1. **State the problem:** Solve the inequality $$\frac{3}{x-4} \leq \frac{2}{x-3}$$ for $x$. 2. **Rewrite the inequality:** Bring all terms to one side: $$\frac{3}{x-4} - \frac{2}{x-3} \leq 0$$ 3. **Find a common denominator:** The common denominator is $(x-4)(x-3)$, so: $$\frac{3(x-3)}{(x-4)(x-3)} - \frac{2(x-4)}{(x-4)(x-3)} \leq 0$$ 4. **Combine the fractions:** $$\frac{3(x-3) - 2(x-4)}{(x-4)(x-3)} \leq 0$$ 5. **Simplify the numerator:** $$3(x-3) - 2(x-4) = 3x - 9 - 2x + 8 = x - 1$$ 6. **Rewrite the inequality:** $$\frac{x - 1}{(x-4)(x-3)} \leq 0$$ 7. **Identify critical points:** The numerator is zero at $x=1$, and the denominator is zero at $x=3$ and $x=4$ (excluded from domain). 8. **Determine sign intervals:** The critical points divide the real line into intervals: $(-\infty,1)$, $(1,3)$, $(3,4)$, $(4,\infty)$. 9. **Test each interval:** - For $x < 1$, numerator $(x-1)<0$, denominator $(x-4)(x-3)>0$ (both negative, product positive), fraction $<0$. - For $1 < x < 3$, numerator $>0$, denominator $(x-4)(x-3)<0$ (one negative, one negative, product positive?), check carefully: - At $x=2$, $(2-4)=-2<0$, $(2-3)=-1<0$, product $(-2)(-1)=2>0$ denominator positive. So fraction positive. - For $3 < x < 4$, numerator $>0$, denominator $(x-4)(x-3)$: - At $x=3.5$, $(3.5-4)=-0.5<0$, $(3.5-3)=0.5>0$, product negative. Fraction negative. - For $x > 4$, numerator $>0$, denominator positive, fraction positive. 10. **Include points where fraction equals zero:** At $x=1$, fraction is zero. 11. **Exclude points where denominator is zero:** $x=3$ and $x=4$ are not in domain. 12. **Solution:** $$(-\infty,1] \cup (3,4)$$ **Final answer:** $$x \in (-\infty,1] \cup (3,4)$$