How do you solve $\frac{x - 3}{2 x - 4} = \frac{x}{x - 2} + 2$ $\frac { x - 3} { 2x - 4} = \frac { x } { x - 2} + 2$ ?

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Answer 1 · 2016-10-27T19:10:15

$x = 1$ $x=1$

Solving the rational equation is computed by finding same denominator for denominators in both sides of the equation

Then,
$if color(red)(a/c=b/c$ then $color(red)(a=c$

Then common denominator is the $L.C.M(2x-4,x-2)$
$color(blue)(2x-4=2(x-2)$
then $color(blue)(2x-4)$ is the common denominator

$rArr(x-3)/(2x-4)=(color(blue)2x)/(color(blue)2(x-2))+(2xxcolor(blue)(2(x-2)))/color(blue)(2(x-2))$

$rArr(x-3)/(2x-4)=(2x+4(x-2))/(2x-4)$

$rArr(x-3)/(2x-4)=(2x+4x-8)/(2x-4)$

$rArr(x-3)/(2x-4)=(6x-8)/(2x-4)$

Then
$color(red)(x-3=6x-8)$

$rArrx-6x=3-8$

$rArr-5x=-5$
$rArrx=(-1)/(-1)$

Therefore, $x=1$

How do you solve \frac { x - 3} { 2x - 4} = \frac { x } { x - 2} + 2x−32x−4=xx−2+2\frac { x - 3} { 2x - 4} = \frac { x } { x - 2} + 2?