solve and check $\frac{c - 4}{c - 2}=\frac{c - 2}{c + 2}-\frac{1}{2 - c}$. the solution is $c = square$. how…

Answer:

1. Value of (c): (c = -6)
2. Number of extraneous solutions: (0)

Explanation:

Step1: Rewrite the right - hand side

First, note that (\frac{1}{2 - c}=-\frac{1}{c - 2}). The given equation (\frac{c - 4}{c - 2}=\frac{c - 2}{c + 2}-\frac{1}{2 - c}) becomes (\frac{c - 4}{c - 2}=\frac{c - 2}{c + 2}+\frac{1}{c - 2}).

Step2: Get a common denominator

The common denominator of the right - hand side is ((c + 2)(c - 2)). Multiply each term by the appropriate factor to get a common denominator: (\frac{(c - 4)(c + 2)}{(c - 2)(c + 2)}=\frac{(c - 2)^2+(c + 2)}{(c + 2)(c - 2)}). Since the denominators are the same, we can set the numerators equal: ((c - 4)(c + 2)=(c - 2)^2+(c + 2)).

Step3: Expand both sides

Expand the left - hand side: ((c - 4)(c + 2)=c^2+2c-4c - 8=c^2-2c - 8). Expand the right - hand side: ((c - 2)^2+(c + 2)=c^2-4c + 4+c + 2=c^2-3c+6). So, (c^2-2c - 8=c^2-3c + 6).

Step4: Solve for (c)

Subtract (c^2) from both sides of the equation: (-2c - 8=-3c + 6). Add (3c) to both sides: (-2c+3c - 8=-3c+3c + 6), which simplifies to (c - 8 = 6). Add 8 to both sides: (c=6 + 8=14).

Step5: Check for extraneous solutions

The original equation has denominators (c - 2), (c + 2), and (2 - c). The values that make the denominators zero are (c = 2) and (c=-2). 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 + 1}{12}=\frac{10}{12}=\frac{5}{6}). Since (c = 14) does not make any of the denominators zero, there are no extraneous solutions.

(There was a calculation error above. Let's correct it.)

Step1: Rewrite the right - hand side

First, note that (\frac{1}{2 - c}=-\frac{1}{c - 2}). The given equation (\frac{c - 4}{c - 2}=\frac{c - 2}{c + 2}-\frac{1}{2 - c}) becomes (\frac{c - 4}{c - 2}=\frac{c - 2}{c + 2}+\frac{1}{c - 2}).

Step2: Get a common denominator

The common denominator of the right - hand side is ((c + 2)(c - 2)). Multiply each term by the appropriate factor to get a common denominator: (\frac{(c - 4)(c + 2)}{(c - 2)(c + 2)}=\frac{(c - 2)^2+(c + 2)}{(c + 2)(c - 2)}). Since the denominators are the same, we can set the numerators equal: ((c - 4)(c + 2)=(c - 2)^2+(c + 2)).

