Question 1120196: The product of the second and third of three consecutive integers is 2 more than 12 times the first number
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! let x = the first number.
since the numbers are consecutive, then your numbers are:
x
x+1
x+2
the product of the second and third numbers is equal to 2 plus 12 * the first number.
the equation for this is (x + 1) * (x + 2) = 2 + 12 * x
perform the multiplication on the left side of the equation to get:
x^2 + 2x + x + 2 = 2 + 12x
combine like terms to get:
x^2 + 3x + 2 = 2 + 12x
subtract (2 + 12x) from both sides of the equation to get:
x^2 + 3x + 2 - (2 + 12x) = 0
simplify to get:
x^2 + 3x + 2 - 2 - 12x = 0
combine like terms to get:
x^2 - 9x = 0
factor to get:
x * (x - 9) = 0
solve for x to get:
x = 0 or x = 9
the number could be 0 or it could be 9.
when the number is 0, the sequence is 0,1,2.
the product of the second and third number is 2.
2 plus 12 * the first number is equal to 2 + 0 which is equal to 2.
the product of the second and third number is equal to 2 plus 12 * the first number is confirmed to be true when the first number is equal to 0.
when the number is 9, the sequence is 9,10,11
the product of the second and third number is 10 * 11 = 110
2 plus 12 times the first number is equal to 2 + 12 * 9 which is equal to 2 + 108 which is equal to 110.
the product of the second and third number is equal to 2 plus 12 * the first number is confirmed to be true.
looks like either 0 or 9 would be a correct solution.
|
|
|
