Solving for x: Clearing Fractions or Finding Common Denominator?

[Dethrone]

Given:

$$\frac{x-2}{x-5}>\frac{x-3}{x-4}$$

How I would normally solve it is to bring everything over to one side and find a common denominator. The answer of which is $x>5$

Now a commentary on this question says to "watch out" for sign changes if you multiply both sides by an expression, so I decided to try it out:

If $x>4$:
$$\frac{(x-2)(x-4)}{x-5}>x-3$$

If $x>5$ and $x>4$:
$$(x-2)(x-4)>(x-3)(x-5)$$

Solving gives us $x>5$, which is the same as above.

Question: Is it just a lucky coincidence that the restriction I applied on the second step $x>4$ just HAPPENED to SATISFY our answer, or could the answer have been say $x<3$ and dividing both sides would have eliminated that solution?

Question 2:

$$\frac{1}{x-2}=\frac{3}{x+2}-\frac{6x}{(x+2)(x-2)}$$

To begin this problem, would you attempt to "clear the fraction" by multiplying by (x-2)(x+2)? I always get paranoid about doing it, because it introduces the solution $x=\pm 2$, which you will then have to replug back into the original equation to confirm, right?

Would attempting to find a common denominator and bring everything to one side be a smarter approach? Or should just clearing the fraction, which is much faster, be preferred? It's just that the latter requires that you resubstitute back into the original equation, whereas the former doesn't require resubstitution. Thoughts?

 

[chisigma]

Given:

$$\frac{x-2}{x-5}>\frac{x-3}{x-4}$$

How I would normally solve it is to bring everything over to one side and find a common denominator. The answer of which is $x>5$

Now a commentary on this question says to "watch out" for sign changes if you multiply both sides by an expression, so I decided to try it out:

If $x>4$:
$$\frac{(x-2)(x-4)}{x-5}>x-3$$

If $x>5$ and $x>4$:
$$(x-2)(x-4)>(x-3)(x-5)$$

Solving gives us $x>5$, which is the same as above.

Question: Is it just a lucky coincidence that the restriction I applied on the second step $x>4$ just HAPPENED to SATISFY our answer, or could the answer have been say $x<3$ and dividing both sides would have eliminated that solution?

What You're saying is true in part. Assuming x <4 the inequality is verified for $\frac{7}{2} < x <4$ ...

Kind regards

$\chi$ $\sigma$

 

[Opalg]

Given:

$$\frac{x-2}{x-5}>\frac{x-3}{x-4}$$

How I would normally solve it is to bring everything over to one side and find a common denominator.

I think that this is definitely the safest strategy. In this case, you would get $$\frac{x-2}{x-5} - \frac{x-3}{x-4} > 0,$$ $$\frac{(x-2)(x-4) - (x-3)(x-5)}{(x-4)(x-5)} > 0,$$ $$\frac{2x-7}{(x-4)(x-5)} > 0.$$ That expression changes sign when any of the linear factors in the numerator or denominator changes sign, telling you that the critical points are when $x = \frac72,\ 4,\ 5$ (as chisigma has pointed out).

Question 2:

$$\frac{1}{x-2}=\frac{3}{x+2}-\frac{6x}{(x+2)(x-2)}$$

To begin this problem, would you attempt to "clear the fraction" by multiplying by (x-2)(x+2)? I always get paranoid about doing it, because it introduces the solution $x=\pm 2$, which you will then have to replug back into the original equation to confirm, right?

This is an equation rather than an inequality, so in this case there is nothing to be lost by multiplying by $(x-2)(x+2)$ (except that you have to remember that $x=\pm 2$ cannot be solutions).