Preserve the solution set

Solve $\frac{x - 2}{x - 5} \leq 0$. The numerator is zero at $2$. The denominator is zero at $5$, which must be excluded.

For $x < 2$, both factors are negative, so the ratio is positive. Between $2$ and $5$, the numerator is positive and the denominator negative, so the ratio is negative. Above $5$, both are positive.

Include $x = 2$ because a zero numerator is allowed. Exclude $x = 5$ because division by zero is never allowed. The complete answer is $2 \leq x < 5$.

Do not multiply by $x - 5$ before considering its sign. Also, $x \in \{2, 5\}$ names two individual values; it does not name the interval between them. The symbol $\cup$ joins alternative intervals.

The same restriction matters in an equation. For $\frac{x^2 - 7x + 10}{x - 5} = 0$, record $x \ne 5$. The numerator factors as $(x - 2)(x - 5)$, suggesting candidates $2$ and $5$. Only $2$ survives: $5$ makes the original denominator zero. Clearing a denominator finds candidates; checking the original decides which ones are solutions.

Now solve $\lvert x - 3 \rvert \geq 2$. Absolute value is distance: $x$ must be at least $2$ units from $3$.

The boundary values are $1$ and $5$. Decide whether the allowed points lie between the boundaries or outside them. Equality is allowed at both boundaries.

When the other side also contains the variable, check its sign before splitting. In $\lvert 3x - 7 \rvert = x + 3$, require $x \geq -3$. The nonnegative branch gives $3x - 7 = x + 3$, so $x = 5$. The negative branch gives $-3x + 7 = x + 3$, so $x = 1$. Substitution confirms $\lvert 8 \rvert = 8$ and $\lvert -4 \rvert = 4$; both roots are valid.