1. **State the problem:** Solve the system of equations algebraically: $$x - 2y + 2 = 0$$ $$x^2 + y^2 = 40$$ 2. **Express one variable in terms of the other:** From the first equation, $$x - 2y + 2 = 0 \implies x = 2y - 2$$ 3. **Substitute into the second equation:** Replace $x$ with $2y - 2$ in the circle equation: $$ (2y - 2)^2 + y^2 = 40 $$ 4. **Expand and simplify:** $$ (2y - 2)^2 = 4y^2 - 8y + 4 $$ So, $$ 4y^2 - 8y + 4 + y^2 = 40 $$ $$ 5y^2 - 8y + 4 = 40 $$ 5. **Bring all terms to one side:** $$ 5y^2 - 8y + 4 - 40 = 0 $$ $$ 5y^2 - 8y - 36 = 0 $$ 6. **Solve the quadratic equation:** Use the quadratic formula $$ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ where $a=5$, $b=-8$, $c=-36$. Calculate the discriminant: $$ \Delta = (-8)^2 - 4 \times 5 \times (-36) = 64 + 720 = 784 $$ So, $$ y = \frac{8 \pm \sqrt{784}}{10} = \frac{8 \pm 28}{10} $$ 7. **Find the two values of $y$:** - For $+$: $$ y = \frac{8 + 28}{10} = \frac{36}{10} = 3.6 $$ - For $-$: $$ y = \frac{8 - 28}{10} = \frac{-20}{10} = -2 $$ 8. **Find corresponding $x$ values:** Recall $x = 2y - 2$. - For $y=3.6$: $$ x = 2(3.6) - 2 = 7.2 - 2 = 5.2 $$ - For $y=-2$: $$ x = 2(-2) - 2 = -4 - 2 = -6 $$ 9. **Final solutions:** $$ (x,y) = (5.2, 3.6) \quad \text{and} \quad (-6, -2) $$