Answer:
Option A) There is only one solution: x = 1. The solution x = 3 is an extraneous solution.
Step-by-step explanation:
We are given the following equation:
[tex]\displaystyle\frac{1}{x} + \frac{1}{x-3} = \frac{x-2}{x-3}[/tex]
Solving the given equation, we have:
[tex]\displaystyle\frac{1}{x} + \frac{1}{x-3} = \frac{x-2}{x-3}\\\\\frac{2x-3}{x(x-3)}=\frac{x-2}{x-2}\\\\2x-3 = x^2 - 2x\\x^2 - 4x + 3 = 0\\x^2 - x-3x + 3 = 0\\x(x-1)-3(x-3) = 0\\(x-3)(x-1) = 0\\x =1, x=3[/tex]
Extraneous solutions:
- An extraneous solution is not a root of the original equation because it was excluded from the domain of the original equation.
- We may arrive to this solution but cannot consider it as a solution because it is not valid.
But x = 3 cannot be a solution to the given equation as when put in the given equation it gives denominator zero.
Thus, x = 1 is the solution of the given equation and x = 3 is an extraneous solution.
