To solve the given equation:
[tex]\[
\frac{1}{x} + \frac{1}{2 - 16} = \frac{x - 2}{x - 10}
\][/tex]
First, simplify the term [tex]\(\frac{1}{2 - 16}\)[/tex]:
[tex]\[
2 - 16 = -14 \\
\frac{1}{2 - 16} = \frac{1}{-14} = -\frac{1}{14}
\][/tex]
Now, substitute this back into the equation:
[tex]\[
\frac{1}{x} - \frac{1}{14} = \frac{x - 2}{x - 10}
\][/tex]
To eliminate the fractions and solve the equation, find a common denominator for the left-hand side:
[tex]\[
\frac{14}{14x} - \frac{x}{14x} = \frac{14 - x}{14x}
\][/tex]
So the left-hand side becomes:
[tex]\[
\frac{14 - x}{14x}
\][/tex]
Now, our equation looks like this:
[tex]\[
\frac{14 - x}{14x} = \frac{x - 2}{x - 10}
\][/tex]
Cross-multiply to solve for [tex]\(x\)[/tex]:
[tex]\[
(14 - x)(x - 10) = 14x(x - 2)
\][/tex]
Expand both sides:
[tex]\[
14x - 140 - x^2 + 10x = 14x^2 - 28x
\][/tex]
Simplify and combine like terms:
[tex]\[
14x - 140 + 10x - x^2 = 14x^2 - 28x \\
24x - 140 - x^2 = 14x^2 - 28x
\][/tex]
Combine all terms on one side of the equation:
[tex]\[
0 = 14x^2 + x^2 - 28x - 24x + 140 \\
0 = 15x^2 - 52x + 140
\][/tex]
This is a quadratic equation of the form [tex]\(ax^2 + bx + c = 0\)[/tex], where [tex]\(a = 15\)[/tex], [tex]\(b = -52\)[/tex], and [tex]\(c = 140\)[/tex]. Solve the quadratic equation using the quadratic formula:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
To find the discriminant:
[tex]\[
b^2 - 4ac = (-52)^2 - 4(15)(140) \\
= 2704 - 8400 = -5696
\][/tex]
Since the discriminant is negative, this indicates that there are no real solutions to the equation. Therefore, we can conclude:
[tex]\[
x = \text{na} \;\text{or}\; x = \text{na}
\][/tex]
If the problem is constrained to real numbers, then there are no real solutions.
Thus, the solution is:
[tex]\[
x = \text{na} \ \text{or} \ x = \text{na}
\][/tex]
