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# Find three consecutive integers such that the product of the first and third is 17 greater that 7 times the second

This would help me a lot thank you.

### 2 Answers by Expert Tutors

Karin S. | Reasonable and Reliable Math Tutor Reasonable and Reliable Math Tutor
4.9 4.9 (838 lesson ratings) (838)
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So the 3 consecutive integers can be denoted as n,n+1 and n+2.

So turn the statement from English to math.
n(n+2)=17+7(n+1). Now simplify and solve

n2+2n=17+7n+7
now bring everything to 1 side.

n2-5n-24=0. The factors of -24 that sum to a -5 are -8 and 3 so this can be factored into
(n-8)(n+3)=0

n=8 or -3. So now since they say integers we need to examine the possibilities of the different integers.

If n is 8 then the numbers are 8,9,10. So let's check the statement. 8(10) compared to 7(9)+17. This checks out.

Now let's check the other possibilities.
-3,-2,-1.
(-3)(-1) compared to 7(-2)+17
3=3 and so there are 2 sets of solutions to this problem.

-3,-2,-1 and 8,9,10

Mark M. | Mathematics Teacher - NCLB Highly QualifiedMathematics Teacher - NCLB Highly Qualif...
4.9 4.9 (187 lesson ratings) (187)
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n, n + 1, n + 2
n(n + 2) = 7(n + 1) + 17

Can you solve for "n?"