Question

solving a rational equation
solve and check: \\(\frac{c - 4}{c - 2} = \frac{c - 2}{c + 2} - \frac{1}{2 - c}\\)
the solution is \\(c = \\)
how many extraneous solutions are there?

Explanation:

Step1: Simplify the Denominator

Notice that \( \frac{1}{2 - c}=-\frac{1}{c - 2} \), so the equation becomes \( \frac{c - 4}{c - 2}=\frac{c - 2}{c + 2}+\frac{1}{c - 2} \).

Step2: Eliminate Denominators

Multiply both sides by \( (c - 2)(c + 2) \) (the least common denominator) to get:
\( (c - 4)(c + 2)=(c - 2)^2+(c + 2) \)

Step3: Expand Each Term

Expand the left side: \( c^2+2c - 4c - 8=c^2 - 2c - 8 \)
Expand the right side: \( c^2-4c + 4+c + 2=c^2-3c + 6 \)

Step4: Solve the Quadratic Equation

Set the left side equal to the right side:
\( c^2 - 2c - 8=c^2-3c + 6 \)
Subtract \( c^2 \) from both sides: \( -2c - 8=-3c + 6 \)
Add \( 3c \) to both sides: \( c - 8 = 6 \)
Add 8 to both sides: \( c = 14 \)

Step5: Check for Extraneous Solutions

Check if \( c = 14 \) makes the original denominators zero. \( c-2=12 eq0 \), \( c + 2 = 16 eq0 \), so it's valid.
Check if there are other potential extraneous solutions from the denominators \( c - 2 = 0\) ( \( c = 2 \)) and \( c + 2 = 0\) ( \( c=-2 \) ). When we solved, we got \( c = 14 \), so the extraneous solutions would be the ones that make denominators zero but aren't solutions, so there are 2 potential extraneous candidates ( \( c = 2 \) and \( c=-2 \) ), but our solution \( c = 14 \) is valid, so the number of extraneous solutions is 0? Wait, no, wait. Wait, when we multiplied by \( (c - 2)(c + 2) \), we assumed \( c eq2 \) and \( c eq - 2 \). Our solution is \( c = 14 \), which is not 2 or -2, so the extraneous solutions are the ones that would make denominators zero but are not solutions. Since our solution is valid, and the only values that make denominators zero are \( c = 2 \) and \( c=-2 \), which were not solutions, so the number of extraneous solutions is 0? Wait, no, let's re - check.

Wait, when we solved the equation, we found \( c = 14 \). Now, let's check the original equation with \( c = 14 \):

Left side: \( \frac{14 - 4}{14 - 2}=\frac{10}{12}=\frac{5}{6} \)

Right side: \( \frac{14 - 2}{14 + 2}-\frac{1}{2 - 14}=\frac{12}{16}+\frac{1}{12}=\frac{3}{4}+\frac{1}{12}=\frac{9 + 1}{12}=\frac{10}{12}=\frac{5}{6} \). So it works.

Now, the values that make the denominator zero are \( c = 2 \) (makes \( c - 2 = 0 \) and \( 2 - c = 0 \)) and \( c=-2 \) (makes \( c + 2 = 0 \)). When we plug \( c = 2 \) into the original equation, the left side is undefined (division by zero), and the right side is also undefined (division by zero). Similarly, \( c=-2 \) makes the right - hand side's first term undefined. So these are not solutions, so the number of extraneous solutions (solutions that we get from the algebraic process but are not valid in the original equation) is 0? Wait, no, actually, when we solve rational equations, extraneous solutions are solutions that come from the solved equation but make the original denominators zero. In our case, the solutions we got from the equation was \( c = 14 \), which is valid. The values \( c = 2 \) and \( c=-2 \) are not solutions of the equation (they make it undefined), so the number of extraneous solutions (solutions that were found but are invalid) is 0? Wait, maybe I made a mistake earlier. Let's re - do the extraneous solution count.

The potential extraneous solutions are the ones that make the denominators zero, i.e., \( c = 2 \) and \( c=-2 \). When we solved the equation, we got \( c = 14 \), which is not equal to 2 or - 2, so there are no extraneous solutions (because extraneous solutions are solutions that are obtained but are invalid, but here we only got one valid solution and the other values that make denominators zero are not solutions of the equation). So the number of extraneous solutions is 0.

