Answer :
Cortes can buy 1 kilo of pork and at least 4 kilos of chicken, or any combination where the total weight is at least 5 kilos and the cost is at most Php 1000.
To solve this problem, let's use algebraic equations. Let's assume x represents the number of kilos of pork and y represents the number of kilos of chicken.
We know that x + y should be greater than or equal to 5, since Cortes wants to buy at least 5 kilos in total.
We also know that 250x + 170y should be less than or equal to 1000, since Cortes can spend at most Php 1000.
To find the possible combinations of pork and chicken, we can set up inequalities:
[tex]x + y ≥ 5[/tex]
[tex]250x + 170y ≤ 1000[/tex]
To solve these inequalities, we can use graphical methods or substitution. For simplicity, let's use substitution.
Let's solve the first equation for y:
y ≥ 5 - x
Now substitute this into the second equation:
[tex]250x + 170(5 - x) ≤ 1000[/tex]
[tex]250x + 850 - 170x ≤ 1000[/tex]
[tex]80x ≤ 150[/tex]
[tex]x ≤ 150/80[/tex]
[tex]x ≤ 1.875[/tex]
Since the number of kilos cannot be a fraction, we can round down x to 1. Therefore, Cortes can buy 1 kilo of pork.
Substituting this value back into the first equation:
[tex]1 + y ≥ 5[/tex]
[tex]y ≥ 4[/tex]
Therefore, Cortes can buy 1 kilo of pork and at least 4 kilos of chicken, or any combination where the total weight is at least 5 kilos and the cost is at most Php 1000.
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