Singularity II

When coming up with this problem, I used several approaches.

1. Same denominators

$\dfrac {x} {x-2} - \dfrac {x+1} {x-1} = \dfrac {x-1} {x-2} - \dfrac {x} {x-1}$

We move fractions with the same denominators to each side respectively:

$\dfrac {x} {x-2} - \dfrac {x-1} {x-2} = \dfrac {x+1} {x-1} - \dfrac {x} {x-1}$

$\dfrac {1} {x-2} = \dfrac {1} {x-1}$

Cross multiplying gives:

$x - 2 = x - 1$

which is not true for any values of $x$ .

2. Combining denominators

$\dfrac {x} {x-2} - \dfrac {x+1} {x-1} = \dfrac {x-1} {x-2} - \dfrac {x} {x-1}$

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

$\dfrac {2} {(x-2)(x-1)} = \dfrac {1} {(x-2)(x-1)}$

$\dfrac {1} {(x-2)(x-1)} = 0$

Since the numerator is $1$ , this is already not true, since the numerator must be $0$ in order for the whole fraction to equal $0$ .

3. Considering vertical asymptotes

Consider the graph of

$y = \dfrac {x} {x-2} - \dfrac {x+1} {x-1}$

$y = \dfrac {x-1} {x-2} - \dfrac {x} {x-1}$

desmos.com

As you can see, $x=2$ and $x=1$ aren't solutions since these two functions are undefined at those points.

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Try Singularity I