Answer:
At x=5 it is extraneous
and at [tex]x=\frac{4}{5}[/tex] is a verified solution.
Step-by-step explanation:
Given : Expression [tex]\frac{5}{x-5}=\frac{x}{x-5}-\frac{5}{4}x[/tex]
To determine : The solution is extraneous or not?
Solution :
We solve the given expression
[tex]\frac{5}{x-5}=\frac{x}{x-5}-\frac{5}{4}x[/tex]
Taking LCM
[tex]\frac{5}{x-5}=\frac{4x-5x(x-5)}{4(x-5)}[/tex]
(x-5) cancel from both sides,
[tex]5=\frac{4x-5x^2+25x}{4}[/tex]
[tex]20=29x-5x^2[/tex]
[tex]5x^2-29x+20=0[/tex]
Apply middle term split,
[tex]5x^2-25x-4x+20=0[/tex]
[tex]5x(x-5)-4(x-5)=0[/tex]
[tex](x-5)(5x-4)=0[/tex]
[tex]\text{Either }(x-5)=0\text{ or }(5x-4)=0[/tex]
[tex]x=5, x=\frac{4}{5}[/tex]
Extraneous is when we get the solution mathematically correct.
Substituting x = 5 gives denominators of 0, which is extraneous.
Substituting [tex]x=\frac{4}{5}[/tex] gives a valid equation, so this is the verified solution.
