3 sandwich choices 2 side item choices 4 beverage choices
Let's say the sandwich choices were PB&J, Ham, and Turkey Let's say the sides were fries and mandarin oranges Let's say the beverages were water, orange juice, milk, soda
PB&J Ham Turkey
Fries Mandarin Oranges
Water OJ Milk Soda
The combinations can be as follows:
PB&J, Fries, Water PB&J, Fries, OJ PB&J, Fries, Milk PB&J, Fries, Soda PB&J, Mandarin Oranges, Water PB&J, Mandarin Oranges, OJ PB&J, Mandarin Oranges, Milk PB&J, Mandarin Oranges, Soda Ham, Fries, Water Ham, Fries, OJ Ham, Fries, Milk Ham, Fries, Soda Ham, Mandarin Oranges, Water Ham, Mandarin Oranges, OJ Ham, Mandarin Oranges, Milk Ham, Mandarin Oranges, Soda Turkey, Fries, Water Turkey, Fries, OJ Turkey, Fries, Milk Turkey, Fries, Soda Turkey, Mandarin Oranges, Water Turkey, Mandarin Oranges, OJ Turkey, Mandarin Oranges, Milk Turkey, Mandarin Oranges, Soda
So your answer is correct: There's 24 lunch combinations that can be made.
Suppose y=0. That means <span>x+z = 18 </span> <span>x-z = 26/5 </span>
<span>x = 58/5 </span> <span>z = 32/5 </span>
<span>So, the largest x can be is 58/5 </span> <span>Now, suppose we toss in y. If we subtract y/2 from both x and z, then the difference between x and z is still 26/5, and the sum is </span>
