Problem 32
Question
Solve the given equation. $$ \frac{4}{x(x-2)}=\frac{2}{x-2} $$
Step-by-Step Solution
Verified Answer
To solve the given equation \(\frac{4}{x(x-2)} = \frac{2}{x-2}\), we identify the common denominator as \(x(x-2)\) and multiply both sides to clear the fractions. This results in the equation \(4 = 2x\), which simplifies to \(x = 2\). However, upon checking for an extraneous solution, the denominator becomes zero when \(x = 2\), so there is no valid solution for the given equation.
1Step 1: Identify common denominator
The given equation is: \(\frac{4}{x(x-2)} = \frac{2}{x-2}\). Here, we can see that the common denominator is \(x(x-2)\).
2Step 2: Clearing the fractions
To clear the fractions, we will multiply both sides of the equation by the common denominator, which is \(x(x-2)\):
\((x(x-2)) \cdot \frac{4}{x(x-2)} = (x(x-2)) \cdot \frac{2}{x-2}\)
Now, we can cancel out the denominators:
\(4 = 2x\)
3Step 3: Solve for x
Now, we solve the equation for x:
\(4 = 2x\)
Divide both sides by 2:
\(x = 2\)
However, we have to check if our solution is valid, since we have a denominator in the given equation.
4Step 4: Check for extraneous solution
Check if the solution x = 2 makes any of the denominators become zero:
\(x(x-2) = 2(2-2) = 0\)
The denominator becomes zero when \(x = 2\). Therefore, our solution is extraneous and there is no valid solution for the given equation.
Key Concepts
Common DenominatorClearing FractionsExtraneous Solution
Common Denominator
When dealing with equations involving fractions, one of the primary steps is finding a common denominator. A common denominator is a shared multiple of the denominators of all the fractions in an equation.
This makes it easier to manipulate and solve the equation because it allows us to combine fractions or eliminate them entirely.
In the given equation \( \frac{4}{x(x-2)} = \frac{2}{x-2} \), we identify \( x(x-2) \) as the common denominator because it includes both individual denominators \( x(x-2) \) and \( x-2 \).
Recognizing this helps us set the stage for clearing fractions, which simplifies the equation into a more straightforward algebraic form.
Remember, finding the common denominator is all about creating a level playing field for each term, which is essential in equations that involve fractions.
This makes it easier to manipulate and solve the equation because it allows us to combine fractions or eliminate them entirely.
In the given equation \( \frac{4}{x(x-2)} = \frac{2}{x-2} \), we identify \( x(x-2) \) as the common denominator because it includes both individual denominators \( x(x-2) \) and \( x-2 \).
Recognizing this helps us set the stage for clearing fractions, which simplifies the equation into a more straightforward algebraic form.
Remember, finding the common denominator is all about creating a level playing field for each term, which is essential in equations that involve fractions.
Clearing Fractions
Clearing fractions from an equation is an effective way to simplify and solve it. This involves multiplying every term in the equation by the common denominator that we previously identified.
By doing this, we effectively "cancel out" the fractions because each term, when multiplied by the common denominator, converts into integer terms.
For instance, in our example, multiplying both sides by \( x(x-2) \) gets rid of the fractions entirely:
From here, solving for \( x \) becomes much simpler.
Clearing fractions eliminates the hassle of working within fractional formats and transforms the equation into a basic algebraic problem, making calculations straightforward and the equation easier to solve.
By doing this, we effectively "cancel out" the fractions because each term, when multiplied by the common denominator, converts into integer terms.
For instance, in our example, multiplying both sides by \( x(x-2) \) gets rid of the fractions entirely:
- On the left side, \( (x(x-2)) \times \frac{4}{x(x-2)} \) simplifies to 4.
- On the right side, \( (x(x-2)) \times \frac{2}{x-2} \) simplifies to \( 2x \).
From here, solving for \( x \) becomes much simpler.
Clearing fractions eliminates the hassle of working within fractional formats and transforms the equation into a basic algebraic problem, making calculations straightforward and the equation easier to solve.
Extraneous Solution
An extraneous solution is a solution to an equation that arises from the process of solving the equation but does not satisfy the original equation.
This can often occur in equations involving fractions, due to multiplying through by terms that assume non-zero values to avoid division by zero.
In our original example, solving \( 4 = 2x \) led us to \( x = 2 \).
However, when substituted back into the original equation, \( x = 2 \) makes the denominator \( x(x-2) \) equal to zero, thus invalidating the solution.
Division by zero is undefined, meaning that \( x = 2 \) cannot be an actual solution.
To avoid accepting extraneous solutions, always verify potential solutions by substituting them back into the original equation.
Ensuring that the solution doesn’t create a division by zero or violate any initial equation restrictions is crucial for confirming the validity of your result.
Checking for extraneous solutions is an essential final step in solving equations that initially include variables in their denominators.
This can often occur in equations involving fractions, due to multiplying through by terms that assume non-zero values to avoid division by zero.
In our original example, solving \( 4 = 2x \) led us to \( x = 2 \).
However, when substituted back into the original equation, \( x = 2 \) makes the denominator \( x(x-2) \) equal to zero, thus invalidating the solution.
Division by zero is undefined, meaning that \( x = 2 \) cannot be an actual solution.
To avoid accepting extraneous solutions, always verify potential solutions by substituting them back into the original equation.
Ensuring that the solution doesn’t create a division by zero or violate any initial equation restrictions is crucial for confirming the validity of your result.
Checking for extraneous solutions is an essential final step in solving equations that initially include variables in their denominators.
Other exercises in this chapter
Problem 32
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State the real number property that iustifies the statement $$ \frac{(2 x+1)(x+3)}{(2 x-1)(x+3)}=\frac{2 x+1}{2 x-1} $$
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