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? $\square$
Answer (2)
Rewrite the equation to have a common denominator.
Combine the terms and cross-multiply.
Simplify the equation and solve for c .
Check for extraneous solutions by plugging the solution back into the original equation. The solution is c = 14 and there are 0 extraneous solutions.
Explanation
Understanding the Problem We are given the equation c − 2 c − 4 = c + 2 c − 2 − 2 − c 1 and we need to solve for c and determine the number of extraneous solutions. Extraneous solutions are solutions that satisfy the transformed equation but not the original equation. They usually arise when we perform operations that are not reversible, such as squaring both sides of an equation or multiplying both sides by an expression that could be zero.
Rewriting the Equation First, we rewrite the equation to eliminate the negative sign in the denominator of the last term: c − 2 c − 4 = c + 2 c − 2 + c − 2 1
Combining Terms Next, we combine the terms on the right side of the equation. To do this, we need to find a common denominator, which is ( c + 2 ) ( c − 2 ) . However, we can also move the c − 2 1 term to the left side to combine with c − 2 c − 4 : c − 2 c − 4 − c − 2 1 = c + 2 c − 2 c − 2 c − 4 − 1 = c + 2 c − 2 c − 2 c − 5 = c + 2 c − 2
Cross-Multiplying Now, we cross-multiply to get rid of the fractions: ( c − 5 ) ( c + 2 ) = ( c − 2 ) ( c − 2 ) c 2 + 2 c − 5 c − 10 = c 2 − 2 c − 2 c + 4 c 2 − 3 c − 10 = c 2 − 4 c + 4
Simplifying the Equation Next, we simplify the equation by subtracting c 2 from both sides: − 3 c − 10 = − 4 c + 4 Add 4 c to both sides: c − 10 = 4 Add 10 to both sides: c = 14
Checking for Extraneous Solutions Now we need to check for extraneous solutions. We plug c = 14 into the original equation: 14 − 2 14 − 4 = 14 + 2 14 − 2 − 2 − 14 1 12 10 = 16 12 − − 12 1 6 5 = 4 3 + 12 1 6 5 = 12 9 + 12 1 6 5 = 12 10 6 5 = 6 5 Since the equation holds true, c = 14 is a valid solution.
Counting Extraneous Solutions Finally, we count the number of extraneous solutions. Since c = 14 is a valid solution, there are no extraneous solutions.
Examples
When solving equations involving rational expressions, it's crucial to identify and exclude values that make the denominator zero. These values are called extraneous solutions because they satisfy the transformed equation but not the original one. For instance, imagine you're designing a bridge and calculating stress distribution using rational functions. Extraneous solutions could lead to incorrect stress values, potentially causing structural failure if not properly accounted for in the design phase. Therefore, always check your solutions against the original equation to ensure they are valid and don't lead to undefined expressions.
The solution to the equation is c = 14 and there are no extraneous solutions. After checking, c = 14 satisfies the original equation without leading to any undefined terms. Therefore, everything checks out perfectly with zero extraneous results.
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