Question

which value(s) of x are solutions) of the equation below?
\frac{1}{x-4}+\frac{x}{x-2}=\frac{2}{x^{2}-6x+8}

Answer 1

To find the values of x that are solutions to the given equation, we first need to find a common denominator for all the fractions. The common denominator in this case will be x^2 - 6x + 8.

After obtaining a common denominator, we can rewrite the equation as follows:

\frac{1(x-2)}{(x-2)(x-4)} + \frac{x(x-4)}{(x-2)(x-4)} = \frac{2}{(x-2)(x-4)}

Simplifying the equation, we get:

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

Combining the like terms in the numerator, we get:

\frac{x^2 - 3x - 2}{(x-2)(x-4)} = \frac{2}{(x-2)(x-4)}

Now we can cross multiply to get:

(x^2 - 3x - 2) = 2

Rearranging the terms, we get a quadratic equation:

x^2 - 3x - 4 = 0

Now, we can factorize the quadratic equation:

(x - 4)(x + 1) = 0

Setting each factor to zero, we get two possible solutions:

x - 4 = 0 => x = 4
x + 1 = 0 => x = -1

Therefore, the values of x that are solutions to the given equation are x = 4 and x = -1.