$$ q+2 p=2 $$ and $$ q^{2}+2 p^{2}=3 q p $$ 2.2 Simplify fully. $$ \frac{8^{n} \cdot 6^{n-3} \cdot 9^{1-n} \cdot 2^{0}}{16^{n-1} \cdot 3^{-n}} $$ 2.3 Solve for $$ r $$. $$ 3^{r}+3^{3-r}=28 $$

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Flynn Byrd · Accepted Answer

### 2.1 Simplify fully. The solutions are $$ (q, p) = (1, \frac{1}{2}) $$ and $$ (q, p) = (\frac{2}{3}, \frac{2}{3}) $$. ### 2.2 Simplify fully. The simplified expression is $$ 8 \cdot 3^{2-n} $$. ### 2.3 Solve for $$ r $$. The solution for $$ r $$ is $$ r = 2 $$.

\( q+2 p=2 \) and \( q^{2}+2 p^{2}=3 q p \) 2.2 Simplify fully. \( \frac{8^{n} \cdot 6^{n-3} \cdot 9^{1-n} \cdot 2^{0}}{16^{n-1} \cdot 3^{-n}} \) 2.3 Solve for \( r \). \( 3^{r}+3^{3-r}=28 \)

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\( q+2 p=2 \) and \( q^{2}+2 p^{2}=3 q p \) 2.2 Simplify fully. \( \frac{8^{n} \cdot 6^{n-3} \cdot 9^{1-n} \cdot 2^{0}}{16^{n-1} \cdot 3^{-n}} \) 2.3 Solve for \( r \). \( 3^{r}+3^{3-r}=28 \)

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