Try this suggested option:
Step-by-step explanation:
1) if Maria has 180USD, then the maximum number of the pants she can buy is 180/15=12 (in this case number of the shirts is 0) and the maximum number of the shirts is 180/9=20 (and in this case number of the pants is 0).
If suppose the number of the pants is 'p' and of the shirts is 's', then
0 ≤p ≤12 - the 1st line in the system;
0 ≤s ≤20 - the 2d in the system.
2) according to the condition Maria has only 180 (USD), she can buy any number of 's' and 'p', this can be written as:
9s+15p = 180 - the 3d line in the system.
3) if to set up the required system, then (note, 's' and 'p' ∈Z):
[tex]\left \{\begin{array}{ccc}9s+15p=180\\0\leq p\leq 12\\0\leq s\leq 20\end{array}\right[/tex]
4) if to solve this system, only five pairs of s&p exist. They are:
0 shirts and 12 paints;
5 shirts and 9 paints;
10 shirts and 6 paints;
15 shirts and 3 paints;
20 shirts and 0 paints.
According to the condition Maria wants to buy s&p for all her 18 cousins, then the pair '15 shirts and 3 paints' is the best choice.
Answer: 15 shirts and 3 paints
PS. for more details see the attached graph, note the five pairs are marked with orange colour as points A, B, C, D and E.
