Let's assume the number of glasses of skim milk to be "x" and the number of quarter pound servings of lean meat to be "y."
We are given the following information:
1 glass of skim milk supplies 0.1 mg of iron and 8.3 g of protein.
1 quarter pound serving of lean meat provides 3.4 mg of iron and 20 g of protein.
The person on the special diet needs to have 7.3 mg of iron and 81.5 g of protein.
Based on this information, we can set up the following system of equations to represent the requirements:
Equation 1: 0.1x + 3.4y = 7.3 (equation for iron)
Equation 2: 8.3x + 20y = 81.5 (equation for protein)
Now, we can solve this system of equations to find the values of "x" and "y."
Step 1: Multiply Equation 1 by 83 and Equation 2 by 10 to eliminate decimals:
83 * (0.1x + 3.4y) = 83 * 7.3
10 * (8.3x + 20y) = 10 * 81.5
Resulting in:
8.3x + 283y = 604.9
83x + 200y = 815
Step 2: Multiply Equation 1 by 10 and Equation 2 by -1 to eliminate x:
10 * (0.1x + 3.4y) = 10 * 7.3
-1 * (8.3x + 20y) = -1 * 81.5
Resulting in:
x + 34y = 73
-8.3x - 20y = -81.5
Step 3: Add the equations obtained in Step 2 to eliminate x:
x + 34y - 8.3x - 20y = 73 - 81.5
-7.3x + 14y = -8.5
Step 4: Solve for y:
14y = -8.5
y = -8.5 / 14
y ≈ -0.607
Step 5: Substitute the value of y into one of the original equations to solve for x (let's use Equation 1):
0.1x + 3.4(-0.607) = 7.3
0.1x - 2.0618 = 7.3
0.1x = 7.3 + 2.0618
0.1x = 9.3618
x ≈ 93.618
Since the number of glasses of milk and servings of meat cannot be negative, we can ignore the negative solutions. Therefore, the person on the special diet should have approximately 93.618 glasses of skim milk and 0 quarter pound servings of lean meat to meet the iron and protein requirements.