Natalie kept track of her calcium intake from three sources for 3 days. The first day she had 2 glasses of milk, 4 servings of ice cream, and 3 calcium supplements in pill form which totaled 2600 mg of calcium. The second day she had 3 glasses of milk, 4 servings of ice cream, and 3 calcium supplements totaling 3200 mg. The third day she had 2 glasses of milk, 5 servings of ice cream, and 3 calcium supplements for a total of 2650 mg. Find the amount of calcium in one glass of milk, in one serving of ice cream, and in one calcium supplement.

Question

Natalie kept track of her calcium intake from three sources for 3 days. The first day she had 2 glasses of milk, 4 servings of ice cream, and 3 calcium supplements in pill form which totaled 2600 mg of calcium. The second day she had 3 glasses of milk, 4 servings of ice cream, and 3 calcium supplements totaling 3200 mg. The third day she had 2 glasses of milk, 5 servings of ice cream, and 3 calcium supplements for a total of 2650 mg. Find the amount of calcium in one glass of milk, in one serving of ice cream, and in one calcium supplement.

Answer 1

Let the amounts of calcium in milk, ice cream and supplements be

x, y, and z respectively

2x + 4y + 3z = 2600
3x + 4y + 3z = 3200
2x + 5y + 3z = 2650

easy to solve:
subtract #1 from #2 ---> x = 600
subtract #2 from #3 ---> -x + y = -550
plug in x = 600
-600 + y = -550 ---> y = 50
back into the first:
2x + 4y + 3z = 2600
1200 + 200 + 3z = 2600

finish it up

Answer 2

To find the amount of calcium in one glass of milk, one serving of ice cream, and one calcium supplement, we need to set up a system of equations using the information given.

Let's assign variables:
- Let's call the amount of calcium in one glass of milk "m."
- Let's call the amount of calcium in one serving of ice cream "i."
- Let's call the amount of calcium in one calcium supplement "s."

Now let's set up the equations based on the information given:

From Day 1:
2m + 4i + 3s = 2600

From Day 2:
3m + 4i + 3s = 3200

From Day 3:
2m + 5i + 3s = 2650

We have a system of three equations with three variables. Now, we can solve this system to find the values of m, i, and s.

To do this, we can use the method of elimination or substitution. Here, let's solve it using the method of substitution:

From the first equation, we can express s in terms of m and i:
s = (2600 - 2m - 4i) / 3

Now, we substitute s in the second and third equations:
3m + 4i + 3((2600 - 2m - 4i) / 3) = 3200
2m + 5i + 3((2600 - 2m - 4i) / 3) = 2650

Simplify and solve for m and i:

3m + 4i + (2600 - 2m - 4i) = 3200
2m + 5i + (2600 - 2m - 4i) = 2650

Now, simplify and combine like terms:
3m + 2600 - 2m = 3200
2m + 2600 - 2m = 2650

Simplify further:
m + 2600 = 3200
2600 = 2650

Subtract 2600 from both sides of the equation:
m = 600
0 = 50

Uh-oh! Since there's no solution for m and i, it seems there may be a mistake in the provided information or calculation. Please double-check the values and try solving the problem again.