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? (Round your answers to 2 decimal places)

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 = 4.16
m = 1.85

Steve I really appreciate the help THANK YOU

To solve this problem, we need to determine the number of glasses of skim milk and quarter-pound servings of lean red meat that would provide the desired amounts of iron and protein. Let's start by setting up some equations:

Let x be the number of glasses of skim milk.
Let y be the number of quarter-pound servings of lean red meat.

The amount of iron provided by x glasses of skim milk is 0.1x mg.
The amount of iron provided by y quarter-pound servings of lean red meat is 3.75y mg.

To find the total amount of iron, we can add these two amounts and set it equal to the desired amount:

0.1x + 3.75y = 7.36 ---(Equation 1)

Similarly, the amount of protein provided by x glasses of skim milk is 8.2x g.
The amount of protein provided by y quarter-pound servings of lean red meat is 21y g.

To find the total amount of protein, we can add these two amounts and set it equal to the desired amount:

8.2x + 21y = 73 ---(Equation 2)

We now have a system of two equations with two variables. Let's solve this system using substitution or elimination.

From Equation 2, we can solve for x in terms of y:
8.2x = 73 - 21y
x = (73 - 21y) / 8.2

Substitute this value of x into Equation 1:
0.1[(73 - 21y) / 8.2] + 3.75y = 7.36

Now, simplify and solve for y:
0.1(73 - 21y) + 3.75y * 8.2 = 7.36 * 8.2
7.3 - 2.1y + 30.75y = 60.232
28.65y = 52.932
y = 1.84

Substitute this value of y back into Equation 2 to find x:
8.2x + 21 * 1.84 = 73
8.2x + 38.64 = 73
8.2x = 73 - 38.64
8.2x = 34.36
x = 4.19

To have 7.36 mg of iron and 73 g of protein, the person on a special diet would need approximately 4.19 glasses of skim milk and 1.84 quarter-pound servings of lean red meat.