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)Multipl y516 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).

2 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. 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) +2(x+ 4)x+ 4 Reduce fractions3 2(x+ 4)5(x+ 4)

3 + 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. Thelargest known Germain prime (at the time of printing) is 183027 2265440 1whichhas 79911 digits!

4 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)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 .

5 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. ( ) 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)

6 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. ( ) - 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. ( )7