1. **State the problem:** Solve the system of linear equations: $$5x = 15 + 3y$$ $$5y + 3x = 10$$ 2. **Rewrite the first equation:** $$5x = 15 + 3y \implies 5x - 3y = 15$$ 3. **Rewrite the system in standard form:** $$\begin{cases} 5x - 3y = 15 \\ 3x + 5y = 10 \end{cases}$$ 4. **Use the elimination method:** Multiply the first equation by 5 and the second by 3 to align coefficients of $y$: $$\begin{cases} 25x - 15y = 75 \\ 9x + 15y = 30 \end{cases}$$ 5. **Add the two equations to eliminate $y$:** $$25x - 15y + 9x + 15y = 75 + 30 \implies 34x = 105$$ 6. **Solve for $x$:** $$x = \frac{105}{34}$$ 7. **Substitute $x$ back into the first equation:** $$5\left(\frac{105}{34}\right) - 3y = 15$$ $$\frac{525}{34} - 3y = 15$$ 8. **Isolate $y$:** $$-3y = 15 - \frac{525}{34} = \frac{510}{34} - \frac{525}{34} = -\frac{15}{34}$$ $$y = \frac{15}{34 \times 3} = \frac{15}{102} = \frac{5}{34}$$ **Final answer:** $$x = \frac{105}{34}, \quad y = \frac{5}{34}$$