The simplified product of [tex]\frac{b- 5}{2b} X \frac{b^{2} + 3b }{b - 5}[/tex] is [tex]\frac{(b+3)}{2}[/tex]
The products can be simplified as follows;
Multiplying fraction,
[tex]\frac{x}{y} . \frac{a}{b} = \frac{xa}{yb}[/tex]
Therefore,
[tex]\frac{b- 5}{2b} X \frac{b^{2} + 3b }{b - 5} = \frac{(b-5)(b^{2} + 3b )}{(2b)(b-5)}[/tex]
Hence,
Let's reduce the fraction by dividing
[tex]\frac{b- 5}{2b} X \frac{b^{2} + 3b }{b - 5} = \frac{(b-5)(b^{2} + 3b )}{(2b)(b-5)} = \frac{(b^{2}+3b )}{(2b)}[/tex]
Therefore,
[tex]\frac{b- 5}{2b} X \frac{b^{2} + 3b }{b - 5} = \frac{(b-5)(b^{2} + 3b )}{(2b)(b-5)} = \frac{(b^{2}+3b )}{(2b)} = \frac{(b(b+3))}{2b}[/tex]
Hence,
[tex]\frac{b- 5}{2b} X \frac{b^{2} + 3b }{b - 5} = \frac{(b-5)(b^{2} + 3b )}{(2b)(b-5)} = \frac{(b^{2}+3b )}{(2b)} = \frac{(b(b+3))}{2b} = \frac{b+3}{2}[/tex]
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