find three consecutive integers so that three times the middle integer is five more than the sum of the first and third.

find three consecutive integers so that three times the middle integer is five more than the sum of the first and third.

$$3a_2=5+a_1+a_3 \quad | \quad a_1 = a_2-1 \qquad a_3=a_2+1\\ 3a_2=5+(a_2-1)+(a_2+1)\\ 3a_2=5+a_2+a_2\\ 3a_2=5+2a_2\\ a_2=5\\ a_1 = a_2-1 = 5-1=4\\ a_3=a_2+1 = 5+1=6$$

4, 5, 6

Let's call our three integers "s", "m", and "l", "s" being the smallest and "l" being the largest.

Before looking at the equation, we know:

s + 1 = m

m + 1 = l

So the equation you gave me was:

3m = s + l + 5

The easiest way to do this is the substitution method: Since we know what l equals, and we know what m equals, we can substitute them in the third equation.

Let me show you:

3m = s + l + 5

We can replace l with m + 1, since that's what it equals.

3m = s + (m + 1) + 5

At this point, since we already have 2 m's, it will be easier to substitue something for s.

We can find out what s equals by modifying the equation s + 1 = m.

Taking one away from each side will make the equation s = m - 1

So now we know what s equals, and we can substitute it as well:

3m = (m - 1) + (m + 1) + 5

Going on:

3m = m - 1 + m + 1 + 5

3m = 2m + 5

Now just need to isolate the variable m. The easiest way to do this is by subtracting 2m from each side:

(1)m = 5

m = 5

Now that we know what m equals, we can substitute in the other two original equations:

s + 1 = m

s + 1 = 5

Subtracting one from each side:

s = 4

And the next one:

m + 1 = l

5 + 1 = l

6 = l

We now know our three numbers: 4, 5, & 6.

Very nice, Mathematician and heureka....!!!