Question 366084
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{5}{5a\ +\ 10}\ -\ \frac{7}{6a\ +\ 12}]


I believe that is the expression that you want to simplify.  Note that I cannot do precisely what you asked to have done.  I cannot simplify the equation.  That is because there is no equation.  An equation has an equals sign in it -- hence the name.  Since you have no equals sign that indicates an equal relationship between two expressions, you do not have an equation.


You have two fractions and you wish to find the sum.  Just like adding any other fractions, you find the Lowest Common Denominator, apply it, and then add the numerators.  The first step in finding an LCD is to determine the factors that the two denominators have in common.  In this case, these two denominators have no factors in common, hence the LCD is simply the product of the two denominators.


Apply the LCD:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{5(6a\ +\ 12)\ -\ 7(5a\ +\ 10)}{(5a\ +\ 10)(6a\ +\ 12)}]


Use the Distributive Property to eliminate the parentheses in numerator (we will find it convenient to leave the denominator alone for the time being).


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{30a\ +\ 60\ -\ 35a\ -\ 70}{(5a\ +\ 10)(6a\ +\ 12)}]


Collect like terms:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{-5a\ -\ 10}{(5a\ +\ 10)(6a\ +\ 12)}]


Factor out a -1:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{-(5a\ +\ 10)}{(5a\ +\ 10)(6a\ +\ 12)}]


Eliminate the factor common to both numerator and denominator:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{-1}{6a\ +\ 12}]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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