Answer: B. $-3$ and $3$

Explanation:1. Understand the problem: We are given the equation $\dfrac{x-4}{5}=\dfrac{x-4}{7}$. We need to determine between which pair of values the value of $x-4$ lies.

2. Solve the equation: Let $A = x-4$. The equation becomes $\dfrac{A}{5} = \dfrac{A}{7}$. To solve for $A$, we can multiply both sides by $35$ (the least common multiple of $5$ and $7$): $35 \cdot \dfrac{A}{5} = 35 \cdot \dfrac{A}{7}$ $7A = 5A$ Subtract $5A$ from both sides: $7A - 5A = 0$ $2A = 0$ $A = 0$

3. Find the value of $x-4$: Since we let $A = x-4$, and we found $A=0$, then $x-4 = 0$.

4. Determine between which pair of values $0$ lies: [A] $-7$ and $-5$: $0$ is not between $-7$ and $-5$. [B] $-3$ and $3$: $0$ is between $-3$ and $3$ (since $-3 < 0 < 3$). [C] $3.5$ and $4.5$: $0$ is not between $3.5$ and $4.5$. [D] $4.75$ and $7.25$: $0$ is not between $4.75$ and $7.25$.

5. Conclusion: The value of $x-4$ is $0$, which lies between $-3$ and $3$.