Question 83088
{{{ (7x)/(3y^2) + (4y)/(6x^2) }}}


In order to add two fractions, you must first find the Least Common Denominator (LCD).  In this case, the LCD = {{{6x^2y^2}}}


In order to get the LCD in the first fraction, you must multiply the numerator and denominator of this first fraction by all the missing factors, which would be {{{2x^2}}}.  In the second fraction, the missing factors are only {{{y^2}}}.


It should look like this:

{{{ (7x)/(3y^2) + (4y)/(6x^2) }}}
{{{ ((7x)/(3y^2))*((2x^2)/(2x^2)) + ((4y)/(6x^2))*((y^2)/(y^2)) }}}

{{{(14x^3)/(6x^2y^2)+(4y^3)/(6x^2y^2) }}}

{{{(14x^3 +4y^3)/(6x^2y^2) }}}


I just noticed that we could have reduced the second fraction FIRST, which would have made this easier.  However, you can reduce the answer now by factoring the numerator:

{{{(2(7x^3 +2y^3))/(6x^2y^2) }}}


Divide out the factors of 2:
{{{(7x^3 +2y^3)/(3x^2y^2) }}}

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Here is what it looks like if you reduce the fraction FIRST:
{{{ (7x)/(3y^2) + (4y)/(6x^2) }}}
{{{ (7x)/(3y^2) + (2y)/(3x^2) }}}


Now the LCD= {{{3x^2y^2}}}
{{{ ((7x)/(3y^2))*((x^2)/(x^2)) + ((2y)/(3x^2))*((y^2)/(y^2))  }}}
{{{ (7x^3)/(3x^2y^2) + (2y^3)/(3x^2y^2)}}}
{{{(7x^3 +2y^3)/(3x^2y^2) }}}


I don't know!!  It's very late in Central Florida, USA, and I may have made a errors on this problem.  If you find any, please let me know!!  If you need more help with LCD problems, click on my tutor name "rapaljer", go to "MATH IN LIVING COLOR" and look for Basic Algebra, Chapter 3.  


R^2 at SCC