Question
Find the simplified product. \( \frac{b-5}{2 b} \cdot \frac{b^{2}+3 b}{b-5} \) \( \frac{b+3}{2} \) \( \frac{2}{b+3} \) \( 2 b+6 \)
Ask by Lindsey Carrillo. in the United States
Mar 10,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
The simplified product is \( \frac{b+3}{2} \).
Solution
Given the expression
\[
\frac{b-5}{2b} \cdot \frac{b^2+3b}{b-5}
\]
we notice that the term \((b-5)\) appears in both the numerator and the denominator, so it cancels out (provided \(b \neq 5\)). The expression then becomes:
\[
\frac{1}{2b} \cdot (b^2+3b)
\]
We can factor the numerator:
\[
b^2+3b = b(b+3)
\]
Substituting, we have:
\[
\frac{1}{2b} \cdot b(b+3)
\]
Cancel the factor \(b\) (with the condition \(b \neq 0\)):
\[
\frac{b+3}{2}
\]
Thus, the simplified product is:
\[
\frac{b+3}{2}
\]
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Simplify this solution The Deep Dive
To simplify the product \( \frac{b-5}{2 b} \cdot \frac{b^{2}+3 b}{b-5} \), we can look for any common factors we can cancel. First, rewrite the product: \[ \frac{b-5}{2b} \cdot \frac{b(b + 3)}{b - 5} \] The \( b-5 \) in the numerator of the first fraction and the denominator of the second fraction cancel out: \[ \frac{1}{2b} \cdot b(b + 3) \] Now, we simplify: \[ \frac{1 \cdot b(b + 3)}{2b} = \frac{b^2 + 3b}{2b} \] Next, we can cancel \( b \) in the numerator and denominator (assuming \( b \neq 0 \)): \[ \frac{b + 3}{2} \] Thus, the simplified product is: \[ \frac{b + 3}{2} \] So the answer is \( \frac{b + 3}{2} \).
