What are 3 cosecutive integers in which three times the first is 9 more than twice the last

Question

Find 3 consecutive integers such that 3 times the first is 9 more than twice the last

Answer 1

First, let's define our variables. We can do this easily due to the piece that says they must be consecutive integers. So, we can let a=x, b=x+1, and c=x+2. Now, we want to find a number such that 3a=9+2c.

By substituting a and c for x and x+2, we have 3x=9+2(x+2). When we distribute on the right side of the equation, we have 3x=9+2x+4. When we subtract 2x from both sides and combine like terms, we find that x=13.

Finally, we want to find the 3 consecutive integers, which simply would be 13,14,15. We can check this by plugging in 13 and 15 into the equation 3a=9+2c, which now gives us 39=39, which is correct.

Answer 2

The integers are n, n+1, and n+2. The relation states that

3(n) = 2(n+2) + 9

3n = 2n + 4 + 9

n = 13 (The lowest integer)

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