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Elena-2011 [213]
1 year ago
11

Help mathlook at pic​

Mathematics
1 answer:
Sloan [31]1 year ago
8 0

Answer: \Large\boxed{(-4,~-3)~\text{and}~(-2,~1)}

Step-by-step explanation:

<u>Given the system of equation</u>

1)~y=-(x+2)^2+1

2)~y=2x+5

<u>Expand the parenthesis of the 1) equation</u>

y=-(x+2)^2+1

y=-(x^2+4x+4)+1

y=-x^2-4x-4+1

y=-x^2-4x-3

<u>Current System</u>

1)~y=-x^2-4x-3

2)~y=2x+5

<u>Substitute the y value of the 1) equation with the 2) equation</u>

2x+5=-x^2-4x-3

<u>Add ( x² + 4x + 3) on both sides</u>

2x+5+(x^2+4x+3)=-x^2-4x-3+(x^2+4x+3)

2x+5+x^2+4x+3=0

<u>Combine like terms</u>

x^2+2x+4x+5+3=0

x^2+6x+8=0

<u>Factorize the quadratic equation</u>

(x+4)(x+2)=0

x=-4~\text{or}~x=-2

<u>Substitute the x values into one of the equations to find the y value</u>

y=2x+5

y=2(-4)+5=-8+5=-3

y=2(-2)+5=-4+5=1

<u>Therefore, the two solutions are:</u>

\Large\boxed{(-4,~-3)~\text{and}~(-2,~1)}

Hope this helps!! :)

Please let me know if you have any questions

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