Yeah I need help.....

1/x  +  1/(x+2)  =  9/40

Multiply each term by the common denominator of  40(x)(x+2):

[1/x](40)(x)(x+2)  +  [1/(x+2)](40)(x)(x+2)   =  [9/40](40)(x)(x+2) 

40(x+2)  +  (40)(x)  =  9(x)(x+2)

40x + 80  +  40x  =  9x² + 18x

                         0  =  9x²  -  62x  -  80

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

Disregarding the answer which is a fraction:

--->     x  =  8;  x + 2  =  10

$\frac{1}{x}+\frac{1}{x+2}=\frac9{40}\\ \frac{x+2+x}{x(x+2)}=\frac{9}{40}\\ \frac{2x+2}{x^2+2x}=\frac{9}{40}\\ 9x^2+18x=80x+80\\ 9x^2-62x-80=0\\ (9x+10)(x−8)=0\\ x_1=\frac{-10}{9}, x_2=8$

only 8 is an integer.