Explanation
[tex]\frac{1}{x}+\frac{1}{x+2}=\frac{9}{40}[/tex]
Adding the two fractions:
[tex]\frac{x+2+x}{x(x+2)}=\frac{9}{40}[/tex][tex]\frac{2x+2}{x^2+2x}=\frac{9}{40}[/tex]
We subtract 9/40 to equal zero:
[tex]\frac{2x+2}{x^{^2}+2x}-\frac{9}{40}=0[/tex]
Subtracting the two fractions:
[tex]\frac{40(2x+2)-9(x^2+2x)}{40(x^2+2x)}=0[/tex][tex]\frac{80x+80-9x^2-18x}{40x^2+80x}=0[/tex][tex]\frac{-9x^2+62x+80}{40x^2+80x}=0[/tex]
Factoring:
[tex]\frac{-(9x+10)(x-8)}{40x(x+2)}=0[/tex]
This equation is zero when:
[tex]\begin{gathered} 9x+10=0 \\ 9x=-10 \\ x=-\frac{10}{9} \end{gathered}[/tex]
or
[tex]\begin{gathered} x-8=0 \\ x=8 \end{gathered}[/tex]
The only integer solution is x=8.
[tex]\begin{gathered} \frac{1}{8}+\frac{1}{10}=\frac{10+8}{80}=\frac{18}{80}=\frac{9}{40} \\ \\ \frac{}{} \\ \end{gathered}[/tex]
Answer
x=8
