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Sagot :
The equations exist [tex]$y =-(x+2)^2+1[/tex] and [tex]y =4x+9[/tex] then the value of
x = -6, x = -2 and y = -15, y = 1.
How to solve the system of equations [tex]$y =-(x+2)^2+1[/tex] and
[tex]y =4x+9[/tex] ?
The given equations are [tex]$y =-(x+2)^2+1[/tex] and [tex]y =4x+9[/tex]
[tex]$-\left(x\:+\:2\right)^2\:+\:1\:=\:4x\:+\:9[/tex]
[tex]$-x^{2}-4 x-3=4 x+9[/tex]
Subtract 9 from both sides, we get
[tex]$-x^{2}-4 x-3-9=4 x+9-9[/tex]
Simplifying the equation, we get
[tex]$-x^{2}-4 x-12=4 x[/tex]
Subtract 4x from both sides
[tex]$-x^{2}-4 x-12-4 x=4 x-4 x[/tex]
[tex]$-x^{2}-8 x-12=0[/tex]
Solve with the quadratic formula
[tex]$x_{1,2}=\frac{-(-8) \pm \sqrt{(-8)^{2}-4(-1)(-12)}}{2(-1)}[/tex]
[tex]$\sqrt{(-8)^{2}-4(-1)(-12)}=4[/tex]
[tex]$x_{1,2}=\frac{-(-8) \pm 4}{2(-1)}[/tex]
Separate the solutions
[tex]$x_{1}=\frac{-(-8)+4}{2(-1)}, x_{2}=\frac{-(-8)-4}{2(-1)}[/tex]
[tex]$x=\frac{-(-8)+4}{2(-1)}=-6[/tex]
[tex]$x=\frac{-(-8)-4}{2(-1)}= \quad-2[/tex]
The solutions to the quadratic equation are x = -6, x = -2
From the above equation [tex]y =4x+9[/tex],
substitute the value of x, then we get
Put, x = -6 then y = 4(-6) + 9 = -15
Put, x = -2 then y = 4(-2) + 9 = 1
The system of equations exists (–2, 1) and (–6, –15).
Therefore, the correct answer is (–2, 1) and (–6, –15).
To learn more about quadratic equations refer to:
https://brainly.com/question/1214333
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