Question

Solve 5 over x minus 5 equals the quantity of x over x minus 5, minus five fourths for x and determine if the solution is extraneous or not. (1 point)

Answer 1

Answer:

At x=5 it is extraneous

and at $x=\frac{4}{5}$ is a verified solution.

Step-by-step explanation:

Given : Expression $\frac{5}{x-5}=\frac{x}{x-5}-\frac{5}{4}x$

To determine : The solution is extraneous or not?

Solution :

We solve the given expression

$\frac{5}{x-5}=\frac{x}{x-5}-\frac{5}{4}x$

Taking LCM

$\frac{5}{x-5}=\frac{4x-5x(x-5)}{4(x-5)}$

(x-5) cancel from both sides,

$5=\frac{4x-5x^2+25x}{4}$

$20=29x-5x^2$

$5x^2-29x+20=0$

Apply middle term split,

$5x^2-25x-4x+20=0$

$5x(x-5)-4(x-5)=0$

$(x-5)(5x-4)=0$

$\text{Either }(x-5)=0\text{ or }(5x-4)=0$

$x=5, x=\frac{4}{5}$

Extraneous is when we get the solution mathematically correct.

Substituting x = 5 gives denominators of 0, which is extraneous.

Substituting $x=\frac{4}{5}$ gives a valid equation, so this is the verified solution.

Answer 2

Assuming the equation is:
5/(x-5) = x/(x-5) - 5x/4
We first multiply by the LCD: 4(x-5)
20 = 4x - 5x(x-5)
20 = 4x - 5x^2 + 25x
5x^2 - 29x + 20 = 0
(5x - 4)(x - 5) = 0
x = 4/5, 5
Substituting x = 5 gives denominators of 0, which is extraneous.
Substituting x = 4/5 gives a valid equation, so this is the only correct solution.