Question 166719
Here is a quick lesson for you.  First, think about fractions you can check for accuracy.  Let's look at adding 2/3 and 3/4. 
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We know that we can not add these two numbers as they are because they do not have the same denominator (the bottom number).  
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So, we need to find a way to get the denominators to be the same.  This is done by finding something called the least common multiple or LCM.  
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The LCM will always contain the denominator of each term.
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The LCM of 2/3 and 3/4 is 12 so you multiply 2/3 by 4/4 and you get 8/12.
multiply 3/4 by 3/3 and you get 9/12.  Now you can add 8/12 and 9/12 together to get 17/12.
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For your problem: the LCM is (k-3)*(k+3)
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Whenever both fractions have the LCM as the denominator then you will be able to add them together.
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In order to make the first term have (k-3)(k+3) you need to multiply the top and bottom (numerator and denominator) by (k+3).  This is equivalent to multiplying the term by 1 because anything divided by itself is 1.
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So you have {{{(2(k+3))/((k-3)(k+3))}}}
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You need to do the same thing with the second fraction
{{{(4(k-3))/((k+3)(k-3))}}}
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now you can add the two fractions together
{{{(2(k+3))/((k-3)(k+3))+(4(k-3))/((k+3)(k-3))}}}
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notice on the sample problem that when adding 8 twelths and 9 twelths you get 17 twelths.  The denominator stays the same when adding.
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So:  {{{(2(k+3))/((k-3)(k+3))+(4(k-3))/((k+3)(k-3))}}} can be written by adding the numerators over the same denominator like this:
{{{((2(k+3))+(4(k-3)))/((k-3)(k+3))}}}
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remember (P)lease Excuse My Dear Aunt Sally for your order of operations
Parenthesis/Exponents
Multiply/Divide
Add/Subtract
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You are probably good once you get to this point so I will just show the final steps.
Distribute the 2 and the 4
{{{(2k+6+4k-12)/((k-3)(k+3))}}}
combine like terms
{{{(6k-6)/((k-3)(k+3))}}}
factor out a 6
{{{(6(k-1))/((k-3)(k+3))}}}
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If you have any questions, email me at justin.sheppard.tech@hotmail.com