1. **State the problem:** We are given several fractions and asked to analyze their equivalence. The reference fraction is $$\frac{4x - 28}{6x - 42}$$ and the other fractions are $$\frac{2(2x - 14)}{3(2x - 13)}, \frac{4}{6}, \frac{1}{2}, \frac{4(x - 7)}{6(x - 7)}, \frac{2}{3}$$. 2. **Simplify the reference fraction:** $$4x - 28 = 4(x - 7)$$ $$6x - 42 = 6(x - 7)$$ So, $$\frac{4x - 28}{6x - 42} = \frac{4(x - 7)}{6(x - 7)}$$ 3. **Simplify the other fractions:** - $$\frac{2(2x - 14)}{3(2x - 13)} = \frac{4x - 28}{6x - 39}$$ (cannot simplify further) - $$\frac{4}{6} = \frac{2}{3}$$ - $$\frac{1}{2}$$ (already simplest) - $$\frac{4(x - 7)}{6(x - 7)}$$ (same as reference fraction) - $$\frac{2}{3}$$ (already simplest) 4. **Check equivalences:** - Reference fraction $$\frac{4(x - 7)}{6(x - 7)}$$ simplifies to $$\frac{2}{3}$$ after canceling $$(x - 7)$$. - $$\frac{4}{6}$$ and $$\frac{2}{3}$$ are equal to $$\frac{2}{3}$$. - $$\frac{1}{2}$$ is not equal to $$\frac{2}{3}$$. - $$\frac{2(2x - 14)}{3(2x - 13)}$$ is not equal to $$\frac{2}{3}$$ because denominator differs. 5. **Conclusion:** The fractions $$\frac{4x - 28}{6x - 42}, \frac{4}{6}, \frac{4(x - 7)}{6(x - 7)}, \frac{2}{3}$$ are equivalent and simplify to $$\frac{2}{3}$$. The fractions $$\frac{2(2x - 14)}{3(2x - 13)}$$ and $$\frac{1}{2}$$ are not equivalent to these. **Final answer:** $$\frac{4x - 28}{6x - 42} = \frac{4(x - 7)}{6(x - 7)} = \frac{4}{6} = \frac{2}{3} \neq \frac{2(2x - 14)}{3(2x - 13)}, \frac{1}{2}$$