Find three consecutive intergers such that the product of the first and the second is two more than three times the third.
To solve this problem, let's first assign variables to the three consecutive integers. Let's call the first integer "x," the second integer "x+1," and the third integer "x+2."
According to the problem, the product of the first and the second integer is two more than three times the third integer. In equation form, this can be written as:
x(x+1) = 3(x+2) + 2
Now let's solve this equation step by step:
1. Distribute 3 to (x+2): 3(x+2) = 3x + 6
The equation becomes:
x(x+1) = 3x + 6 + 2
x^2 + x = 3x + 8
2. Subtract 3x and 8 from both sides:
x^2 + x - 3x - 8 = 0
x^2 - 2x - 8 = 0
3. Factor the quadratic equation:
(x - 4)(x + 2) = 0
Now we have two possibilities for x:
x - 4 = 0 or x + 2 = 0
If x - 4 = 0, then x = 4.
If x + 2 = 0, then x = -2.
So the possible values for the first integer are x = 4 or x = -2.
If x = 4, then the three consecutive integers are:
First integer: x = 4
Second integer: x + 1 = 4 + 1 = 5
Third integer: x + 2 = 4 + 2 = 6
If x = -2, then the three consecutive integers are:
First integer: x = -2
Second integer: x + 1 = -2 + 1 = -1
Third integer: x + 2 = -2 + 2 = 0
So, the three consecutive integers that satisfy the given condition are either 4, 5, and 6 or -2, -1, and 0.