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solve the rational equation,
\\(\frac{3x + 2}{x - 2} + \frac{1}{x} = \frac{-2}{x^2 - 2x}\\)
select one:
a. \\(x = -1\\)
b. no solution
c. \\(x = 0, -1\\)
d. \\(x = 2, -3\\)

Explanation:

Step1: Factor the denominator

The right - hand side denominator \(x^{2}-2x\) can be factored as \(x(x - 2)\). So the equation \(\frac{3x + 2}{x-2}+\frac{1}{x}=\frac{-2}{x^{2}-2x}\) becomes \(\frac{3x + 2}{x-2}+\frac{1}{x}=\frac{-2}{x(x - 2)}\).

Step2: Find the least common denominator (LCD)

The denominators are \(x-2\), \(x\), and \(x(x - 2)\). The LCD of these fractions is \(x(x - 2)\).

Step3: Multiply each term by the LCD

Multiply each term in the equation \(\frac{3x + 2}{x-2}+\frac{1}{x}=\frac{-2}{x(x - 2)}\) by \(x(x - 2)\):
\(x(x - 2)\times\frac{3x + 2}{x-2}+x(x - 2)\times\frac{1}{x}=x(x - 2)\times\frac{-2}{x(x - 2)}\)
Simplify each term:

• For the first term: \(x(3x + 2)=3x^{2}+2x\) (because \(x-2\) in the numerator and denominator cancels out)
• For the second term: \((x - 2)\times1=x - 2\) (because \(x\) in the numerator and denominator cancels out)
• For the third term: \(- 2\) (because \(x(x - 2)\) in the numerator and denominator cancels out)

So we get the equation \(3x^{2}+2x+x - 2=-2\).

Step4: Simplify the equation

Combine like terms on the left - hand side: \(3x^{2}+(2x + x)-2=-2\), which simplifies to \(3x^{2}+3x-2=-2\).
Add 2 to both sides of the equation: \(3x^{2}+3x-2 + 2=-2 + 2\), so \(3x^{2}+3x=0\).
Factor out the common factor \(3x\): \(3x(x + 1)=0\).

Step5: Solve for x

Set each factor equal to zero:

• If \(3x=0\), then \(x = 0\). But if we substitute \(x = 0\) back into the original equation, the denominators \(\frac{1}{x}\) and \(\frac{-2}{x^{2}-2x}\) will be undefined (division by zero). So \(x = 0\) is an extraneous solution.
• If \(x + 1=0\), then \(x=-1\).

We need to check if \(x=-1\) is a valid solution. Substitute \(x=-1\) into the original equation:
Left - hand side: \(\frac{3\times(-1)+2}{-1 - 2}+\frac{1}{-1}=\frac{-3 + 2}{-3}-1=\frac{-1}{-3}-1=\frac{1}{3}-1=\frac{1 - 3}{3}=-\frac{2}{3}\)
Right - hand side: \(\frac{-2}{(-1)^{2}-2\times(-1)}=\frac{-2}{1 + 2}=-\frac{2}{3}\)
Since the left - hand side equals the right - hand side when \(x=-1\), \(x=-1\) is a valid solution.

Answer:

A. \(x=-1\)