Question 143734
<pre><font size = 3 color = "indigo"><b>      
{{{2+3/(1+2/(1-x))}}}

First do only the part boxed in

{{{2+3/(highlight(1+2/(1-x)))}}}

{{{1+2/(1-x)}}}

Write {{{1}}} as {{{1/1}}}

{{{1/1+2/(1-x)}}}

Get the LCD of {{{1-x}}}.  Multiply top and
bottom of {{{1/1}}} by {{{(1-x))}}}

{{{(1(1-x))/(1(1-x))+2/(1-x)}}}

{{{(1-x)/(1-x) + 2/(1-x) }}}

Combine the numerators over the LCD:

{{{(1-x+2)/(1-x)}}}

{{{(3-x)/(1-x)}}}

Now go back to 
{{{2+3/(highlight(1+2/(1-x)))}}}

and replace the boxed-in part by {{{(3-x)/(1-x)}}}

{{{2+3/highlight( (3-x)/(1-x)) }}}

Erasing the box:

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

Now do only the part boxed in here:

{{{2+highlight(3/( (3-x)/(1-x))) }}}

{{{3/( (3-x)/(1-x)) }}}

This can be done by inverting and multiplying:

{{{3* ((1-x)/(3-x)) }}}

Write {{{3}}} as {{{(3/1)}}}:

{{{(3/1)* ((1-x)/(3-x)) }}}

Multiply numerators and denominators:

{{{(3-3x)/(3-x) }}}

Now go back to

{{{2+highlight(3/( (3-x)/(1-x))) }}}

and replace the boxed-in part by {{{(3-3x)/(3-x)}}}

{{{2+highlight( (3-3x)/(3-x)  ) }}}

Erasing the box:

{{{2+ (3-3x)/(3-x)   }}}

Write {{{2}}} as {{{2/1}}}

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

Get the LCD of {{{3-x}}}.  Multiply top and
bottom of {{{2/1}}} by {{{(3-x)}}}

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

{{{(6-2x)/(3-x) + (3-3x)/(3-x) }}}

Combine the numerators over the LCD:

{{{(6-2x+3-3x)/(3-x)}}}

{{{(9-5x)/(3-x)}}}

Edwin</pre>