Question
A restaurant sells burritos and tacos
-one person paid 10.00 for 3 burritos and 2 tacos
-another person paid 6.25 for 1 burrito and 3 tacos
What is the cost of 1 burrito at the restaurant?
-one person paid 10.00 for 3 burritos and 2 tacos
-another person paid 6.25 for 1 burrito and 3 tacos
What is the cost of 1 burrito at the restaurant?
Answers
GPT-4o mini
Let's denote the cost of one burrito as \( b \) and the cost of one taco as \( t \).
We can set up a system of equations based on the information given:
1. From the first person's order, we have:
\[
3b + 2t = 10.00
\]
2. From the second person's order, we have:
\[
1b + 3t = 6.25
\]
Now we can solve this system of equations.
First, we can manipulate the second equation to express \( b \) in terms of \( t \):
\[
b = 6.25 - 3t
\]
Next, we substitute \( b \) into the first equation:
\[
3(6.25 - 3t) + 2t = 10.00
\]
Expanding this gives:
\[
18.75 - 9t + 2t = 10.00
\]
Combining the terms:
\[
18.75 - 7t = 10.00
\]
Now, isolating \( t \):
\[
-7t = 10.00 - 18.75
\]
\[
-7t = -8.75
\]
\[
t = \frac{-8.75}{-7} = 1.25
\]
Now that we have \( t \), we can find \( b \):
\[
b = 6.25 - 3(1.25)
\]
\[
b = 6.25 - 3.75
\]
\[
b = 2.50
\]
Thus, the cost of one burrito at the restaurant is **$2.50**.
We can set up a system of equations based on the information given:
1. From the first person's order, we have:
\[
3b + 2t = 10.00
\]
2. From the second person's order, we have:
\[
1b + 3t = 6.25
\]
Now we can solve this system of equations.
First, we can manipulate the second equation to express \( b \) in terms of \( t \):
\[
b = 6.25 - 3t
\]
Next, we substitute \( b \) into the first equation:
\[
3(6.25 - 3t) + 2t = 10.00
\]
Expanding this gives:
\[
18.75 - 9t + 2t = 10.00
\]
Combining the terms:
\[
18.75 - 7t = 10.00
\]
Now, isolating \( t \):
\[
-7t = 10.00 - 18.75
\]
\[
-7t = -8.75
\]
\[
t = \frac{-8.75}{-7} = 1.25
\]
Now that we have \( t \), we can find \( b \):
\[
b = 6.25 - 3(1.25)
\]
\[
b = 6.25 - 3.75
\]
\[
b = 2.50
\]
Thus, the cost of one burrito at the restaurant is **$2.50**.