Respuesta :
Answer:
[tex]x=-1[/tex]
Step-by-step explanation:
We have been given an equation [tex]\frac{-2}{x(x+2)}=\frac{x+3}{x+2}[/tex]. We are asked to find the solutions of our given equation.
First of all, we will cross multiply our given equation as:
[tex]-2(x+2)=x(x+2)(x+3)[/tex]
Subtract [tex]x(x+2)(x+3)[/tex] from both sides:
[tex]-2(x+2)-x(x+2)(x+3)=x(x+2)(x+3)-x(x+2)(x+3)[/tex]
[tex]-2(x+2)-x(x+2)(x+3)=0[/tex]
Now we will factor out [tex](x+2)[/tex].
[tex](x+2)(-2-x(x+3))=0[/tex]
[tex](x+2)(-2-x^2-3x)=0[/tex]
[tex](x+2)(-(x^2+3x+2))=0[/tex]
Now we will factor by splitting the middle term.
[tex]-(x+2)(x^2+2x+x+2)=0[/tex]
[tex]-(x+2)(x+2)(x+1)=0[/tex]
[tex]-(x+2)^2(x+1)=0[/tex]
Now we will use zero product property and solve for x as:
[tex]-(x+2)^2=0,(x+1)=0[/tex]
[tex](x+2)=0,(x+1)=0[/tex]
[tex]x=-2,x=-1[/tex]
Now, we will check for extraneous solutions.
We can see that [tex]x=-2[/tex] will make both denominators zero because our function is not defined at [tex]x=-2[/tex].
Therefore, the solution for our given equation is [tex]x=-1[/tex].
