To determine the number of solutions to the given system of equations and find their specific values, we need to solve the following system:
1. [tex]\( y = -(x + 4)^2 + 3 \)[/tex]
2. [tex]\( y = -2x - 3 \)[/tex]
Let's follow a step-by-step approach to solve these equations systematically.
1. Set the equations equal to each other:
Since both expressions are equal to [tex]\( y \)[/tex], we can equate them to each other:
[tex]\[
-(x + 4)^2 + 3 = -2x - 3
\][/tex]
2. Rewrite and simplify the equation:
[tex]\[
-(x + 4)^2 + 3 = -2x - 3
\][/tex]
Move all terms to one side of the equation to set it to zero:
[tex]\[
-(x + 4)^2 + 3 + 2x + 3 = 0
\][/tex]
Simplify:
[tex]\[
-(x + 4)^2 + 2x + 6 = 0
\][/tex]
3. Expand and further simplify:
Expand [tex]\(-(x + 4)^2\)[/tex] and then simplify:
[tex]\[
-(x^2 + 8x + 16) + 2x + 6 = 0
\][/tex]
Distribute the negative sign:
[tex]\[
-x^2 - 8x - 16 + 2x + 6 = 0
\][/tex]
Combine like terms:
[tex]\[
-x^2 - 6x - 10 = 0
\][/tex]
4. Solve the quadratic equation:
[tex]\[
-x^2 - 6x - 10 = 0
\][/tex]
To solve the quadratic equation, we rearrange and solve for [tex]\( x \)[/tex] using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex] where [tex]\( a = -1 \)[/tex], [tex]\( b = -6 \)[/tex], and [tex]\( c = -10 \)[/tex].
[tex]\[
x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(-1)(-10)}}{2(-1)}
\][/tex]
Simplify:
[tex]\[
x = \frac{6 \pm \sqrt{36 - 40}}{-2}
\][/tex]
Simplify the expression inside the square root:
[tex]\[
x = \frac{6 \pm \sqrt{-4}}{-2}
\][/tex]
Recognize that [tex]\(\sqrt{-4} = 2i\)[/tex] (where [tex]\( i \)[/tex] is the imaginary unit), so:
[tex]\[
x = \frac{6 \pm 2i}{-2}
\][/tex]
Split the terms:
[tex]\[
x = -3 \mp i
\][/tex]
5. Find corresponding [tex]\( y \)[/tex] values:
Substitute [tex]\( x = -3 - i \)[/tex] and [tex]\( x = -3 + i \)[/tex] into either original equation to find corresponding [tex]\( y \)[/tex] values. Using [tex]\( y = -2x - 3 \)[/tex]:
When [tex]\( x = -3 - i \)[/tex]:
[tex]\[
y = -2(-3 - i) - 3 = 6 + 2i - 3 = 3 + 2i
\][/tex]
When [tex]\( x = -3 + i \)[/tex]:
[tex]\[
y = -2(-3 + i) - 3 = 6 - 2i - 3 = 3 - 2i
\][/tex]
Thus, the solutions to the system of equations are the points:
[tex]\[
(-3 - i, 3 + 2i) \quad \text{and} \quad (-3 + i, 3 - 2i)
\][/tex]
In the specified format, the solutions are:
Solution(s): none and 2 [tex]\( (-3-i,3+2i),(-3+i,3-2i) \)[/tex]
