The product of two consecutive even integers is 10 less than 5 times their sum. Find the two integers.

How do I do this?


'Thank you!

If the first number is 2n then the other is 2n+2 so 2n(2n + 2) = 5[2n +2 + 2n] - 10 so 4n(n+1) = 10(2n + 1) - 10 and that becomes 4n^2 + 4n = 20n + 10 - 10 so 4n^2 - 16n = 0 or 4n(n - 4) = 0 and that gives 4n = 0 or n - 4 = 0; but the first number was 2n so that second one becomes 2n-8 = 0 so either 2n = 0 or 2n = 8so the num,bers would be (0, 2) or (8,10). Checking those against the first line, the product of 2 & 0 is 0; their sum = 2, 5 times that is 10, and 10 less than that = 0 = their product. 2nd pair - 8*10 = 80, and 5(8 + 10) - 10 = 90 - 10 = 80, so both solutions are valid. Just one point - some authorities maintain that 0 is not an even number (blowed if I can temember why!) so if this question is from a book that may explain why they only give 8&10.

x = first even integer
(x+2) = second even integer

x(x+2) = 5(x+(x+2)) -10
x^2+2x = 10x + 10 -10
x^2+2x -10x = 0
x^2-8x = 0
x(x-8) = 0
x =0 or x-8 = 0 => x = 8
Final answer => x1 = 0 ; x2 = 8

x and x+2
x(x+2) = 5(x+(x+2)) - 10 , and solve for x