# Solution to Practice Problem #3

There are multiple different ways to solve this problem, and this is just one of them. Can you find others? That is one way to learn and develop creativity!

\begin{array}{c} \eqalign{ & \frac{1}{x} + \frac{1}{y} = 1 \cr & \frac{3}{x} \,\, – \,\, \frac{2}{y} = 8 \cr & \cr} \end{array} \begin{array}{c} \eqalign{ & \,\,\,\, \left( {\frac{1}{x}} \right) + \,\,\,\, \left( {\frac{1}{y}} \right) = 1 \cr & 3\left( {\frac{1}{x}} \right) \, – \,\, 2\left( {\frac{1}{y}} \right) = 8 \cr & \cr} \end{array} \begin{array}{c} \eqalign{ & Let\,\,\,A = \frac{1}{x}\,\,\,\,\,\,\,\,B = \frac{1}{y} \cr & \cr} \end{array} \begin{array}{c} \eqalign{ & \,\,\,A + \,\,\,B = 1 \cr & 3A \,\, – \, 2B = 8 \cr & \cr} \end{array} \begin{array}{c} \eqalign{ & 2A + 2B = 2 \cr & 3A \,\, – \,\, 2B = 8 \cr & \cr} \end{array} \begin{array}{c} \eqalign{ & 5A + 0B = 10 \cr & A = \,2 \cr & B = \,- 1 \cr & \cr} \end{array} \begin{array}{c} \eqalign{ & A = {1 \over x}\,\,\,\, \to \,\,\,\,\,2 = {1 \over x}\,\,\,\, \to \,\,\,\,x = \,{1 \over 2} \cr & B = {1 \over y}\,\,\,\, \to \,\,\,\, – 1 = {1 \over y}\,\,\,\, \to \,\,\,\,y = \,\, – 1 \cr} \end{array}