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What are the solutions of the system of equations [tex]$y=-(x+2)^2+1$[/tex] and [tex]$y=4x+9$[/tex]?

A. [tex](-2,1)[/tex] and [tex](6,-15)[/tex]
B. [tex](-2,1)[/tex] and [tex](-6,-15)[/tex]
C. [tex](2,1)[/tex] and [tex](6,-15)[/tex]
D. [tex](2,1)[/tex] and [tex](-6,-15)[/tex]



Answer :

GinnyAnswer
To find the solutions of the system of equations given by [tex]\( y = -(x + 2)^2 + 1 \)[/tex] and [tex]\( y = 4x + 9 \)[/tex], we need to determine the points of intersection of these two curves. This involves solving the equations simultaneously.

1. Substitute the expression for [tex]\( y \)[/tex] from the second equation into the first equation:
[tex]\[ 4x + 9 = -(x + 2)^2 + 1 \][/tex]

2. Simplify and expand the quadratic equation:
[tex]\[ 4x + 9 = -(x^2 + 4x + 4) + 1 \][/tex]

[tex]\[ 4x + 9 = -x^2 - 4x - 4 + 1 \][/tex]

[tex]\[ 4x + 9 = -x^2 - 4x - 3 \][/tex]

3. Move all termos to one side of the equation to set it to zero:
[tex]\[ x^2 + 4x + 4x + 9 + 3 = 0 \][/tex]

[tex]\[ x^2 + 8x + 12 = 0 \][/tex]

4. Factor the quadratic equation:
[tex]\[ x^2 + 8x + 12 = (x + 6)(x + 2) = 0 \][/tex]

5. Solve for [tex]\( x \)[/tex]:
[tex]\[ x + 6 = 0 \quad \text{or} \quad x + 2 = 0 \][/tex]

[tex]\[ x = -6 \quad \text{or} \quad x = -2 \][/tex]

6. Substitute these [tex]\( x \)[/tex] values back into either of the original equations to find the corresponding [tex]\( y \)[/tex] values.

For [tex]\( x = -6 \)[/tex]:
[tex]\[ y = 4(-6) + 9 = -24 + 9 = -15 \][/tex]

For [tex]\( x = -2 \)[/tex]:
[tex]\[ y = 4(-2) + 9 = -8 + 9 = 1 \][/tex]

The solutions of the system are the points you have calculated: [tex]\( (-6, -15) \)[/tex] and [tex]\( (-2, 1) \)[/tex].

Therefore, the correct answer is:

[tex]\[ (-2, 1) \text{ and } (-6, -15). \][/tex]

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