The sum of the reciprocals of two consecutive even integers is 9/40. Find the integers. smaller integer larger integer

8 and 10

Step-by-step explanation:

2x and 2x+2 the integers

1/2x+1 / (2x+2) = 9/40 1/x+1 / (x+1) = 9/20 (x+x+1) / (x * (x+1)) = 9/20 9x²+9x = 20 (2x+1) 9x²-31x-20=0 x = 4 integer solution

So the numbers are 8 and 10

Smaller integer is 8, larger one is 10.

Step-by-step explanation:

Let one integer be x then the other is x+2, so from the given information:

1 / x + 1 / x+2 = 9/40

Multiply through by 40x (x+2);

40 (x + 2) + 40x = 9x (x + 2)

80x + 80 = 9x^2 + 18x

9x^2 - 62x - 80 = 0

(9x + 10) (x - 8) = 0

x = 8, - 1.1111

x must be 8 so the other one is 8+2 = 10.