1. **State the problem:** Solve the system of equations using the elimination method. Given system: $$4x + 3y = 6$$ $$-x - 3y = 3$$ 2. **Add the two equations to eliminate $y$:** $$ (4x + 3y) + (-x - 3y) = 6 + 3 $$ $$ 4x - x + 3y - 3y = 9 $$ $$ 3x = 9 $$ 3. **Solve for $x$:** $$ x = \frac{9}{3} $$ $$ x = 3 $$ 4. **Substitute $x=3$ into one of the original equations to find $y$:** Using $$4x + 3y = 6$$: $$ 4(3) + 3y = 6 $$ $$ 12 + 3y = 6 $$ 5. **Isolate $y$:** $$ 3y = 6 - 12 $$ $$ 3y = -6 $$ $$ y = \frac{-6}{3} $$ $$ y = -2 $$ 6. **Final solution:** $$ (x, y) = (3, -2) $$ This means the solution to the system is the point where the two lines intersect, at $(3, -2)$.