Fraction and simplification

Question

solve: $\frac{1}{x(x-1)} + \frac{1}{x} = \frac{1}{x-1}$

What are the possible answers ?

(A) -1 (B) Infinitely Many Solutions (C) No solution (D) 0

The answer from where i've referred this is (B), but when i simplify it I get (D)

My solution:

$$\frac{1}{x(x-1)} + \frac{1}{x} = \frac{1}{x-1}$$

$$ \frac{x +x(x-1)}{x(x-1)\cdot x} = \frac{1}{x-1} \text{ (took l.c.m on l.h.s)}$$

$$ \frac{x + (x^2 -x)}{(x^2 - x)\cdot x}= \frac{1}{x-1}$$

$$\frac{x^2}{x^3 - x^2} = \frac{1}{x-1}$$

$$ x^2(x-1) = x^3 - x^2$$

$$ x^3 - x^2 = x^3 - x^2$$

Have I simplified it correctly?

Answer 1

You can make it much simpler.

First you have to set the domain of validity: you must have $x\ne 0,1$.

Next, on this domain, remove the denominators multiplying both sides by the l.c.m. of the denominators, and simplify; you get: $$1+(x-1)=x\iff x=x.$$ Hence any number $x\ne 0,1$ is a solution.

Answer 2

You have got an identity, L.H.S. = R.H.S. which will hold for all values of X in domain of equation. This implies Infinite solution as domain of the equation is infinite.
In this case $X ={0,1}$ are not in domain of equation.

Answer 3

You can simply multiply by the LCD, $x(x-1)$ on both sides of the equation.

$$1+(x-1)=x$$

$$x=x$$

Therefore, there are infinitely many solutions for $x$, where $x\ne0,1$.

If $x=0, 1$, then the denominator(s) of the original expression would equal $0$, so the expression would be undefined.

In interval notation, the solutions can be expressed as $(-\infty,0)\cup(0,1)\cup(1,\infty)$.