1. **State the problem:** Solve the simultaneous equations: $$x + 2y = 1$$ $$x^2 + y^2 = 10$$ 2. **Use substitution from the linear equation:** From the first equation, express $x$ in terms of $y$: $$x = 1 - 2y$$ 3. **Substitute into the second equation:** $$ (1 - 2y)^2 + y^2 = 10 $$ 4. **Expand and simplify:** $$ (1 - 2y)^2 + y^2 = 1 - 4y + 4y^2 + y^2 = 1 - 4y + 5y^2 $$ So, $$ 1 - 4y + 5y^2 = 10 $$ 5. **Bring all terms to one side:** $$ 5y^2 - 4y + 1 - 10 = 0 $$ $$ 5y^2 - 4y - 9 = 0 $$ 6. **Solve the quadratic equation:** Use the quadratic formula: $$ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ where $a=5$, $b=-4$, $c=-9$. Calculate the discriminant: $$ \Delta = (-4)^2 - 4 \times 5 \times (-9) = 16 + 180 = 196 $$ So, $$ y = \frac{4 \pm \sqrt{196}}{10} = \frac{4 \pm 14}{10} $$ 7. **Find the two values of $y$:** - For $y = \frac{4 + 14}{10} = \frac{18}{10} = 1.8$ - For $y = \frac{4 - 14}{10} = \frac{-10}{10} = -1$ 8. **Find corresponding $x$ values using $x = 1 - 2y$:** - For $y=1.8$: $$ x = 1 - 2(1.8) = 1 - 3.6 = -2.6 $$ - For $y=-1$: $$ x = 1 - 2(-1) = 1 + 2 = 3 $$ 9. **Final solutions:** $$ (x, y) = (-2.6, 1.8) \quad \text{or} \quad (3, -1) $$