# How do you multiply \frac { 7b + 35} { 7} \cdot \frac { b + 8} { b + 5}?

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#### Explanation

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#### Explanation:

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Jan 26, 2018

See a solution process below:

#### Explanation:

First, factor the numerator of the fraction on the left as:

$\frac{\left(7 \cdot b\right) + \left(7 \cdot 5\right)}{7} \cdot \frac{b + 8}{b + 5} \implies$

$\frac{7 \left(b + 5\right)}{7} \cdot \frac{b + 8}{b + 5}$

Next, cancel common terms in the numerators and denominators:

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{7}}} \textcolor{b l u e}{\cancel{\textcolor{b l a c k}{\left(b + 5\right)}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{7}}}} \cdot \frac{b + 8}{\textcolor{b l u e}{\cancel{\textcolor{b l a c k}{b + 5}}}} \implies$

$\frac{1}{1} \cdot \frac{b + 8}{1} \implies$

$b + 8$

However, because we cannot divide by $0$ we need to exclude where:

$b + 5 \ne 0$

Or

$b \ne - 5$

So the solution is: $b + 8$ where $b \ne - 5$

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