Find three consecutive positive integers such that the product of the first and third, minus the second, is 1 more than 3 times the third

Question

Find three consecutive positive integers such that the product of the first and third, minus the second, is 1 more than 3 times the third.

Answer 1

Let x equal the first of 3 consecutive positive integers. That means x+1 and x+2 represent the next two after the first.

 

Now translate the English into Algebra.

 

Product of 1st and 3rd is represented by x(x+2)

minus the 2nd: x(x+2) - (x+1)

IS: =

1 more than 3 times the third: 3(x+2) + 1

 

So we have:

x(x+2) - (x+1) = 3(x+2) + 1

 

x²+2x -x -1 = 3x + 6 + 1

 

x²+x-1 = 3x + 7

 

x²-2x-8 = 0

 

(x-4)(x+2) = 0

 

x=4 or x = -2 (but minus 2 is not positive, so reject)

 

4,5,6

 

(4)(6)-5 =? 1+3(6)

 

check.

Answer 2

Let x be the first number. Then then the other two numbers are x+1 and x+2.

 

Thus X(x+2)-(x+1)= 1+ 3(x+2)

 

x² +2x -x-1 = 1+3x+6

 

x² +x-1 =3x +7

x² -2x -8=0

x² +2x-4x -8=0

x(x+2)-4(x+2)=0

(x+2) (x-4)=0

 

Thus x= -2 or 4 

 

Since the numbers are positive, they are 4, 5, and 6