Question 81187
<pre>

Paul did it one way, but I think the most efficient way
is to clear of fractions:

{{{3/(4x)}}} + {{{2/5}}} = {{{1}}}

Write the 1 as 1/1

{{{3/(4x)}}} + {{{2/5}}} = {{{1/1}}}

Multiply every term by {{{(LCD)/1}}} which is {{{(20x)/1}}}

Write {{{(20x)/1}}}· in front of every term: 

{{{(20x)/1}}}·{{{3/(4x)}}} + {{{(20x)/1}}}·{{{2/5}}} = {{{(20x)/1}}}·{{{1/1}}}

In the first term cancel the 4x into the 20x and get 5.
In the second term, cancel the 5 into the 20 and get 4:

{{{5/1}}}·{{{3/1}}} + {{{(4x)/1}}}·{{{2/1}}} = {{{(20x)/1}}}·{{{1/1}}} 

or

{{{15/1}}} + {{{(8x)/1}}} = {{{(20x)/1}}}

 Erase the 1's on the bottoms:

{{{15}}} + {{{8x}}} = {{{20x}}}

Subtract {{{8x}}} from both sides, getting {{{15}}} on the left
and {{{12x}}} on the right

          {{{15}}} = {{{12x}}}

Divide both sides by {{{12}}}

           {{{15/12}}} = {{{(12x)/12}}}

Reduce fraction:

             {{{5/4}}} = {{{x}}}

Edwin</pre>