1. **State the problem:** Solve the equation $x = xy^2 + 2x - 3x + yx$ for $x$. 2. **Rewrite the equation:** Combine like terms on the right side. $$x = xy^2 + 2x - 3x + yx = xy^2 - x + yx$$ 3. **Group terms involving $x$ on the right:** $$x = x(y^2 - 1 + y)$$ 4. **Rewrite the equation as:** $$x = x(y^2 + y - 1)$$ 5. **Bring all terms to one side:** $$x - x(y^2 + y - 1) = 0$$ 6. **Factor out $x$:** $$x[1 - (y^2 + y - 1)] = 0$$ 7. **Simplify inside the bracket:** $$1 - y^2 - y + 1 = 2 - y^2 - y$$ So the equation becomes: $$x(2 - y^2 - y) = 0$$ 8. **Solve for $x$:** Either $$x = 0$$ or $$2 - y^2 - y = 0$$ 9. **Solve the quadratic in $y$:** $$y^2 + y - 2 = 0$$ 10. **Factor the quadratic:** $$(y + 2)(y - 1) = 0$$ 11. **Find $y$ values:** $$y = -2 \quad \text{or} \quad y = 1$$ **Final answer:** - $x = 0$ for any $y$ - Or $x$ can be any value if $y = -2$ or $y = 1$ (since then $2 - y^2 - y = 0$ and the equation holds for any $x$).