How do you find three consecutive integers such that twice the smallest is six more than the largest?

Answer 1

Three consecutive integers:

n, n + 1, n + 2

2n = n + 2 + 6

Can you solve for n and answer?

Answer 2

Hello, Darwin.

Okay, let's call the 3 consecutive integers:

x, x+1, and x+2

So twice the smallest (x) = 6 more than the largest (x+2)

So:

2(x) = (x + 2) + 6

2x = x + 8

x = 8

So, if x = 8, the other numbers are 9 and 10

So, the answer is:

8, 9, 10

Now check:

2(8) = 10 + 6

16 = 16

The answer is correct.

Answer 3

Let x = the smallest number

x+1 = next number

x+ 2 = largest number

Then, based on the information given 2x = (x+2) + 6

or x = 8

So, x = 8

x+1 = 9

x+2 = 10