Question

solve and check \\(\frac{c - 4}{c - 2}=\frac{c - 2}{c + 2}-\frac{1}{2 - c}\\)
the solution is \\(c = \square\\)
how many extraneous solutions are there?\\(\boldsymbol{\downarrow}\\)

Explanation:

Step1: Simplify the right-hand side

Note that $\frac{1}{2-c} = -\frac{1}{c-2}$, so rewrite the equation:
$$\frac{c-4}{c-2} = \frac{c-2}{c+2} + \frac{1}{c-2}$$

Step2: Eliminate denominators

Multiply all terms by $(c-2)(c+2)$ (the least common denominator, where $c
eq 2, -2$):
$$(c-4)(c+2) = (c-2)^2 + (c+2)$$

Step3: Expand all terms

$$c^2 - 2c - 8 = c^2 - 4c + 4 + c + 2$$

Step4: Simplify both sides

Combine like terms on the right-hand side:
$$c^2 - 2c - 8 = c^2 - 3c + 6$$
Subtract $c^2$ from both sides:
$$-2c - 8 = -3c + 6$$

Step5: Solve for $c$

Add $3c$ to both sides:
$$c - 8 = 6$$
Add 8 to both sides:
$$c = 14$$

Step6: Check for extraneous solutions

Substitute $c=14$ into the original equation:
Left-hand side: $\frac{14-4}{14-2} = \frac{10}{12} = \frac{5}{6}$
Right-hand side: $\frac{14-2}{14+2} - \frac{1}{2-14} = \frac{12}{16} + \frac{1}{12} = \frac{3}{4} + \frac{1}{12} = \frac{9}{12} + \frac{1}{12} = \frac{10}{12} = \frac{5}{6}$
Both sides are equal, so $c=14$ is valid. The values $c=2$ and $c=-2$ make original denominators zero but are not solutions we found, so no extraneous solutions from our solved value.

Answer:

The solution is $c=14$
How many extraneous solutions are there? 0