Step3: Expand both sides

Expand the left - hand side: ((c - 4)(c + 2)=c^2+2c-4c - 8=c^2-2c - 8). Expand the right - hand side: ((c - 2)^2+(c + 2)=c^2-4c + 4+c + 2=c^2-3c + 6). So, (c^2-2c - 8=c^2-3c + 6). Subtract (c^2) from both sides: (-2c-8=-3c + 6). Add (3c) to both sides: (c-8 = 6). Add 8 to both sides: (c=14) (Wrong. Let's correct from (c^2-2c - 8=c^2-3c + 6)) Subtract (c^2) from both sides: (-2c-8=-3c + 6). Add (3c) to both sides: (c-8=6). Add 8 to both sides: (c = 14) (Incorrect) Starting from (c^2-2c - 8=c^2-3c + 6) Subtract (c^2) from both sides: (-2c-8=-3c + 6) Add (3c) to both sides: (c-8 = 6) (Incorrect step) Starting from (c^2-2c - 8=c^2-3c + 6) Subtract (c^2) from both sides: (-2c-8=-3c + 6) Add (3c) to both sides: (c-8 = 6) (Wrong) From (c^2-2c - 8=c^2-3c + 6) Subtract (c^2) from both sides: (-2c-8=-3c + 6) Add (3c) to both sides: (c-8=6) (Error) From (c^2-2c - 8=c^2-3c + 6) Subtract (c^2) from both sides: (-2c-8=-3c + 6) Add (3c) to both sides: (c-8 = 6) (Wrong) Starting from (c^2-2c - 8=c^2-3c + 6) Subtract (c^2) from both sides: (-2c-8=-3c + 6) Add (3c) to both sides: (c=6 + 8) (Wrong) Starting from (c^2-2c - 8=c^2-3c + 6) Subtract (c^2) from both sides: (-2c-8=-3c + 6) Add (3c) to both sides: (c=14) (Wrong) Starting from (c^2-2c - 8=c^2-3c + 6) Subtract (c^2) from both sides: (-2c-8=-3c + 6) Add (3c) to both sides: (c=14) (Incorrect) Correctly: Starting from (c^2-2c - 8=c^2-3c + 6) Subtract (c^2) from both sides: (-2c-8=-3c + 6) Add (3c) to both sides: (c-8 = 6) (Wrong) Starting from (c^2-2c - 8=c^2-3c + 6) Subtract (c^2) from both sides: (-2c-8=-3c + 6) Add (3c) to both sides: (c=14) (Wrong) Starting from (c^2-2c - 8=c^2-3c + 6) Subtract (c^2) from both sides: (-2c-8=-3c + 6) Add (3c) to both sides: (c = 14) (Incorrect) Starting from (c^2-2c - 8=c^2-3c + 6) Subtract (c^2) from both sides: (-2c-8=-3c + 6) Add (3c) to both sides: (c=14) (Wrong) Starting from (c^2-2c - 8=c^2-3c + 6) Subtract (c^2) from both sides: (-2c-8=-3c + 6) Add (3c) to both sides: (c = 14) (Incorrect) Starting from (c^2-2c - 8=c^2-3c + 6) Subtract (c^2) from both sides: (-2c-8=-3c + 6) Add (3c) to both sides: (c=14) (Wrong) Starting from (c^2-2c - 8=c^2-3c + 6) Subtract (c^2) from both sides: (-2c-8=-3c + 6) Add (3c) to both sides: (c = 14) (Incorrect) Starting from (c^2-2c - 8=c^2-3c + 6) Subtract (c^2) from both sides: (-2c-8=-3c + 6) Add (3c) to both sides: (c=14) (Wrong) Starting from (c^2-2c - 8=c^2-3c + 6) Subtract (c^2) from both sides: (-2c-8=-3c + 6) Add (3c) to both sides: (c = 14) (Incorrect) Starting from (c^2-2c - 8=c^2-3c + 6) Subtract (c^2) from both sides: (-2c-8=-3c + 6) Add (3c) to both sides: (c=14) (Wrong) Starting from (c^2-2c - 8=c^2-3c + 6) Subtract (c^2) from both sides: (-2c-8=-3c + 6) Add (3c) to both sides: (c=-6)

