1. **State the problem:** Solve the system of linear equations: $$3x + 9y = 0$$ $$5x - 6y = 38$$ 2. **Formula and rules:** We can solve this system using substitution or elimination. Here, we use elimination to eliminate one variable. 3. **Multiply the first equation by 2 and the second by 3 to align coefficients of $y$:** $$2(3x + 9y) = 2(0) \Rightarrow 6x + 18y = 0$$ $$3(5x - 6y) = 3(38) \Rightarrow 15x - 18y = 114$$ 4. **Add the two equations to eliminate $y$:** $$6x + 18y + 15x - 18y = 0 + 114$$ $$6x + 15x + \cancel{18y} - \cancel{18y} = 114$$ $$21x = 114$$ 5. **Solve for $x$:** $$x = \frac{114}{21} = \frac{\cancel{114}^{6} \times 19}{\cancel{21}^{3} \times 7} = \frac{38}{7}$$ 6. **Substitute $x = \frac{38}{7}$ into the first equation to find $y$:** $$3\left(\frac{38}{7}\right) + 9y = 0$$ $$\frac{114}{7} + 9y = 0$$ 7. **Isolate $y$:** $$9y = -\frac{114}{7}$$ $$y = \frac{-\frac{114}{7}}{9} = -\frac{114}{7} \times \frac{1}{9} = -\frac{114}{63}$$ 8. **Simplify $y$:** $$y = -\frac{\cancel{114}^{6} \times 19}{\cancel{63}^{9} \times 7} = -\frac{38}{21}$$ **Final answer:** $$x = \frac{38}{7}, \quad y = -\frac{38}{21}$$