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?
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.
To find the temperatures of City A and City B, we will solve the given system of equations:
1) 4x = 7 + 3y
2) x − 3y = −5
Let's solve equation (2) for x:
x = 3y - 5
Now, substitute this value of x into equation (1):
4(3y - 5) = 7 + 3y
12y - 20 = 7 + 3y
Combine like terms:
12y - 3y = 7 + 20
9y = 27
Divide both sides by 9:
y = 27 / 9
y = 3
Now substitute the value of y back into equation (2) to find x:
x - 3(3) = -5
x - 9 = -5
Add 9 to both sides:
x = 9 - 5
x = 4
Therefore, the temperature of City A on that day was 4°C, and the temperature of City B was 3°C.
To solve the given system of equations, we can use the method of substitution. Here's how:
Step 1: Solve one of the equations for one variable in terms of the other variable. Let's solve the second equation (x − 3y = −5) for x:
x = 3y - 5
Step 2: Substitute this expression for x into the other equation. Substitute (3y - 5) for x in the first equation (4x = 7 + 3y):
4(3y - 5) = 7 + 3y
Step 3: Simplify and solve for y:
12y - 20 = 7 + 3y
Subtract 3y from both sides:
12y - 3y - 20 = 7
9y - 20 = 7
Add 20 to both sides:
9y = 27
Divide both sides by 9:
y = 3
Step 4: Substitute the value of y back into the second equation (x − 3y = −5) to solve for x:
x − 3(3) = −5
x - 9 = -5
Add 9 to both sides:
x = 4
So, the temperature of City A on that day was 4° C, and the temperature of City B was 3° C.