Answer:
The answer is B). 5 is an extraneous solution because it creates a 0 in the denominator.
Step-by-step explanation:
In order to check Heather's work we need to verify the solutions :
[tex]x=5[/tex] or [tex]x=-1[/tex]
The expression is :
[tex]\frac{1}{x-4}=\frac{x+1}{x-5}-\frac{6}{x^{2}-9x+20}[/tex]
To check the solution [tex]x=5[/tex] we need to replace the ''x'' by 5 in the expression :
[tex]\frac{1}{5-4}=\frac{5+1}{5-5}-\frac{6}{5^{2}-(9).(5)+20}[/tex]
[tex]\frac{1}{1}=\frac{6}{0}-\frac{6}{0}[/tex]
We obtain two zeros in the denominators of the fractions at the right of the equation ⇒ 5 is an extraneous solution because it creates two 0 in the denominators of two fractions in the expression.
B). is a correct option
Given that we have one extraneous solution ⇒ C). is an incorrect option
The other solution [tex]x=-1[/tex] is a correct solution
If we replace ''x'' by -1 in the expression :
[tex]\frac{1}{-1-4}=\frac{-1+1}{-1-5}-\frac{6}{(-1)^{2}-(9).(-1)+20}[/tex]
[tex]-\frac{1}{5}=0-\frac{6}{30}[/tex]
[tex]-\frac{1}{5}=-\frac{1}{5}[/tex]
The solution [tex]x=-1[/tex] verifies the expression ⇒
A) is an incorrect option and D) is an incorrect option
Therefore, B). is the solution to this exercise.
