1. **State the problem:** Solve the system of equations: $$xy=4$$ $$x-y=3$$ 2. **Use substitution or elimination:** From the second equation, express $x$ in terms of $y$: $$x = y + 3$$ 3. **Substitute into the first equation:** $$ (y+3)y = 4 $$ 4. **Simplify:** $$ y^2 + 3y = 4 $$ 5. **Rewrite as a quadratic equation:** $$ y^2 + 3y - 4 = 0 $$ 6. **Factor the quadratic:** $$ (y+4)(y-1) = 0 $$ 7. **Solve for $y$:** $$ y = -4 \quad \text{or} \quad y = 1 $$ 8. **Find corresponding $x$ values:** - For $y = -4$: $$ x = -4 + 3 = -1 $$ - For $y = 1$: $$ x = 1 + 3 = 4 $$ 9. **Final solutions:** $$ (x,y) = (-1,-4) \quad \text{or} \quad (4,1) $$