Posted by on Monday, November 1, 2010 at 8:00am.
so now change each fraction to have that LCD
[ (a+2)(a+3) - (3a+1)(a+2) ]/[(a-1)(a+2)(a+3)]
expand the top and collect like terms, leave the bottom alone
=(a^2 + 5a + 6 - 3a^2 - 7a - 2)/[(a-1)(a+2)(a+3)]
= (-2a^2 - 2a + 4)/[(a-1)(a+2)(a+3)]
= -2(a^2 + a - 2)/[(a-1)(a+2)(a+3)]
= -2(a+2)(a-1)/[(a-1)(a+2)(a+3)]
= -2/(a+3)
Thanks!
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