+0

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

0
514
3

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

Guest Feb 3, 2015

#2
+20526
+10

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

heureka  Feb 3, 2015
#1
+1090
+10

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.

Mathematician  Feb 3, 2015
#2
+20526
+10

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

heureka  Feb 3, 2015
#3
+92365
+3

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

CPhill  Feb 3, 2015