Question

which value (s) of x are sulution (s) of the equation below?

Answer 1

Complete question:

Which value(s) of x are the solution(s) of the equation below?

$\frac{1}{x-4} +\frac{x}{x-2} = \frac{2}{x^2 -6x +8}$

A) x = -1

B) x = -1, 4

C) x = 2, 4

D) x = 2

Answer:

The values of x are -1 and 4.

Step-by-step explanation:

Given,

The equation:

$\frac{1}{x-4} +\frac{x}{x-2} = \frac{2}{x^2 -6x +8}$

To find,

The values of the variable 'x'.

Calculation,

$\frac{1}{x-4} +\frac{x}{x-2} = \frac{2}{x^2 -6x +8}$

$\implies \frac{(x-2) + x(x-4)}{(x-2)(x-4)} = \frac{2}{x^2-6x+8}$....(1)  (Performed L.C.M on L.H.S)

Considering the denominator part of R.H.S and factorizing

i.e. x² - 6x + 8

⇒ x² -4x -2x + 8

⇒ x(x - 4) -2(x - 4)

⇒ (x - 2)(x - 4)

Replacing the denominator of R.H.S of equation (1) with (x -2)(x - 4)

we get:

$\implies \frac{(x-2) + x(x-4)}{(x-2)(x-4)} = \frac{2}{(x -2)(x-4)}$

⇒ (x - 2) + (x² - 4x) = 2  (Here x ≠ 4 and x ≠ 2)

⇒ x² - 3x -4 = 0

⇒ x² - 4x + x - 4 = 0

⇒ x(x - 4) +1(x - 4) = 0

⇒ (x + 1)(x - 4) = 0

⇒ x = -1, 4

Therefore, the values of x are -1 and 4.

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Answer 2

Answer:

idk

Step-by-step explanation:

i cant figure it out