Question

1. solve for all values of x.

\\(\frac{9x}{x - 2}-\frac{8}{x^{2}-2x}=-\frac{1}{x - 2}\\) factor
lcd: x(x - 2)
make the bottom number the same
\\(\frac{9x}{x - 2}-\frac{8}{x(x - 2)}=-\frac{1}{x - 2}\\)
\\(\frac{9xcdot x}{x(x - 2)}-\frac{8}{x(x - 2)}=-\frac{1cdot x}{x(x - 2)}\\)
\\(\frac{9x^{2}-8}{x(x - 2)}=-\frac{x}{x(x - 2)}\\)
9x^{2}-8=-x
9x^{2}+x - 8 = 0
\\(9x - 8)(x + 1)=0\\)

Explanation:

Step1: Identify the LCD

The given equation is $\frac{9x}{x - 2}-\frac{8}{x^{2}-2x}=-\frac{1}{x - 2}$. Factor $x^{2}-2x=x(x - 2)$. The least - common denominator (LCD) of the fractions is $x(x - 2)$.

Step2: Multiply each term by the LCD

Multiply each term in the equation $\frac{9x}{x - 2}-\frac{8}{x(x - 2)}=-\frac{1}{x - 2}$ by $x(x - 2)$:
$x(x - 2)\times\frac{9x}{x - 2}-x(x - 2)\times\frac{8}{x(x - 2)}=-x(x - 2)\times\frac{1}{x - 2}$
$9x\times x-8=-x$
$9x^{2}-8=-x$

Step3: Rearrange to form a quadratic equation

Move all terms to one side to get a quadratic equation:
$9x^{2}+x - 8=0$

Step4: Factor the quadratic equation

Factor the quadratic equation $9x^{2}+x - 8=(9x - 8)(x + 1)=0$

Step5: Solve for x

Set each factor equal to zero:
If $9x-8 = 0$, then $9x=8$, and $x=\frac{8}{9}$
If $x + 1=0$, then $x=-1$
We need to check for extraneous solutions by substituting these values back into the original equation. The original equation has denominators $x - 2$ and $x(x - 2)$. When $x=\frac{8}{9}$, $x-2=\frac{8}{9}-2=\frac{8 - 18}{9}=-\frac{10}{9}
eq0$ and $x(x - 2)=\frac{8}{9}\times(-\frac{10}{9})
eq0$. When $x=-1$, $x - 2=-1-2=-3
eq0$ and $x(x - 2)=(-1)\times(-3)
eq0$.

Answer:

$x=-1,\frac{8}{9}$