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

Question

This would help me a lot thank you. 

Answer 1

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

 

 

Answer 2

n, n + 1, n + 2

n(n + 2) = 7(n + 1) + 17

 

Can you solve for "n?"