How do you solve $(x - 1) ( x + 3+ \frac { 3} { x - 1} ) = ( \frac { 4- x } { x - 1} ) ( x - 1)$ ?

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Answer 1 · 2017-04-11T15:45:00

$x_1=-4$ ; $cancel(color(red)(x_2=1))$

You would multiply and simplify $x-1$ and state that $color(red)(x!=1)$ :

$x^2+3xcancel(-x)cancel(-3)cancel(+3)=4cancel(-x)$

that's

$x^2+3x-4=0$

$x=(-3+-sqrt(9+16))/2$

$x=(-3+-5)/2$

and the solutions are:

$x_1=-4$ ; $cancel(color(red)(x_2=1))$

How do you solve (x - 1) ( x + 3+ \frac { 3} { x - 1} ) = ( \frac { 4- x } { x - 1} ) ( x - 1)(x - 1) ( x + 3+ \frac { 3} { x - 1} ) = ( \frac { 4- x } { x - 1} ) ( x - 1)?