To solve the equation:
[tex]\[
\frac{1}{x-2} + \frac{6x}{x^2-4} = \frac{3}{x+2},
\][/tex]
we'll break it down step-by-step.
First, note that [tex]\(x^2 - 4\)[/tex] can be factored:
[tex]\[
x^2 - 4 = (x-2)(x+2).
\][/tex]
This allows us to rewrite the equation as:
[tex]\[
\frac{1}{x-2} + \frac{6x}{(x-2)(x+2)} = \frac{3}{x+2}.
\][/tex]
To solve this equation, we need to find a common denominator for the fractions. The common denominator is [tex]\((x-2)(x+2)\)[/tex]. Rewrite each term with this common denominator:
[tex]\[
\frac{(x+2)}{(x-2)(x+2)} + \frac{6x}{(x-2)(x+2)} = \frac{3(x-2)}{(x-2)(x+2)}.
\][/tex]
Combine all the terms over the common denominator:
[tex]\[
\frac{x+2 + 6x}{(x-2)(x+2)} = \frac{3(x-2)}{(x-2)(x+2)}.
\][/tex]
Simplify the numerators:
[tex]\[
\frac{7x+2}{(x-2)(x+2)} = \frac{3x-6}{(x-2)(x+2)}.
\][/tex]
Since the denominators are now the same, we can equate the numerators:
[tex]\[
7x + 2 = 3x - 6.
\][/tex]
Solve for [tex]\(x\)[/tex]:
[tex]\[
7x + 2 = 3x - 6,
\][/tex]
[tex]\[
7x - 3x = -6 - 2,
\][/tex]
[tex]\[
4x = -8,
\][/tex]
[tex]\[
x = -2.
\][/tex]
However, if we look closer, [tex]\(x = -2\)[/tex] makes the original denominators zero (the terms [tex]\(\frac{1}{x-2}\)[/tex] and [tex]\(\frac{3}{x+2}\)[/tex] both become undefined). Thus, [tex]\(x = -2\)[/tex] is not a valid solution. We must consider potential solutions:
Reviewing the relationship, it is evident that there is no value of [tex]\(x\)[/tex] that would satisfy this equation without causing a division by zero or conflicting terms, i.e., [tex]\(x - 2\)[/tex] or [tex]\(x + 2\)[/tex] cannot be zero simultaneously while satisfying the equation's equality.
Thus, ultimately, the equation has:
[tex]\[
\text{No solution.}
\][/tex]
The only scenario we encounter makes the equation undefined, resulting in no solution.
