Respuesta :
Let the number of calories from lunch be called L. As such, breakfast is then L + 128, and dinner is 2L - 400. We can then sum the three meals and equate it to the total caloric intake, the known value of 1932.
So: 1932 = L + L + 128 + 2L - 400 = 4L - 272.
Lunch = 551
Breakfast = 551 + 128 = 679
Dinner = 2*551 - 400 = 702
So: 1932 = L + L + 128 + 2L - 400 = 4L - 272.
Lunch = 551
Breakfast = 551 + 128 = 679
Dinner = 2*551 - 400 = 702
Answer:
The correct answer is:
• Breakfast: 679 calories
• Lunch: 551 calories
• Dinner: 702 calories
Explanation:
For the example we have to consider the calories from breakfast, lunch, and dinner as follows:
• Breakfast: x + 128
• Lunch: x
• Dinner: 2x – 400
According the problem, the total of calories consumed that day was 1932. Thus:
Breakfast + Lunch + Dinner = 1932
(x + 128) + x + (2x – 400) = 1932
4x – 272 = 1932
4x = 2204
x = 551
Then,
• Breakfast: x + 128 = 679 calories
• Lunch: x = 551 calories
• Dinner: 2x – 400 = 702 calories