How do you solve #x/(x-3)=2-2/(x-3)#?

1 Answer
Jul 6, 2017

Answer:

The solution is #x=8#.

Explanation:

In an equality you have that the left side of the #=# is the same as the right side of the #=#. So the trick is to continue to maintain them equal multiplying or dividing by the same quantities.

In this case

$$\frac{x}{x-3}=2-\frac{2}{x-3}$$

we start multiplying left and right for #x-3#

$$(x-3)\frac{x}{x-3}=(x-3)\left(2-\frac{2}{x-3}\right)$$

$$(x-3)\frac{x}{x-3}=2(x-3)-(x-3)\frac{2}{x-3}$$
we simplify the #x-3#
$$x=2(x-3)-2$$
$$x=2x-6-2$$
$$x=2x-8$$
now we add left and right $-2x$
$$x-2x=2x-2x-8$$
$$-x=-8$$
and finally we multiply for #-1# left and right
$$-1(-x)=-1(-8)$$
$$x=8.$$

In order to verify if our result is correct we substitute it in the initial equation and we must obtain an identity.
$$\frac{x}{x-3}=2-\frac{2}{x-3}$$
$$\frac{8}{8-3}=2-\frac{2}{8-3}$$
$$\frac{8}{5}=2-\frac{2}{5}$$
$$\frac{8}{5}=\frac{2\times 5-2}{5}$$
$$\frac{8}{5}=\frac{10-2}{5}$$
$$\frac{8}{5}=\frac{8}{5}$$
so we are sure that our solution is correct.