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 8Á C more than 3 times the temperature of City B. The temperature of City A minus twice the temperature of City B was _3Á C. What was the temperature of City A and City B on that day?

City A was 5Á C, and City B was 4Á C.
City A was 3Á C, and City B was _1Á C.
City A was 8Á C, and City B was _3Á C.
City A was 5Á C, and City B was _5Á C.

Answer 1

Let A be the temperature of city A and B be the temperature of city B.
4A = 3B + 8
Also,
A - 2B = -3
Making A the subject in the second equation:
A = 2B - 3
Then we substitute this into the first equation:
4(2B - 3) = 3B + 8
8B - 12 = 3B + 8
5B = 20;
B = 4 °C
A = 2(4) - 3
A = 5 °C
The first option is correct

Answer 2

The answer is City A was 5Á C, and City B was 4Á C.

a - the temperature of City A
b - the temperature of City B

Four times the temperature of City A was 8°C more than 3 times thetemperature of City B: 4a = 8 + 3b
The temperature of City A minus twice the temperature of City B was -3°C:
a - 2b = -3

Now we have the system of two equations:
4a = 3b + 8
a - 2b = -3
___________
Rearrange the first equation:
4a - 3b = 8
a - 2b = -3
___________
Multiply the second equation by (-4):
4a - 3b = 8
-4a + 8b = 12
___________
Add these two equations:
4a - 3b - 4a + 8b = 8 + 12
___________
4a can be canceled out:
8b - 3b = 8 + 12
5b = 20
b = 20/5
b = 4

Since we know the temperature of City B, we will calculate the temperature of City A by using the equation from above:
4a = 8 + 3b
4a = 8 + 3·4
4a = 8 + 12
4a = 20
a = 20/4
a = 5

Therefore, the temperature of City A was 5°C and the temperature of City B was 4°C