1. **State the problem:** We have two lines $j$ and $k$ cut by two transversals $m$ and $n$. We want to find which relationship between $x$ and $y$ is sufficient to prove that $j \parallel k$. 2. **Identify given angles:** - Angle between $m$ and $j$ is $(x + y)^\circ$. - Angle between $m$ and $k$ is $(x + 30)^\circ$. - Angle between $n$ and $j$ is $(3x - y - 60)^\circ$. 3. **Recall the rule for parallel lines:** Lines $j$ and $k$ are parallel if corresponding angles or alternate interior angles formed by the transversals are equal. 4. **Check angles on transversal $m$:** Corresponding angles on $m$ are $(x + y)^\circ$ and $(x + 30)^\circ$. For $j \parallel k$, these must be equal: $$x + y = x + 30$$ Subtract $x$ from both sides: $$\cancel{x} + y = \cancel{x} + 30$$ So, $$y = 30$$ 5. **Check angles on transversal $n$:** Angles on $n$ are $(3x - y - 60)^\circ$ and the angle corresponding to $(x + 30)^\circ$ on $k$ (not given explicitly, so we focus on the given options). 6. **Test each answer choice:** - $x=60$: Substitute into $(x + y)$ and $(x + 30)$, no direct equality unless $y$ is specified. - $y=30$: From step 4, this satisfies the equality of corresponding angles on $m$. - $y = x - 30$: Substitute into $(x + y)$ gives $x + (x - 30) = 2x - 30$, which generally does not equal $(x + 30)$. - $y = 150 - 2x$: Substitute into $(x + y)$ gives $x + 150 - 2x = 150 - x$, which generally does not equal $(x + 30)$. 7. **Conclusion:** The relationship $y=30$ is sufficient to prove $j \parallel k$ because it makes the corresponding angles on transversal $m$ equal. **Final answer:** $y=30$