Tim has $20 to buy snacks for 12 people in an office. Each person will get one snack. Tim is buying bags of pretzels that cost $1.50 per bag and bags of crackers that cost $2.00 per bag.

4 bags of crackers and 8 bags of pretzels.

Step-by-step explanation:

Let number of crackers=c

Let number of pretzels=p

Tim is buying snacks for 12 people and each person gets only one snacks. Therefore:

c+p=12. (Equation 1)

Bags of pretzels costs $1.50 per bag.

Bags of crackers that cost $2.00 per bag.

Tim has $20 to buy snacks.

1.50p+2c=20. (Equation 2)

We then solve the two equations to find out how many bag of each type of snacks can be bought.

c+p=12.

1.50p+2c=20

From Equation (1), c=12-p

Substitute c=12-p into equation (2).

1.50p+2(12-p)=20

Opening the bracket

1.50p+24-2p=20

Next, I collect like terms

1.50p-2p=20-24

-0.5p=-4

Divide both sides by -0.5 to solve for p.

p=8.

Recall from Equation 1.

c=12-p =12-8=4

Therefore, Tim can get 4 bags of crackers and 8 bags of pretzels.

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