find three consecutive even integers such that the product of the second and the third exceed the product of the first and third by 20.
P.S: The Checking part also needs to be correct...
To solve this problem, let's represent the three consecutive even integers as "n, n + 2, and n + 4." The key information we have is:
The product of the second (n + 2) and the third (n + 4) exceeds the product of the first (n) and the third (n + 4) by 20.
This can be written as an equation: (n + 2) * (n + 4) = n * (n + 4) + 20.
Now, let's solve the equation step by step:
Expanding the equation:
(n^2 + 6n + 8) = (n^2 + 4n) + 20
Simplifying the equation:
n^2 + 6n + 8 = n^2 + 4n + 20
Removing the common terms (n^2) from both sides:
6n + 8 = 4n + 20
Moving the constants to the right side and variables to the left side:
6n - 4n = 20 - 8
2n = 12
Dividing both sides by 2 to solve for n:
n = 12/2
n = 6
So the first even integer is 6, and the next two consecutive even integers are 6 + 2 = 8 and 6 + 4 = 10.
Therefore, the three consecutive even integers that satisfy the given conditions are 6, 8, and 10.
let the middle number be x, then the first is x-2 and the third is x+2
x(x+2) - (x-2)(x+2) = 20
x^2 + 2x - (x^2 - 4) = 20
your turn to finish it