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To determine how many solutions the equation [tex]\(\frac{x+2}{x-4} - \frac{1}{x} = \frac{4}{x^2-4x}\)[/tex] has, we need to analyze it step-by-step:
1. Simplify and understand the given equation:
[tex]\[ \frac{x+2}{x-4} - \frac{1}{x} = \frac{4}{x^2-4x} \][/tex]
We notice that the term [tex]\(\frac{4}{x^2-4x}\)[/tex] matches [tex]\( \frac{4}{x(x-4)} \)[/tex] since [tex]\(x^2-4x\)[/tex] can be factored as [tex]\(x(x-4)\)[/tex].
2. Rewrite the equation with the common denominator:
Convert [tex]\( \frac{x+2}{x-4} \)[/tex] and [tex]\( \frac{1}{x} \)[/tex] to have a common denominator of [tex]\( x(x-4) \)[/tex]:
[tex]\[ \frac{(x+2)x - (x-4)}{x(x-4)} = \frac{4}{x(x-4)} \][/tex]
Simplify the numerator:
[tex]\[ \frac{x^2 + 2x - x + 4}{x(x-4)} = \frac{4}{x(x-4)} \][/tex]
Which further simplifies to:
[tex]\[ \frac{x^2 + x + 4}{x(x-4)} = \frac{4}{x(x-4)} \][/tex]
3. Set the numerators equal since the denominators are the same:
Since the denominators [tex]\(x(x-4)\)[/tex] don't affect solutions other than causing [tex]\(x \neq 0\)[/tex] and [tex]\(x \neq 4\)[/tex]:
[tex]\[ x^2 + x + 4 = 4 \][/tex]
Simplify this equation:
[tex]\[ x^2 + x + 4 - 4 = 0 \][/tex]
[tex]\[ x^2 + x = 0 \][/tex]
4. Solve the quadratic equation:
Factor the quadratic equation:
[tex]\[ x(x + 1) = 0 \][/tex]
The solutions to this equation are:
[tex]\[ x = 0 \quad \text{or} \quad x = -1 \][/tex]
5. Evaluate and reject extraneous solutions:
From earlier, we noticed [tex]\(x \neq 0\)[/tex] and [tex]\(x \neq 4\)[/tex] to avoid division by zero. Thus, [tex]\( x = 0 \)[/tex] is not a valid solution. Therefore, only the solution [tex]\( x = -1 \)[/tex] remains.
6. Count the valid solutions:
Considering [tex]\(x = -1\)[/tex] as the valid solution, we see there is only one valid solution.
Therefore, the equation [tex]\(\frac{x+2}{x-4} - \frac{1}{x} = \frac{4}{x^2-4 x}\)[/tex] has exactly:
[tex]\[ \boxed{1} \][/tex]
Thus, the correct answer is B. 1.
1. Simplify and understand the given equation:
[tex]\[ \frac{x+2}{x-4} - \frac{1}{x} = \frac{4}{x^2-4x} \][/tex]
We notice that the term [tex]\(\frac{4}{x^2-4x}\)[/tex] matches [tex]\( \frac{4}{x(x-4)} \)[/tex] since [tex]\(x^2-4x\)[/tex] can be factored as [tex]\(x(x-4)\)[/tex].
2. Rewrite the equation with the common denominator:
Convert [tex]\( \frac{x+2}{x-4} \)[/tex] and [tex]\( \frac{1}{x} \)[/tex] to have a common denominator of [tex]\( x(x-4) \)[/tex]:
[tex]\[ \frac{(x+2)x - (x-4)}{x(x-4)} = \frac{4}{x(x-4)} \][/tex]
Simplify the numerator:
[tex]\[ \frac{x^2 + 2x - x + 4}{x(x-4)} = \frac{4}{x(x-4)} \][/tex]
Which further simplifies to:
[tex]\[ \frac{x^2 + x + 4}{x(x-4)} = \frac{4}{x(x-4)} \][/tex]
3. Set the numerators equal since the denominators are the same:
Since the denominators [tex]\(x(x-4)\)[/tex] don't affect solutions other than causing [tex]\(x \neq 0\)[/tex] and [tex]\(x \neq 4\)[/tex]:
[tex]\[ x^2 + x + 4 = 4 \][/tex]
Simplify this equation:
[tex]\[ x^2 + x + 4 - 4 = 0 \][/tex]
[tex]\[ x^2 + x = 0 \][/tex]
4. Solve the quadratic equation:
Factor the quadratic equation:
[tex]\[ x(x + 1) = 0 \][/tex]
The solutions to this equation are:
[tex]\[ x = 0 \quad \text{or} \quad x = -1 \][/tex]
5. Evaluate and reject extraneous solutions:
From earlier, we noticed [tex]\(x \neq 0\)[/tex] and [tex]\(x \neq 4\)[/tex] to avoid division by zero. Thus, [tex]\( x = 0 \)[/tex] is not a valid solution. Therefore, only the solution [tex]\( x = -1 \)[/tex] remains.
6. Count the valid solutions:
Considering [tex]\(x = -1\)[/tex] as the valid solution, we see there is only one valid solution.
Therefore, the equation [tex]\(\frac{x+2}{x-4} - \frac{1}{x} = \frac{4}{x^2-4 x}\)[/tex] has exactly:
[tex]\[ \boxed{1} \][/tex]
Thus, the correct answer is B. 1.
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