To find the solutions to the system of equations:
[tex]\[ y = -(x + 2)^2 + 1 \][/tex]
[tex]\[ y = 3x + 7 \][/tex]
we need to find the points [tex]\((x, y)\)[/tex] where both equations are satisfied simultaneously.
1. Substitute [tex]\( y \)[/tex] from the second equation into the first equation:
Since both equations equal [tex]\( y \)[/tex], set them equal to each other:
[tex]\[ 3x + 7 = -(x + 2)^2 + 1 \][/tex]
2. Simplify the equation:
Expand and simplify the quadratic equation:
[tex]\[ 3x + 7 = -(x^2 + 4x + 4) + 1 \][/tex]
[tex]\[ 3x + 7 = -x^2 - 4x - 4 + 1 \][/tex]
[tex]\[ 3x + 7 = -x^2 - 4x - 3 \][/tex]
3. Move all terms to one side to form a standard quadratic equation:
[tex]\[ 3x + 7 + x^2 + 4x + 3 = 0 \][/tex]
[tex]\[ x^2 + 7x + 10 = 0 \][/tex]
4. Factor the quadratic equation:
To factor [tex]\( x^2 + 7x + 10 = 0 \)[/tex], we look for two numbers that multiply to 10 and add up to 7. Those numbers are 2 and 5:
[tex]\[ (x + 2)(x + 5) = 0 \][/tex]
5. Solve for [tex]\( x \)[/tex]:
Set each factor equal to zero:
[tex]\[ x + 2 = 0 \quad \text{or} \quad x + 5 = 0 \][/tex]
[tex]\[ x = -2 \quad \text{or} \quad x = -5 \][/tex]
6. Find the corresponding [tex]\( y \)[/tex] values for [tex]\( x \)[/tex]:
Plug [tex]\( x = -2 \)[/tex] into one of the original equations to find [tex]\( y \)[/tex]:
[tex]\[ y = 3(-2) + 7 = -6 + 7 = 1 \][/tex]
So, one solution is [tex]\((-2, 1)\)[/tex].
Plug [tex]\( x = -5 \)[/tex] into the same equation to find [tex]\( y \)[/tex]:
[tex]\[ y = 3(-5) + 7 = -15 + 7 = -8 \][/tex]
So, another solution is [tex]\((-5, -8)\)[/tex].
Therefore, the solutions to the system of equations are:
[tex]\[ (-2, 1) \quad \text{and} \quad (-5, -8) \][/tex]
So, the correct answer is:
[tex]\[ \boxed{(-2,1) \text{ and } (-5,-8)} \][/tex]
