Question

What are three consecutive multiples of 3 if 2/3 of the sum of the first 2 numbers is 1 greater than the third number?

[please help]

Answer 1

Let's start by using algebra to represent the problem.

Let x be the first multiple of 3, then the next two consecutive multiples of 3 are x + 3 and x + 6.

We know that 2/3 of the sum of the first two numbers is 1 greater than the third number, so we can write:

2/3(x + x + 3) = (x + 6) + 1

Simplifying, we get:

4x/3 + 2 = x + 7

Multiplying both sides by 3 to get rid of the fraction, we get:

4x + 6 = 3x + 21

Subtracting 3x and then simplifying, we get:

x = 15

Therefore, the first multiple of 3 is 15, and the next two consecutive multiples of 3 are 18 and 21.

So the three consecutive multiples of 3 are 15, 18, and 21.

Explanation: