To subtract the expression \( \frac{x}{3x+7} - \frac{x}{9x+21} \), we will first find a common denominator and then perform the subtraction.
Step 1: Identify the denominators
The denominators are:
- \( 3x + 7 \)
- \( 9x + 21 \)
We can factor \( 9x + 21 \) as: \[ 9x + 21 = 9(x + \frac{7}{3}) = 3(3x + 7) \] However, the common denominator for the two fractions is the product of the distinct linear factors: \[ (3x + 7)(9x + 21) = (3x + 7)(3(3x + 7)) = 3(3x + 7)^2 \]
Step 2: Rewrite each term with the common denominator
Now, rewrite each fraction:
- \( \frac{x}{3x+7} \) must be multiplied by \( \frac{9x+21}{9x+21} \)
- \( \frac{x}{9x+21} \) must be multiplied by \( \frac{3x+7}{3x+7} \)
Step 3: Rewrite the expressions
The first term becomes: \[ \frac{x(9x + 21)}{(3x + 7)(9x + 21)} = \frac{9x^2 + 21x}{(3x + 7)(9x + 21)} \] The second term becomes: \[ \frac{x(3x + 7)}{(3x + 7)(9x + 21)} = \frac{3x^2 + 7x}{(3x + 7)(9x + 21)} \]
Step 4: Subtract the two fractions
Now we can subtract them: \[ \frac{9x^2 + 21x - (3x^2 + 7x)}{(3x + 7)(9x + 21)} = \frac{(9x^2 - 3x^2) + (21x - 7x)}{(3x + 7)(9x + 21)} = \frac{6x^2 + 14x}{(3x + 7)(9x + 21)} \]
Step 5: Factor the numerator
The numerator can be factored: \[ 6x^2 + 14x = 2x(3x + 7) \] Putting this back into our expression gives: \[ \frac{2x(3x + 7)}{(3x + 7)(9x + 21)} \]
Step 6: Cancel common factors
Since \( 3x + 7 \) appears in both the numerator and the denominator, we can cancel it (assuming \( 3x + 7 \neq 0 \)): \[ \frac{2x}{9x + 21} \]
Final Result
The result is: \[ \frac{2x}{9x + 21} \]
So from the given options, the answer is: \[ \boxed{\frac{2x}{9x + 21}} \]
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