Rational Expressions - Complex Fractions

1 Expressions - Complex FractionsObjective: Simplify Complex Fractions by multiplying eachterm by theleast common Fractions have Fractions in either the numerator, or denominator, or usu-ally both. These Fractions can be simplified in one of two ways. This will be illus-trated first with integers, then we will consider how the process can be expandedto include Expressions with first method uses order of operations to simplify the numerator and denomi-nator first, then divide the two resulting Fractions by multiplying by the 1456+12 Get common denominator in top and bottom fractions812 31256+36 Add and subtract Fractions ,reducing solutions51243To divide Fractions we multiply by the reciprocal(512)(34)Reduce(54)(14)

2 Multiply516 Our SolutionThe process above works just fine to simplify, but between getting commondenominators, taking reciprocals, and reducing, it can be avery involved we prefer a different method, to multiply the numerator and denomi-nator of the large fraction (in effect each term in the complexfraction) by theleast common denominator (LCD). This will allow us to reduceand clear thesmall Fractions . We will simplify the same problem using this second 1456+12 LCD is 12,multiply each term12(12)3 1(12)45(12)6+1(12)2 Reduce each fraction2(4) 1(3)5(2) + 1(6)Multiply8 310+ 6 Add and subtract516 Our SolutionClearly the second method is a much cleaner and faster methodto arrive at oursolution.

3 It is the method we will use when simplifying with variables as well. Wewill first find the LCD of the small Fractions , and multiply each term by this LCDso we can clear the small Fractions and 1x21 1xIdentify LCD(use highest exponent)LCD=x2 Multiply each term by LCD1(x2) 1(x2)x21(x2) 1(x2)xReduce Fractions (subtract exponents)1(x2) 11(x2) xMultiplyx2 1x2 xFactor(x+ 1)(x 1)x(x 1)Divide out(x 1)factorx+ 1xOur SolutionThe process is the same if the LCD is a binomial, we will need todistribute3x+ 4 25 +2x+ 4 Multiply each term by LCD,(x+ 4)23(x+ 4)x+ 4 2(x+ 4)5(x+ 4)

4 +2(x+ 4)x+ 4 Reduce fractions3 2(x+ 4)5(x+ 4) + 2 Distribute3 2x 85x+20+ 2 Combine like terms 2x 55x+22 Our SolutionThe more Fractions we have in our problem, the more we repeat the same 3ab3+1ab4a2b+ab 1abIdenfity LCD(highest exponents)LCD=a2b3 Multiply each term by LCD2(a2b3)ab2 3(a2b3)ab3+1(a2b3)ab4(a2b3)a2b+ab(a2b3) 1(a2b3)abReduce each fraction(subtract exponents)2ab 3a+ab24b2+a3b4 ab2 Our SolutionWorld View Note:Sophie Germain is one of the most famous women in mathe-matics, many primes, which are important to finding an LCD, carry her primes are prime numbers where one more than double the primenumber is also prime, for example 3 is prime and so is2 3 + 1 = 7prime.

5 Thelargest known Germain prime (at the time of printing) is 183027 2265440 1whichhas 79911 digits!Some problems may require us to FOIL as we simplify. To avoid sign errors, ifthere is a binomial in the numerator, we will first distributethe negative throughthe 3x+ 3 x+ 3x 3x 3x+ 3+x+ 3x 3 Distribute the subtraction to numeratorx 3x+ 3+ x 3x 3x 3x+ 3+x+ 3x 3 Identify LCD3 LCD= (x+ 3)(x 3)Multiply each term by LCD(x 3)(x+ 3)(x 3)x+ 3+( x 3)(x+ 3)(x 3)x 3(x 3)(x+ 3)(x 3)x+ 3+(x+ 3)(x+ 3)(x 3)x 3 Reduce Fractions (x 3)(x 3) + ( x 3)(x+ 3)(x 3)(x 3) + (x+ 3)(x+ 3)

6 FOILx2 6x+ 9 x2 6x 9x2 6x+ 9 +x2+ 6x 9 Combine like terms 12x2x2+18 Factor out2in denominator 12x2(x2+ 9)Divide out common factor2 6xx2 9 Our SolutionIf there are negative exponents in an expression we will haveto first convert thesenegative exponents into Fractions . Remember, the exponentis only on the factorit is attached to, not the whole 2+ 2m 1m+ 4m 2 Make each negative exponent intoafraction1m2+2mm+4m2 Multiply each term by LCD, m21(m2)m2+2(m2)mm(m2) +4(m2)m2 Reduce the fractions1 + 2mm3+ 4 Our SolutionOnce we convert each negative exponent into a fraction, the problem solvesexactly like the other Complex fraction and Intermediate Algebra by Tyler Wallace is licensed under a Creative CommonsAttribution Unported License.

7 ( ) Practice - Complex )1 +1x1 1x23)a 24a a5)1a2 1a1a2+1a7)2 4x+ 25 10x+ 29)32a 3+ 2 62a 3 411)xx+ 1 1xxx+ 1+1x13)3x9x215)a2 b24a2ba+b16ab217)1 3x 10x21 +11x+18x219)1 2x3x 4x 323x 421)x 1 +2x 4x+ 3 +6x 42)1y2 11 +1y4)25a a5 +a6)1b+124b2 18)4 +122x 35 +152x 310) 5b 5 310b 5+ 612)2aa 1 3a 6a 1 414)x3x 2x9x2 416)1 1x 6x21 4x+3x218)15x2 2x 14x2 5x+ 420)1 123x+10x 83x+1022)x 5 18x+ 2x+ 7 +6x+ 2523)x 4 +92x+ 3x+ 3 52x+ 325)2b 5b+ 33b+3b+ 327)2b2 5ab 3a22b2+7ab+3a229)yy+ 2 yy 2yy+ 2+yy 224)1a 3a 22a+5a 226)1y2 1xy 2x21y2 3xy+2x228)x 1x+ 1 x+ 1x 1x 1x+ 1+x+ 1x 130)x+ 1x 1 1 x1 +x1(x+ 1)2+1(x 1)2 Simplify each of the following fractional )x 2 y 2x 1+y 133)x 3y xy 3x 2 y 235)x 2 6x 1+ 9x2 932)x 2y+xy 2x 2 y 234)4 4x 1+x 24 x 236)x 3+y 3x 2 x 1y 1+y 2 Beginning and Intermediate Algebra by Tyler Wallace is licensed under a Creative CommonsAttribution Unported License.

8 ( ) - Complex Fractions1)xx 12)1 yy3) aa+ 24)5 aa5) a 1a+ 16)b3+ 2b b 28b7)258)459) 1210) 1211)x2 x 1x2+x+ 112)2a2 3a+ 3 4a2 2a13)x314)3x+ 215)4b(a b)a16)x+ 2x 117)x 5x+ 918) (x 3)(x+ 5)4x2 5x+ 419)13x+ 820)1x+ 421)x 2x+ 222)x 7x+ 523)x 3x+ 424) 2a 23a 425) b 22b+ 326)x+yx y27)a 3ba+ 3b28) 2xx2+ 129) 2y30)x2 131)y xxy32)x2 xy+y2y x33)x2+y2xy34)2x 12x+ 135)(1 3x)2x2(x+ 3)(x 3)36)x+yxyBeginning and Intermediate Algebra by Tyler Wallace is licensed under a Creative CommonsAttribution Unported License.

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