Question

Help mathlook at pic​

Answer 1

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

Step-by-step explanation:

Given the system of equation

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

$2)~y=2x+5$

Expand the parenthesis of the 1) equation

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

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

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

$y=-x^2-4x-3$

Current System

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

$2)~y=2x+5$

Substitute the y value of the 1) equation with the 2) equation

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

Add ( x² + 4x + 3) on both sides

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

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

Combine like terms

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

$x^2+6x+8=0$

Factorize the quadratic equation

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

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

Substitute the x values into one of the equations to find the y value

$y=2x+5$

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

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

Therefore, the two solutions are:

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

Hope this helps!! :)

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