Algebra/Trig Review - Pauls Online Math Notes - Lamar University

Algebra/Trig ReviewSolutionAll of these problems make use of one or more of the following properties.nn m n+ m p n−m1p p = p = p =mm−np pn( )mnm0p p p p( pq)p−n= = 1, provided ≠ 0n n n⎛ p⎞p= p q⎜ ⎟ =⎝ q ⎠ q1 1 n= = pn−npp−n⎛ p⎞ ⎛ q ⎞ q⎜ ⎟ = ⎜ ⎟ =⎝ q ⎠ ⎝ p⎠pThis particular problem only uses the first property.nnn1 3 1 35−4 3 19 3 4 4 19 3 3 −−− − − − 4 −15 −3122x y x + y y = 2x y + y = 2x y + yRemember that the y’s in the last two terms can’t be combined! You can onlycombine terms that are products or quotients. Also, while this would be an acceptableand often preferable answer in a calculus class an algebra class would probably wantyou to get rid of the negative exponents as well. In this case your answer would be.1 3 5−−4 −3 −19 3 4 −15 −32 1122xy x + yy = 2x y + y = +15 3 5x y12yThe 2 will stay in the numerator of the first term because it doesn’t have a negativeexponent.nnn2.3 15 2 2xxx −Solution3 1 3 1 6 20 5 21− + 2− + −5 2 2 5 2 10 10 10 10xxx = x = x = xNot much to this solution other than just adding the exponents.3.1−3xx2x5Solution1 2 2 13−−3 3 3 −53xx x x x x 1= = = = x5 52x2x2 2 2Note that you could also have done the following (probably is easier….).13−3© 2006 Paul Dawkins 4http://tutorial.math.lamar.edu/terms.aspx