Question
What is the quotient? \( \frac{n+3}{2 n-6} \div \frac{n+3}{3 n-9} \)
Ask by Guzman Wood. in the United States
Mar 10,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
The quotient is \( \frac{3}{2} \).
Solution
We start with
\[
\frac{n+3}{2n-6} \div \frac{n+3}{3n-9}.
\]
1. Rewrite the division as multiplication by the reciprocal:
\[
\frac{n+3}{2n-6} \times \frac{3n-9}{n+3}.
\]
2. Factor the expressions where possible:
\[
2n-6 = 2(n-3) \quad \text{and} \quad 3n-9 = 3(n-3).
\]
This gives
\[
\frac{n+3}{2(n-3)} \times \frac{3(n-3)}{n+3}.
\]
3. Cancel the common factors \(n+3\) and \(n-3\) (provided \(n \neq -3\) and \(n \neq 3\)):
\[
\frac{\cancel{n+3}}{2\cancel{(n-3)}} \times \frac{3\cancel{(n-3)}}{\cancel{n+3}} = \frac{3}{2}.
\]
Thus, the quotient is
\[
\frac{3}{2}.
\]
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To find the quotient, we first rewrite the division of fractions as multiplication by the reciprocal. Thus, we have: \[ \frac{n+3}{2n-6} \div \frac{n+3}{3n-9} = \frac{n+3}{2n-6} \times \frac{3n-9}{n+3} \] Next, we can simplify this expression. Notice that \(n+3\) in the numerator of the second fraction will cancel with \(n+3\) in the first fraction: \[ = \frac{1}{2n-6} \times (3n-9) \] Now we can simplify \(3n-9\) as \(3(n-3)\) and \(2n-6\) as \(2(n-3)\). This gives us: \[ = \frac{3(n-3)}{2(n-3)} \] We can once again cancel \(n-3\) (assuming \(n \neq 3\)): \[ = \frac{3}{2} \] Thus, the final quotient is: \[ \frac{3}{2} \]
