Question 1094535
<pre>

{{{(x+1)/(3x-6) = (5x)/6 + 1/(x-2)}}} 

Factor the first denominator by factoring out 3

{{{(x+1)/(3(x-2)) = (5x)/6 + 1/(x-2)}}}

The LCD is 6(x-2) so we multiply every term
through by {{{red(6(x-2)/1)}}}

{{{red(6(x-2)/1)((x+1)/(3(x-2))) = red(6(x-2)/1)((5x)/6) + red(6(x-2)/1)(1/(x-2))}}}

We cancel whatever will cancel:

{{{red(""^2cross(6)(cross(x-2))/1)((x+1)/(cross(3)(cross(x-2)))) = red(cross(6)(x-2)/1)((5x)/cross(6)) + red(6(cross(x-2))/1)(1/(cross(x-2)))}}}

What we have left is:

{{{2(x+1)=(x-2)(5x)+6}}}

{{{2x+2=5x(x-2)+6}}}
{{{2x+2=5x^2-10x+6}}}

Get 0 on the right:

{{{-5x^2+8x-4=0}}}

Multiply through by -1 so x<sup>2</sup> term will be positive:

{{{5x^2-8x+4=0}}}

Factor:

{{{(5x-2)(x-2)=0}}}

5x-2 = 0;   x-2 = 0
  5x = 2      x = 2
   x = {{{2/5}}}

Two answers are {{{2/5}}} and {{{2}}}.

However, the original problem has denominators that contain 
variables.  Therefore we must check for extraneous answers 
that are not solutions because they cause the denominator to 
equal 0.  No denominator can ever equal to 0.  

The denominators in the original equation are 3x-6 and x-2

Substituting {{{2/5}}} for x in denominator 3x-6 is

{{{3(2/5)-6}}}{{{""=""}}}{{{6/5-6}}}{{{""=""}}}{{{6/5-30/5}}}{{{""=""}}}{{{-24/5}}} which is not 0

Substituting {{{2/5}}} for x in denominator x-2 is

{{{2/5-2)}}}{{{""=""}}}{{{2/5-10/5}}}{{{""=""}}}{{{-8/5}}} which is not 0 
either, so {{{2/5}}} is a solution..

Substituting {{{2}}} for x in denominator 3x-6 is

{{{3(2)-6}}}{{{""=""}}}{{{6-6}}}{{{""=""}}}{{{0}}} so we must

discard the answer x = 2, for it is extraneous and is not a
solution.   Therefore there is only one solution, x = {{{2/5}}}

----------------------------------------

The second one:

{{{(3x)/(x+1) - 5/(2x) = 3/(2x)}}}

Add {{{""+5/(2x)}}} to both sides:

{{{(3x)/(x+1) = 3/(2x) + 5/(2x)}}}

Since both terms on the right have the same denominator
we can just add the numerators and place it over their
common denominator:

{{{(3x)/(x+1) = 8/(2x)}}}

Simplify the right side by dividing top and bottom by 4:

{{{(3x)/(x+1) = 4/x}}}

We could get an LCD as we did in the other one but since 
there is only a fraction on each side, we can just 
cross-multiply:

{{{3x(x) = 4(x+1)}}}

{{{3x^2 = 4x+4}}}

Get 0 on the right:

{{{3x^2-4x-4=0}}}

{{{(3x+2)(x-2)=0}}}

3x+2 = 0;    x-2 = 0
  3x = -2;     x = 2
   x = {{{-2/3}}}

Two answers are {{{-2/3}}} and {{{2}}}.

However, the original problem also has denominators that contain 
variables.  Therefore we must check for extraneous answers 
that are not solutions because they cause the denominator to 
equal 0.  No denominator can ever equal to 0.  

The denominators in the original equation are x-1 and 2x.

Substituting {{{-2/3}}} for x in denominator x-1 is

{{{-2/3-1}}}{{{""=""}}}{{{-2/3-3/3}}}{{{""=""}}}{{{-5/3}}}
which is not 0

Substituting {{{-2/3}}} for x in denominator 2x is

{{{2(-2/3)}}}{{{""=""}}}{{{-4/3}}} which is not 0 either, so 
{{{-2/3}}} is a solution.

Substituting {{{2}}} for x in denominator x-1 is

{{{2-1}}}{{{""=""}}}{{{1}}}which is not 0

Substituting {{{2}}} for x in denominator 2x is

{{{2(2)}}}{{{""=""}}}{{{4}}} which is not 0 either, so 
{{{2}}} is also a solution.

So there are two solutions  {{{-2/3}}} and 2. 

But as you see we must make sure that any answer we get
really is a solution by substituting into each denominator
to show that no answer causes any denominator in the original
equation to be 0.

Edwin</pre>