Respuesta :
The solution of given rational equation [tex]\frac{2}{n} +\frac{n+2}{n+1}=\frac{-2}{n^2+n}[/tex] is n = -2
In this question,
We have been given a rational equation.
[tex]\frac{2}{n} +\frac{n+2}{n+1}=\frac{-2}{n^2+n}[/tex]
We need to find the solution of given rational equation.
First we simplify the LHS of given equation.
[tex]\frac{2}{n} +\frac{n+2}{n+1}\\\\=\frac{2(n+1)+n(n+2)}{n(n+1)} \\\\=\frac{2n+2+n^2+2n}{n^2+n}\\\\ =\frac{n^2+4n+2}{n^2+n}[/tex]
So, given equation would be,
[tex]\frac{n^2+4n+2}{n^2+n}=\frac{-2}{n^2+n}[/tex]
Here, the denominator is same on the both sides.
⇒ n² + 4n + 2 = -2
⇒ n² + 4n + 4 = 0
⇒ (n + 2)² = 0
⇒ n + 2 = 0
⇒ n = -2
We need to check above solution.
so, substitute n = -2 in LHS of given equation.
[tex]\frac{2}{-2} +\frac{-2+2}{-2+1}\\\\=-1+0\\\\=-1[/tex]
Now, substitute n = -2 in RHS of given equation.
[tex]\frac{-2}{-2^2-2}\\\\=\frac{-2}{4-2}\\\\= \frac{-2}{2} \\\\-1[/tex]
For n = -2, LHS = RHS
Therefore, the solution of given rational equation [tex]\frac{2}{n} +\frac{n+2}{n+1}=\frac{-2}{n^2+n}[/tex] is n = -2
Learn more about the rational equation here:
brainly.com/question/20850120
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