Final answer:
The equation in standard form that models the possible combinations of breakfasts and lunches a student can afford for 10 days, with $x$ representing breakfasts and $y$ representing lunches, is 3x + 6y = 90.
Step-by-step explanation:
The subject of this question is using algebra to model a budget constraint, which is a common concept in consumer economics. The student is asking us to create an equation that models the possible combinations of breakfasts and lunches they can buy with a certain amount of money. To begin, we calculate the total cost for 10 days of breakfast and lunch. The cost for one breakfast is $1.50 and the cost for one lunch is $3.00. Therefore, the total cost for 10 days would be 10 breakfasts times $1.50 plus 10 lunches times $3.00.
Let x represent the number of breakfasts and y represent the number of lunches. The equation representing the total amount of money spent for 10 days is:
1.50x + 3.00y = Total cost for 10 days
Since the student wants to add exactly enough money to their meal account to pay for 10 breakfasts and 10 lunches, we replace the total cost with the actual amount:
1.50x + 3.00y = (1.50 × 10) + (3.00 × 10)
Now we calculate the right side of the equation:
1.50x + 3.00y = 15 + 30
Finally, we simplify it to:
1.50x + 3.00y = 45
To express this equation in standard form Ax + By = C, where A, B, and C are integers, we multiply the entire equation by 2 to eliminate the decimals and get:
3x + 6y = 90
This is the equation in standard form that models the possible combinations of breakfasts and lunches the student can afford to buy for 10 days, with x representing the number of breakfasts and y representing the number of lunches.