Answer: 10 tables and 10 chairs
Step-by-step explanation: Because we know neither x or y can be equal to 0 we can eliminate Option 2 and 3 and that leaves us with 1 and 4. Looking back at the formula 5x+3y is less than or equal to 80 if we plug in 20 for both x (tables) and y(chairs) 5(20)+3(20)= 160 which is much greater than 80 so only Option is valid with 10 tables and 10 chairs or 5(10)+3(10) = 80 which makes our equation valid :)
