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To solve the equation [tex]\( \frac{x}{7} + \frac{2}{x-2} = \frac{x}{x-2} \)[/tex] for [tex]\( x \)[/tex], let's follow the steps and check for valid solutions.
First, observe the given equation:
[tex]\[ \frac{x}{7} + \frac{2}{x-2} = \frac{x}{x-2} \][/tex]
The first step is to combine the terms on the left-hand side into a single fraction:
[tex]\[ \frac{x}{7} + \frac{2}{x-2} \][/tex]
Next, we find a common denominator for the fractions [tex]\(\frac{x}{7}\)[/tex] and [tex]\(\frac{2}{x-2}\)[/tex]. The common denominator is [tex]\(7(x-2)\)[/tex]:
[tex]\[ \frac{x(x-2)}{7(x-2)} + \frac{2 \cdot 7}{7(x-2)} = \frac{x}{x-2} \][/tex]
Combine the fractions:
[tex]\[ \frac{x(x-2) + 14}{7(x-2)} = \frac{x}{x-2} \][/tex]
Now we have:
[tex]\[ \frac{x^2 - 2x + 14}{7(x-2)} = \frac{x}{x-2} \][/tex]
To clear the denominators, we multiply both sides of the equation by [tex]\(7(x-2)\)[/tex] (note that [tex]\(x \neq 2\)[/tex] to avoid division by zero):
[tex]\[ x^2 - 2x + 14 = 7x \][/tex]
Rearrange the equation to set it to zero:
[tex]\[ x^2 - 2x + 14 - 7x = 0 \][/tex]
[tex]\[ x^2 - 9x + 14 = 0 \][/tex]
Next, we solve the quadratic equation [tex]\(x^2 - 9x + 14 = 0\)[/tex] using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 1\)[/tex], [tex]\(b = -9\)[/tex], and [tex]\(c = 14\)[/tex]:
[tex]\[ x = \frac{-(-9) \pm \sqrt{(-9)^2 - 4(1)(14)}}{2(1)} \][/tex]
[tex]\[ x = \frac{9 \pm \sqrt{81 - 56}}{2} \][/tex]
[tex]\[ x = \frac{9 \pm \sqrt{25}}{2} \][/tex]
[tex]\[ x = \frac{9 \pm 5}{2} \][/tex]
This gives us two potential solutions:
[tex]\[ x = \frac{9 + 5}{2} = 7 \][/tex]
[tex]\[ x = \frac{9 - 5}{2} = 2 \][/tex]
However, we must check for any extraneous solutions by substituting back into the original equation. Recall that [tex]\(x = 2\)[/tex] will make the original equation undefined since it involves division by zero.
Therefore, [tex]\(x = 2\)[/tex] is not a valid solution. So the only valid solution is:
[tex]\[ x = 7 \][/tex]
Thus, the correct answer is:
C. [tex]\( x = 7 \)[/tex]
First, observe the given equation:
[tex]\[ \frac{x}{7} + \frac{2}{x-2} = \frac{x}{x-2} \][/tex]
The first step is to combine the terms on the left-hand side into a single fraction:
[tex]\[ \frac{x}{7} + \frac{2}{x-2} \][/tex]
Next, we find a common denominator for the fractions [tex]\(\frac{x}{7}\)[/tex] and [tex]\(\frac{2}{x-2}\)[/tex]. The common denominator is [tex]\(7(x-2)\)[/tex]:
[tex]\[ \frac{x(x-2)}{7(x-2)} + \frac{2 \cdot 7}{7(x-2)} = \frac{x}{x-2} \][/tex]
Combine the fractions:
[tex]\[ \frac{x(x-2) + 14}{7(x-2)} = \frac{x}{x-2} \][/tex]
Now we have:
[tex]\[ \frac{x^2 - 2x + 14}{7(x-2)} = \frac{x}{x-2} \][/tex]
To clear the denominators, we multiply both sides of the equation by [tex]\(7(x-2)\)[/tex] (note that [tex]\(x \neq 2\)[/tex] to avoid division by zero):
[tex]\[ x^2 - 2x + 14 = 7x \][/tex]
Rearrange the equation to set it to zero:
[tex]\[ x^2 - 2x + 14 - 7x = 0 \][/tex]
[tex]\[ x^2 - 9x + 14 = 0 \][/tex]
Next, we solve the quadratic equation [tex]\(x^2 - 9x + 14 = 0\)[/tex] using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 1\)[/tex], [tex]\(b = -9\)[/tex], and [tex]\(c = 14\)[/tex]:
[tex]\[ x = \frac{-(-9) \pm \sqrt{(-9)^2 - 4(1)(14)}}{2(1)} \][/tex]
[tex]\[ x = \frac{9 \pm \sqrt{81 - 56}}{2} \][/tex]
[tex]\[ x = \frac{9 \pm \sqrt{25}}{2} \][/tex]
[tex]\[ x = \frac{9 \pm 5}{2} \][/tex]
This gives us two potential solutions:
[tex]\[ x = \frac{9 + 5}{2} = 7 \][/tex]
[tex]\[ x = \frac{9 - 5}{2} = 2 \][/tex]
However, we must check for any extraneous solutions by substituting back into the original equation. Recall that [tex]\(x = 2\)[/tex] will make the original equation undefined since it involves division by zero.
Therefore, [tex]\(x = 2\)[/tex] is not a valid solution. So the only valid solution is:
[tex]\[ x = 7 \][/tex]
Thus, the correct answer is:
C. [tex]\( x = 7 \)[/tex]
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