and 60 micrograms of vitamin A.

An ounce of orange juice contains 40 milligrams of calcium and 10 micrograms of
vitamin A. How many ounces of milk and orange juice should she drink each day to
provide exactly 2200 milligrams of calcium and 1100 micrograms of vitamin A?

Answer 1

10 ounces of milk and 50 ounces of orange juice

Step-by-step explanation:

Let the amount of milk be x ounces and the amount of orange juice be y ounces.

1. An ounce of milk contains 20 milligrams of calcium and 60 micrograms of vitamin A. Then x ounces of milk contain 20x milligrams of calcium and 60x micrograms of vitamin A.

2. An ounce of orange juice contains 40 milligrams of calcium and 10 micrograms of vitamin A. Then y ounces of orange juice contain 40y milligrams of calcium and 10y micrograms of vitamin A.

Calcium:

20x milligrams in milk

40y milligrams in orange juice

2,200 milligrams in total

Hence,

Vitamin A:

60x micrograms in milk

10y micrograms in orange juice

1,100 micrograms in total

Hence,

Solve the system of two equations:

Divide each equation by 10:

From the second equation:

Substitute it into the first equation:

Answer 2

10 ounces of milk and 50 ounces of orange juice

Explanation:

Let the amount of milk be x ounces and the amount of orange juice be y ounces.

1. An ounce of milk contains 20 milligrams of calcium and 60 micrograms of vitamin A. Then x ounces of milk contain 20x milligrams of calcium and 60x micrograms of vitamin A.

2. An ounce of orange juice contains 40 milligrams of calcium and 10 micrograms of vitamin A. Then y ounces of orange juice contain 40y milligrams of calcium and 10y micrograms of vitamin A.

Calcium:

20x milligrams in milk

40y milligrams in orange juice

2,200 milligrams in total

Hence,

`20x+40y=2,200`

Vitamin A:

60x micrograms in milk

10y micrograms in orange juice

1,100 micrograms in total

Hence,

`60x+10y=1,100`

Solve the system of two equations:

`\left\{\begin{array}{l}20x+40y=2,200\\ \\60x+10y=1,100\end{array}\right.`

Divide each equation by 10:

`\left\{\begin{array}{l}2x+4y=220\\ \\6x+y=110\end{array}\right.`

From the second equation:

`y=110-6x`

Substitute it into the first equation:

`2x+4(110-6x)=220\\ \\2x+440-24x=220\\ \\2x-24x=220-440\\ \\-22x=-220\\ \\22x=220\\ \\x=10\\ \\y=110-6\cdot 10=110-60=50`