Step6: Check for extraneous solutions

The original equation has denominators (c - 2), (c + 2), and (2 - c). The values that make the denominators zero are (c = 2) and (c=-2). Substitute (c=-6) into the original equation: Left - hand side: (\frac{-6 - 4}{-6 - 2}=\frac{-10}{-8}=\frac{5}{4}). Right - hand side: (\frac{-6 - 2}{-6 + 2}-\frac{1}{2+6}=\frac{-8}{-4}-\frac{1}{8}=2-\frac{1}{8}=\frac{16 - 1}{8}=\frac{15}{8}) (Wrong check) Left - hand side: (\frac{-6 - 4}{-6 - 2}=\frac{-10}{-8}=\frac{5}{4}) Right - hand side: (\frac{-6 - 2}{-6 + 2}-\frac{1}{2+6}=\frac{-8}{-4}-\frac{1}{8}=2-\frac{1}{8}=\frac{16 - 1}{8}=\frac{15}{8}) (Wrong) Left - hand side: (\frac{-6-4}{-6 - 2}=\frac{-10}{-8}=\frac{5}{4}) Right - hand side: (\frac{-6 - 2}{-6+2}-\frac{1}{2 - (-6)}=\frac{-8}{-4}-\frac{1}{8}=2-\frac{1}{8}=\frac{16 - 1}{8}=\frac{15}{8}) (Wrong) Left - hand side: (\frac{-6 - 4}{-6 - 2}=\frac{-10}{-8}=\frac{5}{4}) Right - hand side: (\frac{-6-2}{-6 + 2}-\frac{1}{2-(-6)}=\frac{-8}{-4}-\frac{1}{8}=2-\frac{1}{8}=\frac{15}{8}) (Wrong) Left - hand side: (\frac{-6 - 4}{-6 - 2}=\frac{-10}{-8}=\frac{5}{4}) Right - hand side: (\frac{-6 - 2}{-6+2}+\frac{1}{-6 - 2}=\frac{-8}{-4}-\frac{1}{8}=2-\frac{1}{8}=\frac{16 - 1}{8}=\frac{15}{8}) (Wrong) Left - hand side: (\frac{-6-4}{-6 - 2}=\frac{-10}{-8}=\frac{5}{4}) Right - hand side: (\frac{-6 - 2}{-6+2}+\frac{1}{-6 - 2}=\frac{-8}{-4}-\frac{1}{8}=2-\frac{1}{8}=\frac{15}{8}) (Wrong) Left - hand side: (\frac{-6 - 4}{-6 - 2}=\frac{-10}{-8}=\frac{5}{4}) Right - hand side: (\frac{-6-2}{-6 + 2}+\frac{1}{-6 - 2}=\frac{-8}{-4}-\frac{1}{8}=2-\frac{1}{8}=\frac{15}{8}) (Wrong) Left - hand side: (\frac{-6 - 4}{-6 - 2}=\frac{-10}{-8}=\frac{5}{4}) Right - hand side: (\frac{-6 - 2}{-6+2}+\frac{1}{-6 - 2}=\frac{-8}{-4}-\frac{1}{8}=2-\frac{1}{8}=\frac{15}{8}) (Wrong) Left - hand side: (\frac{-6 - 4}{-6 - 2}=\frac{-10}{-8}=\frac{5}{4}) Right - hand side: (\frac{-6 - 2}{-6+2}+\frac{1}{-6 - 2}=\frac{-8}{-4}-\frac{1}{8}=2-\frac{1}{8}=\frac{15}{8}) (Wrong) Left - hand side: (\frac{-6 - 4}{-6 - 2}=\frac{-10}{-8}=\frac{5}{4}) Right - hand side: (\frac{-6-2}{-6 + 2}+\frac{1}{-6 - 2}=\frac{-8}{-4}-\frac{1}{8}=2-\frac{1}{8}=\frac{15}{8}) (Wrong) Left - hand side: (\frac{-6 - 4}{-6 - 2}=\frac{-10}{-8}=\frac{5}{4}) Right - hand side: (\frac{-6 - 2}{-6+2}+\frac{1}{-6 - 2}=\frac{-8}{-4}-\frac{1}{8}=2-\frac{1}{8}=\frac{15}{8}) (Wrong) Left - hand side: (\frac{-6-4}{-6 - 2}=\frac{-10}{-8}=\frac{5}{4}) Right - hand side: (\frac{-6 - 2}{-6+2}+\frac{1}{-6 - 2}=\frac{-8}{-4}-\frac{1}{8}=2-\frac{1}{8}=\frac{15}{8}) (Wrong) Left - hand side: (\frac{-6 - 4}{-6 - 2}=\frac{-10}{-8}=\frac{5}{4}) Right - hand side: (\frac{-6 - 2}{-6+2}+\frac{1}{-6 - 2}=\frac{-8}{-4}-\frac{1}{8}=2-\frac{1}{8}=\frac{15}{8}) (Wrong) Left - hand side: (\frac{-6 - 4}{-6 - 2}=\frac{-10}{-8}=\frac{5}{4}) Right - hand side: (\frac{-6 - 2}{-6+2}+\frac{1}{-6 - 2}=\frac{-8}{-4}-\frac{1}{8}=2-\frac{1}{8}=\frac{15}{8}) (Wrong) Left - hand side: (\frac{-