Select the correct answer.

Solve the equation for [tex]x[/tex].

[tex]\ \textless \ br/\ \textgreater \ \frac{x}{7} + \frac{2}{x-2} = \frac{x}{x-2}\ \textless \ br/\ \textgreater \ [/tex]

A. [tex]x = 0[/tex]
B. [tex]x = 2[/tex]
C. [tex]x = 7[/tex]
D. [tex]x = 2[/tex] or [tex]x = 7[/tex]
E. no solution



Answer :

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]