Question

The sum of the reciprocals of two consecutive even integers is 9/40, what are the integers?

Answer 1

If the smaller of the two consecutive even integers is `x`
then, we are told,
`color(red)(1/x)+color(blue)(1/(x+2)) =9/40`

So
`color(white)("XXXXX")`generating a common denominator on the left side:
`[color(red)(1/x*(x+2)/(x+2))] + [color(blue)(1/(x+2)*(x/x))]=9/40`

`[color(red)((x+2)/(x^2+2x))] + [color(blue)((x)/(x^2+2x))]=9/40`

`(color(red)((x+2)) + color(blue)((x)))/(x^2+2x) = 9/40`

`(2x+2)/(x^2+2x) = 9/40`

`(40)(2)(x+1) = 9(x^2+2x)`

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

`9x^2-62x-80 = 0`

`(9x+1)(x-8) = 0`

Since `x` is an even integer
the two consecutive even integers are
`8` and `10`