A glass of skim milk supplies 0.1 mg of iron and 8.2 g of protein. A quarter pound of lean red meat provides 3.75 mg of iron and 21 g of protein. If a person on a special diet is to have 7.36 mg of iron and 73 g of protein, how many glasses of skim milk and how many quarter-pound servings of meat would provide this?

If there are g glasses of skim and m servings of meat

.1g + 3.75m = 7.36
8.2g + 21m = 73

Now just solve and you get

g = 4.1
m = 1.8

So, 5 glasses of mile and 2 servings of meat will provide the desired amounts

To calculate the number of glasses of skim milk and quarter-pound servings of lean red meat needed to provide 7.36 mg of iron and 73 g of protein, we need to set up a system of equations.

Let's assign the variables:
x = number of glasses of skim milk
y = number of quarter-pound servings of lean red meat

Based on the given information, we can write the following equations:

Equation 1: 0.1x + 3.75y = 7.36 (equation for iron content)
Equation 2: 8.2x + 21y = 73 (equation for protein content)

Now, we need to solve this system of equations.

By using any method (such as substitution, elimination, or matrix), we can find the values of x and y.

For simplicity, let's solve using the substitution method:

From Equation 1, we can solve for x in terms of y:
0.1x = 7.36 - 3.75y
x = (7.36 - 3.75y) / 0.1
x = 73.6 - 37.5y

Now we substitute this expression for x in Equation 2:
8.2(73.6 - 37.5y) + 21y = 73
602.72 - 305y + 21y = 73
602.72 - 305y + 21y = 73
602.72 - 284y = 73
602.72 - 73 = 284y
529.72 = 284y
y ≈ 1.86

Now we have the value of y, which represents the number of quarter-pound servings of lean red meat. Since we can't have a fraction for the servings, the number of servings needs to be rounded to the nearest whole number.

Therefore, we need approximately 2 quarter-pound servings of lean red meat.

To find the number of glasses of skim milk (x), substitute this value of y into Equation 1:
x = (7.36 - 3.75y) / 0.1
x = (7.36 - 3.75 * 1.86) / 0.1
x ≈ 22.6

So, we need approximately 23 glasses of skim milk.

Therefore, to provide 7.36 mg of iron and 73 g of protein, approximately 23 glasses of skim milk and 2 quarter-pound servings of lean red meat would be needed.

To solve this problem, we need to set up a system of equations using the given information.

Let's represent the number of glasses of skim milk as "m" and the number of quarter-pound servings of meat as "r".

The equation for the iron content is:
0.1m + 3.75r = 7.36

The equation for the protein content is:
8.2m + 21r = 73

Now, we can solve this system of equations using either substitution or elimination method.

Let's use the elimination method to solve this system:

Multiply the first equation by 10 to eliminate the decimals:
1m + 37.5r = 73.6

Multiply the second equation by 5:
41m + 105r = 365

Next, multiply the first equation by 41 and the second equation by 1 to create a system of equations with equal coefficients of "m":
41m + 1875r = 3017.6
41m + 105r = 365

Now, subtract the second equation from the first equation:
(41m + 1875r) - (41m + 105r) = 3017.6 - 365
1770r = 2652.6

Divide both sides of the equation by 1770:
r = 1.5

Now, substitute the value of "r" back into either equation. Let's use the first equation:
0.1m + 3.75(1.5) = 7.36
0.1m + 5.625 = 7.36
0.1m = 7.36 - 5.625
0.1m = 1.735

Divide both sides of the equation by 0.1:
m = 1.735 / 0.1
m = 17.35

Since we cannot have a fraction of a serving, we need to round our answers.

Therefore, 1 quarter-pound serving of meat and 17 glasses of skim milk would provide 7.36 mg of iron and 73 g of protein.