To find the solutions to the system of equations:
[tex]\[
y = (x + 4)^2 - 1
\][/tex]
[tex]\[
y = -3x - 13
\][/tex]
we need to find the points [tex]\((x, y)\)[/tex] that satisfy both equations simultaneously. First, let's set these equations equal to each other:
[tex]\[
(x + 4)^2 - 1 = -3x - 13
\][/tex]
Now we need to solve for [tex]\(x\)[/tex]. Expand the left side and simplify:
[tex]\[
(x+4)^2 - 1 = x^2 + 8x + 16 - 1 = x^2 + 8x + 15
\][/tex]
So we have the equation:
[tex]\[
x^2 + 8x + 15 = -3x - 13
\][/tex]
Now, move all terms to one side to set the equation to zero:
[tex]\[
x^2 + 8x + 15 + 3x + 13 = 0
\][/tex]
Combine like terms:
[tex]\[
x^2 + 11x + 28 = 0
\][/tex]
Next, factor the quadratic equation:
[tex]\[
(x + 7)(x + 4) = 0
\][/tex]
This gives us the solutions for [tex]\(x\)[/tex]:
[tex]\[
x + 7 = 0 \quad \text{or} \quad x + 4 = 0
\][/tex]
[tex]\[
x = -7 \quad \text{or} \quad x = -4
\][/tex]
Now, we need to find the corresponding [tex]\(y\)[/tex]-values by substituting these [tex]\(x\)[/tex]-values back into either of the original equations. Let's use the second equation [tex]\( y = -3x - 13 \)[/tex].
For [tex]\( x = -7 \)[/tex]:
[tex]\[
y = -3(-7) - 13
\][/tex]
[tex]\[
y = 21 - 13
\][/tex]
[tex]\[
y = 8
\][/tex]
This gives us the point [tex]\((-7, 8)\)[/tex].
For [tex]\( x = -4 \)[/tex]:
[tex]\[
y = -3(-4) - 13
\][/tex]
[tex]\[
y = 12 - 13
\][/tex]
[tex]\[
y = -1
\][/tex]
This gives us the point [tex]\((-4, -1)\)[/tex].
So, the solutions to the system of equations are:
[tex]\[
(-7, 8) \quad \text{and} \quad (-4, -1)
\][/tex]
Therefore, the solutions to this system of equations are [tex]\((-7, 8)\)[/tex] and [tex]\((-4, -1)\)[/tex].
