Question 161326
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I will use {{{NUM}}} to stand for "numerator" and {{{DEN}}} to
stand for "denominator", so for instance, {{{NUM1}}} stands for 
the first numerator, {{{DEN1}}} stands for the first denominator,
{{{NUM2}}} stands for the second numerator, etc.

Suppose your problem is of this type:

{{{((NUM1/DEN1)+(NUM2/DEN2))/((NUM3/DEN3)+(NUM4/DEN4))}}}

Get LCD of all denominators DEN1, DEN2, DEN3, and DEN4.

Form the fraction {{{LCD/1}}} and put it in parentheses
{{{(LCD/1)}}}

Multiply all four fractions by {{{(LCD/1)}}}

{{{((LCD/1)(NUM1/DEN1)+(LCD/1)(NUM2/DEN2))/((LCD/1)(NUM3/DEN3)+(LCD/1)(NUM4/DEN4))}}}

Now every one of those denominators will cancel 
entirely into the LCD, and the end result will be
much simpler, like this:

{{{(TERM1 + TERM2)/(TERM3 + TERM4)}}}

The smae principle works well regardless of how many fractions
you have on top or bottom.
--------------------------------------

1+2 over x divided by 7 over x

{{{(1+2/x)/  (7/x)  }}}

Write the {{{1}}} as {{{1/1}}} so that
everything will be a fraction, and put parentheses around
every fraction:

{{{((1/1)+(2/x))/  ((7/x))  }}}

Get LCD of all denominators 1, x, and x.

Then LCD = x 

Form the fraction {{{x/1}}} and put it in parentheses
{{{(x/1)}}}

Multiply all three fractions by {{{(x/1)}}}

{{{((x/1)(1/1)+(x/1)(2/x))/  ((x/1)(7/x))  }}}

Now every one of those denominators will cancel 
entirely into the LCD, and the end result will be
much simpler, like this:

{{{((x/1)(1/1)+(cross(x)/1)(2/cross(x)))/  ((cross(x)/1)(7/cross(x)))  }}}

and that is simply:

{{{(x+2)/7}}}

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

a over b divided by a- 5a over b 

{{{(a/b)/(a-(5a)/b)}}}

Write the {{{a}}} as {{{a/1}}} so everything will
be a fraction and put parentheses around everything:

{{{((a/b))/((a/1)-((5a)/b))}}}

Get LCD of all denominators {{{b}}}, {{{1}}}, and {{{b}}}.

Form the fraction {{{b/1}}} and put it in parentheses
{{{(b/1)}}}

Multiply all four fractions by {{{(b/1)}}}

{{{((b/1)(a/b))/((b/1)(a/1)- (b/1)(5a/b) )}}}

Now every denominator will cancel 
entirely into the LCD, and the end result will be
much simpler, like this:

{{{((cross(b)/1)(a/cross(b)))/((b/1)(a/1)- (cross(b)/1)(5a/cross(b)) )}}}

which is simply

{{{a/(ba-5a)}}}

Now we can factor out {{{a}}} on the bottom:

{{{a/(a(b-5))}}}

Now the {{{a}}}'s will cancel, leaving a {{{1}}} on top:

   {{{1}}}
{{{cross(a)/(cross(a)(b-5))}}}

Answer:  {{{1/(b-5)}}}

If I have time I'll come back and do the others.
But they're all done the same way.

Edwin</pre>