1. **State the problem:** Solve the system of equations: $$\sqrt{x} + 2y = 2$$ $$y = -\frac{x}{2} + 1$$ 2. **Substitute the expression for $y$ from the second equation into the first:** $$\sqrt{x} + 2\left(-\frac{x}{2} + 1\right) = 2$$ 3. **Simplify the equation:** $$\sqrt{x} - x + 2 = 2$$ 4. **Subtract 2 from both sides:** $$\sqrt{x} - x = 0$$ 5. **Rewrite as:** $$\sqrt{x} = x$$ 6. **Square both sides to eliminate the square root:** $$x = x^2$$ 7. **Rearrange to standard quadratic form:** $$x^2 - x = 0$$ 8. **Factor the quadratic:** $$x(x - 1) = 0$$ 9. **Solve for $x$:** $$x = 0 \quad \text{or} \quad x = 1$$ 10. **Find corresponding $y$ values using $y = -\frac{x}{2} + 1$:** - For $x=0$: $$y = -\frac{0}{2} + 1 = 1$$ - For $x=1$: $$y = -\frac{1}{2} + 1 = \frac{1}{2}$$ 11. **Check solutions in the original equation $\sqrt{x} + 2y = 2$:** - For $(0,1)$: $$\sqrt{0} + 2(1) = 0 + 2 = 2$$ (valid) - For $(1, \frac{1}{2})$: $$\sqrt{1} + 2\left(\frac{1}{2}\right) = 1 + 1 = 2$$ (valid) **Final solutions:** $$\boxed{(0,1) \text{ and } (1, \frac{1}{2})}$$