The question provided in the image involves solving the given equations to determine a point of intersection or a specific solution pair $(x, y)$.

Given equations:

1. $x^2 + y^2 + 4x - 12y = 0$
2. $x + y = 6$

Steps to solve:

1. Substitute $y = 6 - x$ from the second equation into the first equation.

   $x^2 + (6 - x)^2 + 4x - 12(6 - x) = 0$

2. Expand the equation:

   • Expand $(6 - x)^2$:

     $x^2 + (36 - 12x + x^2) + 4x - 12(6 - x) = 0$

   • Simplify terms:

     $x^2 + 36 - 12x + x^2 + 4x - 72 + 12x = 0$

3. Combine like terms:

   • Combine $x^2$ terms: $x^2 + x^2 = 2x^2$
   • Combine $x$-terms: $-12x + 4x + 12x = 4x$
   • Combine constants: $36 - 72 = -36$

   This simplifies to:

   $2x^2 + 4x - 36 = 0$

4. Simplify the quadratic equation:

   Divide through by 2:

   $x^2 + 2x - 18 = 0$

5. Solve for $x$ using factorization:

   Factorize the equation $x^2 + 2x - 18 = 0$:

   $(x + 6)(x - 3) = 0$

   Therefore:

   $x = -6 \quad \text{or} \quad x = 3$

6. Find corresponding $y$-values:

   • If $x = -6$, then $y = 6 - (-6) = 12$.
   • If $x = 3$, then $y = 6 - 3 = 3$.

7. Solutions:

   The two solutions are:

   $(-6, 12) \quad \text{and} \quad (3, 3)$

Verify solutions:

• Substituting both points into the first equation $x^2 + y^2 + 4x - 12y = 0$:

  • For $(-6, 12)$:

    $(-6)^2 + (12)^2 + 4(-6) - 12(12) = 36 + 144 - 24 - 144 = 0$

    ✅ Satisfied.

  • For $(3, 3)$:

    $(3)^2 + (3)^2 + 4(3) - 12(3) = 9 + 9 + 12 - 36 = 0$

    ✅ Satisfied.

Final Answer:

The solutions are:

$\boxed{(-6, 12) \text{ and } (3, 3)}$

Do you want a deeper explanation of any step? Let me know!

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Tip: Substitution is a powerful method when solving systems with one linear and one non-linear equation. Always simplify the quadratic carefully!