1. **State the problem:** Solve the system of equations: $$ (5x)^2 + (5y)^2 = 148 $$ $$ x + y = -2 $$ 2. **Rewrite the first equation:** $$ 25x^2 + 25y^2 = 148 $$ Divide both sides by 25: $$ \cancel{25}x^2 + \cancel{25}y^2 = \frac{148}{\cancel{25}} $$ which simplifies to: $$ x^2 + y^2 = \frac{148}{25} $$ 3. **Use the second equation to express $y$ in terms of $x$:** $$ y = -2 - x $$ 4. **Substitute $y$ into the first equation:** $$ x^2 + (-2 - x)^2 = \frac{148}{25} $$ Expand the square: $$ x^2 + (x^2 + 4 + 4x) = \frac{148}{25} $$ Simplify: $$ 2x^2 + 4x + 4 = \frac{148}{25} $$ 5. **Multiply both sides by 25 to clear the fraction:** $$ 25(2x^2 + 4x + 4) = 148 $$ $$ 50x^2 + 100x + 100 = 148 $$ 6. **Bring all terms to one side:** $$ 50x^2 + 100x + 100 - 148 = 0 $$ $$ 50x^2 + 100x - 48 = 0 $$ 7. **Divide entire equation by 2 to simplify:** $$ \cancel{2} \times 25x^2 + \cancel{2} \times 50x - \cancel{2} \times 24 = 0 $$ $$ 25x^2 + 50x - 24 = 0 $$ 8. **Use the quadratic formula:** $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ where $a=25$, $b=50$, $c=-24$. Calculate discriminant: $$ \Delta = 50^2 - 4 \times 25 \times (-24) = 2500 + 2400 = 4900 $$ $$ \sqrt{4900} = 70 $$ 9. **Find the roots:** $$ x = \frac{-50 \pm 70}{2 \times 25} = \frac{-50 \pm 70}{50} $$ Two solutions: $$ x_1 = \frac{-50 + 70}{50} = \frac{20}{50} = \frac{2}{5} $$ $$ x_2 = \frac{-50 - 70}{50} = \frac{-120}{50} = -\frac{12}{5} $$ 10. **Find corresponding $y$ values:** For $x_1 = \frac{2}{5}$: $$ y = -2 - \frac{2}{5} = -\frac{10}{5} - \frac{2}{5} = -\frac{12}{5} $$ For $x_2 = -\frac{12}{5}$: $$ y = -2 - \left(-\frac{12}{5}\right) = -2 + \frac{12}{5} = -\frac{10}{5} + \frac{12}{5} = \frac{2}{5} $$ **Final solutions:** $$ \left( \frac{2}{5}, -\frac{12}{5} \right) \quad \text{and} \quad \left( -\frac{12}{5}, \frac{2}{5} \right) $$