1. **State the problem:** Solve the simultaneous equations: $$5x + 3y = 0$$ $$11x + 3y = 48$$ 2. **Formula and rules:** To solve simultaneous linear equations, we can use the elimination or substitution method. Here, elimination is convenient because both equations have the term $3y$. 3. **Eliminate $y$ by subtracting the first equation from the second:** $$ (11x + 3y) - (5x + 3y) = 48 - 0 $$ $$ 11x + 3y - 5x - 3y = 48 $$ $$ (11x - 5x) + (3y - 3y) = 48 $$ $$ 6x + \cancel{0} = 48 $$ $$ 6x = 48 $$ 4. **Solve for $x$:** $$ x = \frac{48}{6} $$ $$ x = 8 $$ 5. **Substitute $x=8$ into the first equation to find $y$:** $$ 5(8) + 3y = 0 $$ $$ 40 + 3y = 0 $$ $$ 3y = -40 $$ 6. **Solve for $y$:** $$ y = \frac{-40}{3} $$ 7. **Final answer:** $$ x = 8, \quad y = -\frac{40}{3} $$