Question
If a and b are different positive primes such that$(\text{a}+\text{b})^{-1}(\text{a}^{-1}+\text{b}^{-1})=\text{a}^\text{x}\text{b}^\text{y},$ find $\text{x}+\text{y}+2.$
✓
Answer
$(\text{a}+\text{b})^{-1}(\text{a}^{-1}+\text{b}^{-1})=\text{a}^\text{x}\text{b}^\text{y}$$\Rightarrow\frac{1}{\text{a}+\text{b}}\times\Big(\frac{1}{\text{a}}+\frac{1}{\text{b}}\Big)=\text{a}^\text{x}\text{b}^\text{y}$
$\Rightarrow\frac{1}{\text{a}+\text{b}}\times\frac{\text{a}+\text{b}}{\text{ab}}=\text{a}^\text{x}\text{b}^\text{x}$
$\Rightarrow\frac{1}{\text{ab}}=\text{a}^\text{x}\text{b}^\text{y}$
$\Rightarrow(\text{ab})^{-1}=\text{a}^\text{x}\text{b}^\text{y}$
$\Rightarrow\text{a}^{-1}\times\text{b}^{-1}=\text{a}^\text{x}\times\text{b}^\text{y}$
$\Rightarrow\text{x}=-1$ and $\text{y}=-1$
$\therefore\text{x}+\text{y}+2=-1-1+2=0$
$\Rightarrow\frac{1}{\text{a}+\text{b}}\times\frac{\text{a}+\text{b}}{\text{ab}}=\text{a}^\text{x}\text{b}^\text{x}$
$\Rightarrow\frac{1}{\text{ab}}=\text{a}^\text{x}\text{b}^\text{y}$
$\Rightarrow(\text{ab})^{-1}=\text{a}^\text{x}\text{b}^\text{y}$
$\Rightarrow\text{a}^{-1}\times\text{b}^{-1}=\text{a}^\text{x}\times\text{b}^\text{y}$
$\Rightarrow\text{x}=-1$ and $\text{y}=-1$
$\therefore\text{x}+\text{y}+2=-1-1+2=0$
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