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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. 

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Karin S. | Reasonable and Reliable Math Tutor Reasonable and Reliable Math Tutor
5.0 5.0 (785 lesson ratings) (785)
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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 (169 lesson ratings) (169)
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n, n + 1, n + 2
n(n + 2) = 7(n + 1) + 17
 
Can you solve for "n?"