1. **State the problem:** Solve the system of equations simultaneously: $$y = 4$$ $$y = x^2 + 3x$$ 2. **Set the equations equal to each other:** Since both expressions equal $y$, we can set them equal: $$4 = x^2 + 3x$$ 3. **Rewrite the equation:** Move all terms to one side to form a quadratic equation: $$x^2 + 3x - 4 = 0$$ 4. **Factor the quadratic:** Find two numbers that multiply to $-4$ and add to $3$. $$x^2 + 4x - x - 4 = 0$$ $$x(x + 4) - 1(x + 4) = 0$$ $$(x - 1)(x + 4) = 0$$ 5. **Solve for $x$:** Set each factor equal to zero: $$x - 1 = 0 \Rightarrow x = 1$$ $$x + 4 = 0 \Rightarrow x = -4$$ 6. **Find corresponding $y$ values:** Substitute $x$ back into $y = 4$ (already given), so $y=4$ for both. 7. **Final solutions:** $$(x, y) = (1, 4) \text{ and } (-4, 4)$$ These are the points where the parabola $y = x^2 + 3x$ intersects the line $y=4$.