To find the solutions of the given system of equations:
[tex]\[
\left\{\begin{array}{l}
y = -6x - 6 \\
y = x^2 - 5x - 6
\end{array}\right.
\][/tex]
we need to determine the points [tex]\((x, y)\)[/tex] where both equations are satisfied simultaneously.
### Step 1: Equate the expressions for [tex]\( y \)[/tex]
Since both equations equal [tex]\( y \)[/tex], we can set the right-hand sides equal to each other:
[tex]\[
-6x - 6 = x^2 - 5x - 6
\][/tex]
### Step 2: Move all terms to one side of the equation to set it to zero
Rearrange the equation:
[tex]\[
x^2 - 5x - 6 - (-6x - 6) = 0
\][/tex]
[tex]\[
x^2 - 5x - 6 + 6x + 6 = 0
\][/tex]
[tex]\[
x^2 + x = 0
\][/tex]
### Step 3: Factor the quadratic equation
Factor out the common term [tex]\( x \)[/tex]:
[tex]\[
x(x + 1) = 0
\][/tex]
### Step 4: Solve for [tex]\( x \)[/tex]
Set each factor equal to zero:
[tex]\[
x = 0
\][/tex]
[tex]\[
x + 1 = 0 \implies x = -1
\][/tex]
### Step 5: Find corresponding [tex]\( y \)[/tex] values for each [tex]\( x \)[/tex]
Substitute [tex]\( x \)[/tex] back into either of the original equations to get the corresponding [tex]\( y \)[/tex].
#### For [tex]\( x = 0 \)[/tex]:
Substitute [tex]\( x = 0 \)[/tex] into [tex]\( y = -6x - 6 \)[/tex]:
[tex]\[
y = -6(0) - 6 = -6
\][/tex]
So, one solution is:
[tex]\[
(0, -6)
\][/tex]
#### For [tex]\( x = -1 \)[/tex]:
Substitute [tex]\( x = -1 \)[/tex] into [tex]\( y = -6x - 6 \)[/tex]:
[tex]\[
y = -6(-1) - 6 = 6 - 6 = 0
\][/tex]
So, the other solution is:
[tex]\[
(-1, 0)
\][/tex]
### Conclusion
The solutions to the system of equations are:
[tex]\[
\boxed{(-1, 0) \text{ and } (0, -6)}
\][/tex]
