$$ \left.\begin{array}{l} ext { When solving }\left\{\begin{array}{l}x^{2}-4 y=12 \ x+y=-3\end{array} ext { by the substitution method, we obtain } x=-4 ext { or } x=0 ext {, so the solution set is } ight. \ ext { Fill in the blank. } \ ext { When solving }\left\{\begin{array}{l}x^{2}-4 y=12 \ x+y=-3\end{array} ext { by the substitution method, we obtain } x=-4 ext { or } x=0 ext {, so the solution set is }\{ ight.\end{array} ight\} $$.

Question

$$ \left.\begin{array}{l}	ext { When solving }\left\{\begin{array}{l}x^{2}-4 y=12 \ x+y=-3\end{array} 	ext { by the substitution method, we obtain } x=-4 	ext { or } x=0 	ext {, so the solution set is }ight. \ 	ext { Fill in the blank. } \ 	ext { When solving }\left\{\begin{array}{l}x^{2}-4 y=12 \ x+y=-3\end{array} 	ext { by the substitution method, we obtain } x=-4 	ext { or } x=0 	ext {, so the solution set is }\{ ight.\end{array}ight\} $$.

UpStudy AI · Accepted Answer

The original system is  
$$
\begin{cases}
x^2 - 4y = 12, \
x + y = -3.
\end{cases}
$$

**Step 1.** Solve the second equation for $$ y $$:  
$$
y = -3 - x.
$$

**Step 2.** Substitute $$ y = -3 - x $$ into the first equation:  
$$
x^2 - 4(-3 - x) = 12.
$$

**Step 3.** Simplify and solve for $$ x $$:  
$$
x^2 + 12 + 4x = 12.
$$
Subtract $$12$$ from both sides:  
$$
x^2 + 4x = 0.
$$
Factor out $$ x $$:  
$$
x(x + 4) = 0.
$$

Thus,  
$$
x = 0 \quad \text{or} \quad x = -4.
$$

**Step 4.** Find the corresponding $$ y $$ for each $$ x $$:

- When $$ x = 0 $$:  
  $$
  y = -3 - 0 = -3.
  $$
  
- When $$ x = -4 $$:  
  $$
  y = -3 - (-4) = 1.
  $$

**Step 5.** Write the solution set:  
$$
\{(0, -3), (-4, 1)\}.
$$

So the filled blank is $$\{(0, -3), (-4, 1)\}$$.
The solution set is $$\{(0, -3), (-4, 1)\}$$.

.

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