Let's simplify the expression step by step. The expression given is:
\[ \frac{8y + 8}{y - 2} + \frac{7y + 9}{y - 2} \]
Since both fractions have the same denominator, we can combine them:
\[ \frac{(8y + 8) + (7y + 9)}{y - 2} \]
Now, simplify the numerator:
\[ 8y + 8 + 7y + 9 = 15y + 17 \]
So, the expression becomes:
\[ \frac{15y + 17}{y - 2} \]
Next, we need to state any restrictions on the variable \( y \). The only restriction comes from the denominator, which cannot be equal to zero:
\[ y - 2 \neq 0 \implies y \neq 2 \]
Thus, our simplified expression is:
\[ \frac{15y + 17}{y - 2} \quad \text{for all } y \neq 2. \]
Now regarding the second part of the question, we see that you might have meant to ask about products or division, but the given equation had a + sign, not a division.
Assuming you want to simplify the following expression as well:
\[ \frac{\frac{8y + 8}{y - 2}}{\frac{7y + 9}{y - 2}} \]
This can also be simplified. When you divide two fractions, you multiply by the reciprocal:
\[ \frac{8y + 8}{y - 2} \div \frac{7y + 9}{y - 2} = \frac{8y + 8}{y - 2} \cdot \frac{y - 2}{7y + 9} \]
The \(y - 2\) in the numerator and denominator cancel out (as long it does not equal zero):
\[ \frac{8y + 8}{7y + 9} \]
Now we set restrictions for this case again. The restriction remains \(y \neq 2\). Additionally, we should check the denominator \(7y + 9\):
\[ 7y + 9 \neq 0 \implies y \neq -\frac{9}{7} \]
Thus, the final answer for the second expression is:
\[ \frac{8y + 8}{7y + 9} \quad \text{for } y \neq 2, -\frac{9}{7} \]
To summarize:
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\(\frac{8y + 8}{y - 2} + \frac{7y + 9}{y - 2} = \frac{15y + 17}{y - 2} \text{ for all } y \neq 2\)
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\(\frac{\frac{8y+8}{y-2}}{\frac{7y+9}{y-2}} = \frac{8y+8}{7y+9} \text{ for } y \neq 2, -\frac{9}{7}\)