Quantcast XP Math - Forums - View Single Post - ALGEBRA FRACTIONS!!! how can i do 2/n(n+1)(n+2) - 2/(n+1)(n+2)(n+3) HELPPPPPP?
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Old 08-04-2007   #2
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When subtracting fractions you need to get a common denominator. Since both fractions have an (n + 1) and an (n + 2) term you don't need to worry about that now.The first denominator needs an (n + 3) term, so you are going to multiply it by (n+3)/(n+3) so your first fraction becomes:2(n+3) / n(n+1)(n+2)(n+3)The second fraction needs an n term so you are going to multiply it by n/n to get:2n / n(n+1)(n+2)(n+3)Now that your fractions have the same denominator you can combine them into one fraction.2(n+3) / n(n+1)(n+2)(n+3) - 2n / n(n+1)(n+2)(n+3) = [2(n+3) - 2n] / [n(n+1)(n+2)(n+3)]Simplify the numerator by distributing and combining like terms. 2(n+3) - 2n = 2n + 6 - 2n = 6So you final answer is 6 / n(n+1)(n+2)(n+3)