Question

what is the smallest of 3 consecutive positive integers if the product of smaller two interfere is 5 less than 5 times the larger?

Answer 1

Answer:

Two options:

5,6, and 7

-1,0,and 1

Step-by-step explanation:

Three consecutive natural numbers can be represented as n, n+1, and n+2.

The product of the smaller two would be n(n+1). The less than 5 times the larger is 5(n+2)-5.

Set them equal and solve by factoring:

$n(n+1) = 5(n+2) -5\\n^2+n = 5n+10 -5\\n^2+n = 5n+5\\n^2 -4n-5=0\\(n-5)(n+1)=0$

Set each factor equal to 0.

n-5==0 so n=5.

n+1=0 so n=-1.

This means the 3 consecutive numbers would be 5,6, and 7

OR

-1, 0 or 1.