How do I simplify #(2x)/(x-3)-(x)/(x+3)#?

1 Answer
Aug 9, 2015

#(x(x+9))/((x-3)(x+3))#

Explanation:

Think of how you simplify fractions without algebraic terms. (e.g. #1/3 + 2/5#. Determine the lowest common multiple between #3# and #5#, which is #15#.

For the first fraction multiply #5# on both numerator and denominator to get the denominator to #15# and for the second fraction multiply #3# on both numerator and denominator to get the denominator to #15#.

Thus, #1/3 + 2/5 = 5/15 + 6/15 = 11/15#

The principle for fractions with algebraic terms is the same. The questions asks to simplify #(2x)/(x-3) - x/(x+3)#. Determine the lowest common multiple between #x-3# and #x+3#, which is #(x-3)(x+3)#.

For the first fraction multiply #x+3# on both numerator and denominator to get the denominator to #(x-3)(x+3)# and for the second fraction multiply #x-3# on both numerator and denominator to get the denominator to #(x-3)(x+3)#.

Thus,
#(2x)/(x-3) - x/(x+3) =(2x(x+3))/((x-3)(x+3))-(x(x-3))/((x-3)(x+3)) =(2x^2+6x)/((x-3)(x+3))-(x^2-3x)/((x-3)(x+3)) =(2x^2+6x-x^2+3x)/((x-3)(x+3)) =(x^2+9x)/((x-3)(x+3)) =(x(x+9))/((x-3)(x+3))#

Always factorise your answer in the end unless the questions asks otherwise.