To solve the equation [tex]\(\frac{8}{x} + 2 = \frac{x + 4}{x - 6}\)[/tex], we need to follow several algebraic steps ensuring that we respect the given condition [tex]\(x \neq \{0, 6\}\)[/tex].
1. Identify the domain: The condition [tex]\(x \neq \{0, 6\}\)[/tex] tells us to exclude [tex]\(x = 0\)[/tex] and [tex]\(x = 6\)[/tex] from our solutions right from the start.
2. Multiply through by [tex]\(x(x-6)\)[/tex]: This will clear the denominators.
[tex]\[
x(x-6) \left(\frac{8}{x} + 2\right) = x(x-6) \cdot \frac{x+4}{x-6}
\][/tex]
3. Simplify the equation:
[tex]\[
8(x-6) + 2x(x-6) = (x+4)x
\][/tex]
[tex]\[
8x - 48 + 2x^2 - 12x = x^2 + 4x
\][/tex]
4. Combine like terms: Rearrange and combine all the terms on one side of the equation.
[tex]\[
2x^2 - 4x - 48 = x^2 + 4x
\][/tex]
[tex]\[
2x^2 - 4x - 48 - x^2 - 4x = 0
\][/tex]
[tex]\[
x^2 - 8x - 48 = 0
\][/tex]
5. Solve the quadratic equation:
[tex]\[
x^2 - 8x - 48 = 0
\][/tex]
We can solve this using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 1\)[/tex], [tex]\(b = -8\)[/tex], and [tex]\(c = -48\)[/tex].
[tex]\[
x = \frac{8 \pm \sqrt{64 + 192}}{2}
\][/tex]
[tex]\[
x = \frac{8 \pm \sqrt{256}}{2}
\][/tex]
[tex]\[
x = \frac{8 \pm 16}{2}
\][/tex]
Therefore, we get two solutions:
[tex]\[
x = \frac{24}{2} = 12 \quad \text{and} \quad x = \frac{-8}{2} = -4
\][/tex]
6. Exclude invalid solutions: Ensure that our solutions do not violate the condition [tex]\(x \neq \{0,6\}\)[/tex]. Both [tex]\(x = 12\)[/tex] and [tex]\(x = -4\)[/tex] are valid as they do not equal 0 or 6.
Thus, the solution to the equation [tex]\(\frac{8}{x} + 2 = \frac{x + 4}{x - 6}\)[/tex] is:
[tex]\[
\{-4, 12\}
\][/tex]
