Math 017 Materials With Exercises

Example 5.10 Simplify the following expressions, if possible. If not possible, explain why it is notpossible.m 2mnxyv 4za)b) c)22mx y 3xy4zSolution:a) One needs to find all common factors of both the denominator and thenumerator. In this case the common factor is m . We will factor m in the numerator andthen cancel it with m in the denominator and arrive at our result..m 2mnm(1 2n) 12nm mb) The common factor is xy . Factor xy in the denominator and cancel it.xy xy 1 x2 y 3xy2 xy(x 3y)x 3yc) It cannot be simplified. In the numerator 4 z is used only as a term, not as afactor.Application of the factorization of −1 to the simplification of algebraic fractions x yConsider , x y 0 .x yThere are no common factors that could be canceled, but we should notice that signs in the numeratorare exactly the opposite of signs in the denominator (x follows a minus sign in the numerator but a plussign in the denominator; y is preceded by a plus sign in the numerator but by a minus sign in thedenominator). Factoring 1 either in the numerator or in the denominator (the choice is arbitrary) willreverse the signs and allow the simplification. x y 1(x y)1(x y)= 1x y x y x yExample 5.11 Simplify the following expressiona 4d. 4d aSolution:One needs to notice „the reversed signs‟ (all terms in the numerators follow a plus sign,while all terms in the denominator follow a minus sign). This requires factorization of 1 (either in the numerator or denominator). We will factor 1 in the denominator.a 4da 4da 4d 1 4d a 1(4d a)1(a 4d)55