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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?

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JcAlmighty
The answer is the temperature of City A was 5°C and the temperature of City B was 4°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 the temperature 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
___________
If we express the second equation as a = 2b - 3, we can substitute a in the first equation.
4(2b - 3) = 3b + 8
4 · 2b - 4 · 3 = 3b + 8
8b - 12 = 3b + 8
8b - 3b = 12 + 8
5b = 20
⇒ b = 20/5 = 4

Now, calculate a from the equation a = 2b - 3:
a = 2 · 4 - 3 = 8 - 3 = 5

So: b = 4, a = 5


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

Answer:

City A was 5°C, and City B was 4°C.

Step-by-step explanation:

I just took the test and this question was correct. Hope this helped and good luck!

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