To solve the equation [tex]\(\frac{c-4}{c-2} = \frac{c-2}{c+2} - \frac{1}{2-c}\)[/tex], follow these steps:
### Step 1: Simplify and combine terms
First, notice that [tex]\(\frac{1}{2-c}\)[/tex] can be rewritten by recognizing that [tex]\(2 - c = -(c - 2)\)[/tex]:
[tex]\[
\frac{1}{2-c} = -\frac{1}{c-2}
\][/tex]
So our equation becomes:
[tex]\[
\frac{c-4}{c-2} = \frac{c-2}{c+2} + \frac{1}{c-2}
\][/tex]
### Step 2: Get a common denominator on the right-hand side
Combine the terms on the right side by getting a common denominator. The right side becomes:
[tex]\[
\frac{(c-2)^2 + 1}{(c+2)(c-2)}
\][/tex]
Simplify [tex]\((c-2)^2 + 1\)[/tex] to get:
[tex]\[
c^2 - 4c + 4 + 1 = c^2 - 4c + 5
\][/tex]
So, we now have:
[tex]\[
\frac{c-4}{c-2} = \frac{c^2 - 4c + 5}{(c+2)(c-2)}
\][/tex]
### Step 3: Cross-multiply and simplify
Since [tex]\((c-2)\)[/tex] appears on both sides, multiply both sides by [tex]\((c-2)(c+2)\)[/tex]:
[tex]\[
(c-4)(c+2) = c^2 - 4c + 5
\][/tex]
Expand each side:
[tex]\[
c^2 + 2c - 8 = c^2 - 4c + 5
\][/tex]
### Step 4: Set up the simplified equation
Cancel out [tex]\(c^2\)[/tex] on both sides of the equation:
[tex]\[
2c - 8 = -4c + 5
\][/tex]
### Step 5: Solve for [tex]\(c\)[/tex]
Combine like terms:
[tex]\[
2c + 4c = 5 + 8
\][/tex]
[tex]\[
6c = 13
\][/tex]
Solve for [tex]\(c\)[/tex]:
[tex]\[
c = \frac{13}{6}
\][/tex]
### Step 6: Verify the solution and check for extraneous solutions
In the original problem, we had restrictions on [tex]\(c\)[/tex] based on denominators. Specifically, [tex]\(c \neq 2\)[/tex], [tex]\(c \neq -2\)[/tex], and [tex]\(c \neq 2\)[/tex].
Upon further examination and careful re-evaluation:
Solving:
• For [tex]\(c = 14\)[/tex] directly check:
[tex]\[
\frac{14-4}{14-2} = \frac{10}{12} = \frac{5}{6}
\][/tex]
And
[tex]\[
\frac{14-2}{14+2}-\frac{1}{2-14} = \frac{12}{16} - \frac{1}{-12} = \frac{3}{4} + \frac{1}{12}
= \frac{3*3}{12} - (-1) = 1 = \frac{5}{6}
Correctly matches up, verifing finalised.
### Conclusion
There is 1 valid solution:
\[
c = 14
\][/tex]
There are no extraneous solutions here because:
All tested consistently checks out verify against initial temporary assumptions leading reliably verification,
Thus:
The number of extraneous solutions is zero.
So, the solution to the equation is [tex]\(c = 14\)[/tex] and there are zero extraneous solutions.
