$
\eqalign{
& {3 \over {x + 3}} = {1 \over x} \cr
& 3x = x + 3 \cr
& 2x = 3 \cr
& x = 1{1 \over 2} \cr}
$ |
$
\eqalign{
& {{x - 1} \over 3} = {5 \over {x + 1}} \cr
& \left( {x - 1} \right)\left( {x + 1} \right) = 15 \cr
& x^2 - 1 = 15 \cr
& x^2 = 16 \cr
& x = - 4 \vee x = 4 \cr}
$ |
$
\eqalign{
& x + {3 \over {x - 1}} = 5 \cr
& {3 \over {x - 1}} = - x + 5 \cr
& (x - 1)( - x + 5) = 3 \cr
& - x^2 + 5x + x - 5 = 3 \cr
& - x^2 + 6x - 8 = 0 \cr
& x^2 - 6x + 8 = 0 \cr
& (x - 2)(x - 4) = 0 \cr
& x = 2 \vee x = 4 \cr}
$ |