Wait, but let's check the equation again. Wait, the original equation:

\( \frac{c - 4}{c - 2}=\frac{c - 2}{c + 2}-\frac{1}{2 - c} \)

We rewrote \( -\frac{1}{2 - c}=\frac{1}{c - 2} \), so the equation is \( \frac{c - 4}{c - 2}=\frac{c - 2}{c + 2}+\frac{1}{c - 2} \)

Subtract \( \frac{1}{c - 2} \) from both sides: \( \frac{c - 4-1}{c - 2}=\frac{c - 2}{c + 2} \)

\( \frac{c - 5}{c - 2}=\frac{c - 2}{c + 2} \)

Cross - multiply: \( (c - 5)(c + 2)=(c - 2)^2 \)

\( c^2+2c-5c - 10=c^2-4c + 4 \)

\( c^2-3c - 10=c^2-4c + 4 \)

\( -3c - 10=-4c + 4 \)

\( c = 14 \). Oh! I made a mistake in step 2 earlier. Let's correct that.

So step 2 correction:

Original equation: \( \frac{c - 4}{c - 2}=\frac{c - 2}{c + 2}-\frac{1}{2 - c} \)

Since \( \frac{1}{2 - c}=-\frac{1}{c - 2} \), so \( -\frac{1}{2 - c}=\frac{1}{c - 2} \)

So the equation becomes \( \frac{c - 4}{c - 2}=\frac{c - 2}{c + 2}+\frac{1}{c - 2} \)

Subtract \( \frac{1}{c - 2} \) from both sides: \( \frac{c - 4-1}{c - 2}=\frac{c - 2}{c + 2} \)

\( \frac{c - 5}{c - 2}=\frac{c - 2}{c + 2} \)

Cross - multiply: \( (c - 5)(c + 2)=(c - 2)^2 \)

Expand left: \( c^2+2c-5c - 10=c^2-3c - 10 \)

Expand right: \( c^2-4c + 4 \)

Set equal: \( c^2-3c - 10=c^2-4c + 4 \)

Subtract \( c^2 \): \( -3c - 10=-4c + 4 \)

Add \( 4c \): \( c - 10 = 4 \)

Add 10: \( c = 14 \). Okay, that's correct. Then check \( c = 14 \):

Left: \( (14 - 4)/(14 - 2)=10/12 = 5/6 \)

Right: \( (14 - 2)/(14 + 2)-1/(2 - 14)=12/16+1/12 = 3/4+1/12 = 9/12+1/12 = 10/12 = 5/6 \). Correct.

Now, the denominators are zero when \( c = 2 \) (makes \( c - 2 = 0 \) and \( 2 - c = 0 \)) and \( c=-2 \) (makes \( c + 2 = 0 \)). These are the values that would be extraneous if they were solutions, but our solution is \( c = 14 \), so the number of extraneous solutions is 0? Wait, no. Wait, extraneous solutions are solutions that are obtained from the solved equation but make the original denominators zero. In our case, the solved equation gave \( c = 14 \), which does not make any denominator zero. The values \( c = 2 \) and \( c=-2 \) are not solutions of the equation (they make it undefined), so the number of extraneous solutions (solutions that are invalid) is 0? Wait, maybe the question is asking how many extraneous solutions are there, i.e., how many solutions did we get that are extraneous. Since we got one solution \( c = 14 \) which is valid, and the other potential solutions (from denominators) are not solutions, so the number of extraneous solutions is 0? Wait, no, let's think again. When we solve a rational equation, we can get solutions that when plugged back in, make the denominator zero. In this case, our solution \( c = 14 \) does not make any denominator zero. The values that make denominators zero ( \( c = 2 \) and \( c=-2 \)) are not solutions of the equation (because when we plug them in, the equation is undefined, not false). So the number of extraneous solutions (solutions that are obtained but are invalid) is 0.

Answer:

The solution is \( c=\boxed{14} \), and the number of extraneous solutions is \( \boxed{0} \)