Respuesta :

Answer:

[tex]x \neq-2 \text { and } x \neq-1[/tex]

Step-by-step explanation:

The restricted value is a value that makes the denominator of a function equal to zero. Thus, solving these function results in an undefined solution.

Thus,[tex]x \neq-2[/tex] and [tex]x \neq-1[/tex] are the restricted values for [tex]\frac{x+5}{x+2}[/tex] divided by  [tex]\frac{x-3}{x+1}[/tex]

Because, substituting[tex]x=-2[/tex] in [tex]\frac{x+5}{x+2}[/tex]

Now, substituting [tex]x=-2[/tex], we get,

[tex]\begin{aligned}\frac{x+5}{x+2} &=\frac{x+5}{-2+2} \\&=\frac{x+5}{0}\end{aligned}[/tex]

Thus, results the solution undefined.

Also, substituting [tex]x=-1[/tex] in [tex]\frac{x-3}{x+1}[/tex]  

[tex]\begin{aligned}\frac{x-3}{x+1} &=\frac{x-3}{-1+1} \\&=\frac{x-3}{0}\end{aligned}[/tex]

Thus results the solution undefined.

Hence, the restricted values are  [tex]x \neq-2 \text { and } x \neq-1[/tex]

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