Question
City A and City B had two different temperatures on a particular day. On that day, four times the temperature of City A was 7° C more than three times the temperature of City B. The temperature of City A minus three times the temperature of City B was −5° C. The following system of equations models this scenario:
4x = 7 + 3y
x − 3y = −5
What was the temperature of City A and City B on that day?
4x = 7 + 3y
x − 3y = −5
What was the temperature of City A and City B on that day?
Answers
Bot
To solve the system of equations, we can use substitution. Rearranging the second equation, we get x = 3y - 5. Substituting this into the first equation, we get:
4(3y - 5) = 7 + 3y
12y - 20 = 7 + 3y
9y = 27
y = 3
Substituting y = 3 into x = 3y - 5, we get:
x = 3(3) - 5 = 4
Therefore, the temperature of City A was 4°C and the temperature of City B was 3°C on that day.
4(3y - 5) = 7 + 3y
12y - 20 = 7 + 3y
9y = 27
y = 3
Substituting y = 3 into x = 3y - 5, we get:
x = 3(3) - 5 = 4
Therefore, the temperature of City A was 4°C and the temperature of City B was 3°C on that day.