What is the smallest of 3 consecutive positive integers if the product of the smaller two integers is 5 less than 5 t imes the largest integer?

The answer is x= 5 
Hope this helps!

Consecutive positive integers will be three numbers right in a row on the number line.

x= 1st integer
x+1= 2nd integer
x+2= 3rd integer

product= multiply
less than= subtraction
is= equal sign

FIRST INTEGER
(x)(x + 1)= (5)(x + 2) - 5
multiply both sets of parentheses 1st

(x*x) + (x*1)= (5*x) + (5*2) - 5
multiply within parentheses

x^2 + x= 5x + 10 - 5
combine like terms

x^2 + x= 5x + 5
subtract 5x from both sides

x^2 - 4x= 5
subtract 5 from both sides

x^2 - 4x - 5= 0
factor

(x + 1)(x - 5)= 0
set each parentheses equal to 0

x + 1= 0
x= -1

x - 5= 0
x= 5 first integer

***since the integers are positive, x has to be equal to 5, not -1

SECOND INTEGER
= x + 1
= 5 + 1
= 6 second integer

THIRD INTEGER
= x + 2
= 5 + 2
= 7 third integer

ANSWER: The smallest of three consecutive positive integers is 5.

Hope this helps! :)

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