Answer
x = 1 or x = n.a.
Explanation
![(1)/(x)+(1)/(x-10)=(x-9)/(x-10)](https://img.qammunity.org/2023/formulas/mathematics/high-school/3ovha01860yuyi9beh3qr3kaoz8uh5st2t.png)
Multiplying by (x - 10) at both sides of the equation:
![\begin{gathered} (x-10)((1)/(x)+(1)/(x-10))=(x-9)/(x-10)(x-10) \\ \text{ Distributing and simplifying:} \\ (x-10)/(x)+(x-10)/(x-10)=x-9 \\ (x-10)/(x)+1=x-9 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/1gxtjs4zqlidz7cr51rcxedb9mnu6ek52j.png)
Multiplying by x at both sides of the equation:
![\begin{gathered} x((x-10)/(x)+1)=x(x-9) \\ \text{ Distributing and simplifying:} \\ (x(x-10))/(x)+x=x^2-9x \\ x-10+x=x^2-9x \\ 2x-10=x^2-9x \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/fx042pkkr3lj2ahfafkldijg5kd6h9ys9z.png)
Subtracting 2x and adding 10 at both sides of the equation:
![\begin{gathered} 2x-10-2x+10=x^2-9x-2x+10 \\ 0=x^2-11x+10 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/tuiiibjptl2dewt6rcqz4hkqgnmltifsho.png)
We can solve this equation with the help of the quadratic formula with the coefficients a = 1, b = -11, and c = 10, as follows:
![\begin{gathered} x_(1,2)=\frac{-b\pm{}√(b^2-4ac)}{2a} \\ x_(1,2)=\frac{11\pm\sqrt{(-11)^2-4\cdot1\operatorname{\cdot}10}}{2\operatorname{\cdot}1} \\ x_(1,2)=(11\pm√(81))/(2) \\ x_1=(11+9)/(2)=10 \\ x_2=(11-9)/(2)=1 \end{gathered}]()
The solution x = 10 is not possible because it makes zero the denominator in 2 of the rational expressions of the original equation. In consequence, it must be discarded.