1. **State the problem:** We need to find the value of $k$ such that the line joining points $A(3,k)$ and $B(-2,5)$ is parallel to the line given by the equation $5y + 2x = 10$. 2. **Find the slope of the given line:** Rewrite $5y + 2x = 10$ in slope-intercept form $y = mx + b$. $$5y = -2x + 10$$ $$y = -\frac{2}{5}x + 2$$ So, the slope $m_1 = -\frac{2}{5}$. 3. **Find the slope of the line joining points $A$ and $B$:** The slope formula is $$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{5 - k}{-2 - 3} = \frac{5 - k}{-5} = -\frac{5 - k}{5}$$ 4. **Set the slopes equal for parallel lines:** $$m = m_1$$ $$-\frac{5 - k}{5} = -\frac{2}{5}$$ 5. **Solve for $k$:** Multiply both sides by 5: $$-(5 - k) = -2$$ Multiply both sides by -1: $$5 - k = 2$$ Subtract 2 from both sides: $$5 - 2 = k$$ $$k = 3$$ **Final answer:** $k = 3$