Mark M. answered • 09/12/17

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The consecutive integers:

n, n+1, and n+2

11(n+2) ≥ n(n+1) + 46

Can you solve for n and answer?

Emily A.

asked • 09/12/17I am in algebra 2 and my teacher gave us this and is moving on tomorrow and I still dont understand how to do it

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Mark M. answered • 09/12/17

Tutor

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(243)
Mathematics Teacher - NCLB Highly Qualified

The consecutive integers:

n, n+1, and n+2

11(n+2) ≥ n(n+1) + 46

Can you solve for n and answer?

Rich G. answered • 09/12/17

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Experienced Algebra Tutor at High School and College Level

So you have 3 consecutive integers - n, n+1, n+2. The product of 11(n+2) is at least 46 more than the product of n(n+1).

Our equation would look like this:

11(n+2) ≥ 46 + n(n+1)

Multiply it out:

11n + 22 ≥ 46 + n^{2} + n

Since you have a square term, set the inequality in terms of zero:

0 ≥ n^{2} -10n + 24

factor:

0 ≥ (n-6)(n-4)

Our numbers of interest are n = 6 and n = 4. When n > 6 the inequality won't be true (for example, when n = 7 the result will be (7-6)(7-4) = 3 which is not less than 0). When n < 4 it will also not be true since we'll have a negative times a negative, which will be >0 (example, (3-6)(3-4) = 3) so we can rule out when n <4. Therefore the only time it will be true is when n is between 4 and 6

n = 6

n+1 = 7

n+2 = 8

8(11) ≥ 46 + 6*7

n = 5

n+1 = 6

n+2 = 7

7(11) ≥ 46 + 5*6

n = 4

n+1 = 5

n+2 = 6

6(11) ≥ 46 + 4*5

These all check out so all the triples are:

4,5,6

5,6,7

6,7,8

It's a bit wordy but I hope it helps

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