# 7.6 Complex Fractions

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1 Section 7.6 Comple Fractions Comple Fractions In tis section we learn ow to simplify wat are called comple fractions, an eample of wic follows Note tat bot te numerator and denominator are fraction problems in teir own rigt, lending credence to wy we refer to suc a structure as a comple fraction. Tere are two very different tecniques we can use to simplify te comple fraction (. Te first tecnique is a natural coice. ( Simplifying Comple Fractions First Tecnique. To simplify a comple fraction, proceed as follows:. Simplify te numerator. 2. Simplify te denominator. 3. Simplify te division problem tat remains. Let s follow tis outline to simplify te comple fraction (. First, add te fractions in te numerator as follows Secondly, add te fractions in te denominator as follows Substitute te results from (2 and (3 into te numerator and denominator of (, respectively Te rigt-and side of (4 is equivalent to Tis is a division problem, so invert and multiply, factor, ten cancel common factors (2 (3 (4 Copyrigted material. See: ttp://msenu.redwoods.edu/intalgtet/

2 696 Capter 7 Rational Functions Here is an arrangement of te work, from start to finis, presented witout comment. Tis is a good template to emulate wen doing your omework Now, let s look at a second approac to te problem. We saw tat simplifying te numerator in (2 required a common denominator of 6. Simplifying te denominator in (3 required a common denominator of 2. So, let s coose anoter common denominator, tis one a common denominator for bot numerator and denominator, namely, 2. Now, multiply top and bottom (numerator and denominator of te comple fraction ( by 2, as follows ( ( 2 2 Distribute te 2 in bot numerator and denominator and simplify. (5

3 Section 7.6 Comple Fractions 697 ( ( 2 2 ( 2 ( ( 3 ( Let s summarize tis second tecnique. Simplifying Comple Fractions Second Tecnique. To simplify a comple fraction, proceed as follows:. Find a common denominator for bot numerator and denominator. 2. Clear fractions from te numerator and denomaintor by multiplying eac by te common denominator found in te first step. Note tat for tis particular problem, te second metod is muc more efficient. It saves bot space and time and is more aestetically pleasing. It is te tecnique tat we will favor in te rest of tis section. Let s look at anoter eample. Eample 6. Use bot te First and Second Tecniques to simplify te epression 2. (7 State all restrictions. Let s use te first tecnique, simplifying numerator and denominator separately before dividing. First, make equivalent fractions wit a common denominator for te subtraction problem in te numerator of (7 and simplify. Do te same for te denominator. 2 Net, invert and multiply, ten factor ( + ( Let s invoke te sign cange rule and negate two parts of te fraction ( /, numerator and fraction bar, ten cancel te common factors.

4 698 Capter 7 Rational Functions 2 2 ( + ( ( + ( Hence, +. 2 Now, let s try te problem a second time, multiplying numerator and denominator by 2 to clear fractions from bot te numerator and denominator. ( ( 2 2 ( 2 2 ( 2 ( ( Te order in te numerator of te last fraction intimates tat a sign cange would be elpful. Negate te numerator and fraction bar, factor, ten cancel common factors ( ( ( + ( ( + ( + Tis is precisely te same answer found wit te first tecnique. To list te restrictions, we must make sure tat no values of make any denominator equal to zero, at te beginning of te problem, in te body of our work, or in te final answer. In te original problem, if 0, ten bot / and / 2 are undefined, so 0 is a restriction. In te body of our work, te factors + and found in various denominators make and restrictions. No oter denominators supply restrictions tat ave not already been listed. Hence, for all oter tan, 0, and, te left-and side of + 2 is identical to te rigt-and side. Again, te calculator s table utility provides ample evidence of tis fact in te screensots sown in Figure. Note te ERR (error messages at eac of te restricted values of, but also note te perfect agreement of Y and Y2 at all oter values of. (8 Let s look at anoter eample, an important eample involving function notation.

5 Section 7.6 Comple Fractions 699 (a (b (c Figure. Using te table feature of te graping calculator to ceck te identity in (8. Eample 9. Given tat f(, simplify te epression List all restrictions. f( f(2. 2 Remember, f(2 means substitute 2 for. f(2 /2, so f( f(2 2 Because f( /, we know tat 2 2. To clear te fractions from te numerator, we d use a common denominator of 2. Tere are no fractions in te denominator tat need clearing, so te common denominator for numerator and denominator is 2. Multiply numerator and denominator by 2. ( ( ( f( f(2 2 ( 22 ( 22 Negate te numerator and fraction bar, ten cancel common factors. f( f( ( 2 2 2( ( 2 In te original problem, we ave a denominator of 2, so 2 is a restriction. If te body of our work, tere is a fraction /, wic is undefined wen 0, so 0 is also a restriction. Te remaining denominators provide no oter restrictions. Hence, for all values of ecept 0 and 2, te left-and side of is identical to te rigt-and side. f( f(2 2 2

6 700 Capter 7 Rational Functions Let s look at anoter eample involving function notation. Eample 0. Given simplify te epression List all restrictions. f( 2, f( + f(. ( Te function notation f( + is asking us to replace eac instance of in te formula / 2 wit +. Tus, f( + /( + 2. Here is anoter way to tink of tis substitution. Suppose tat we remove te from so tat it reads f( 2, f( ( 2. (2 Now, if you want to compute f(2, simply insert a 2 in te blank area between parenteses. In our case, we want to compute f( +, so we insert an + in te blank space between parenteses in (2 to get f( + ( + 2. Wit tese preliminary remarks in mind, let s return to te problem. interpret te function notation as in our preliminary remarks and write f( + f( ( First, we Te common denominator for te numerator is found by listing eac factor to te igest power tat it occurs. Hence, te common denominator is 2 ( + 2. Te denominator as no fractions to be cleared, so it suffices to multiply bot numerator and denominator by 2 ( + 2. f( + f( ( ( ( ( ( ( ( ( ( ( + 2 ( 2 2 ( + 2

7 Section 7.6 Comple Fractions 70 We will now epand te numerator. Don t forget to use parenteses and distribute tat minus sign. f( + f( 2 ( ( ( ( + 2 Finally, factor a out of te numerator in opes of finding a common factor to cancel. f( + f( (2 + 2 ( + 2 (2 + 2 ( + 2 (2 + 2 ( + 2 We must now discuss te restrictions. In te original question (, te in te denominator must not equal zero. Hence, 0 is a restriction. In te final simplified form, te factor of 2 in te denominator is undefined if 0. Hence, 0 is a restriction. Finally, te factor of ( + 2 in te final denominator is undefined if + 0, so is a restriction. Te remaining denominators provide no additional restrictions. Hence, provided 0, 0, and, for all oter combinations of and, te left-and side of is identical to te rigt-and side. f( + f( (2 + 2 ( + 2 Let s look at one final eample using function notation. Eample 3. If f( + (4 simplify f(f(. We first evaluate f at, ten evaluate f at te result of te first computation. Tus, we work te inner function first to obtain ( f(f( f. +

8 702 Capter 7 Rational Functions Te notation f(/( + is asking us to replace eac occurrence of in te formula /( + wit te epression /( +. Confusing? Here is an easy way to tink of tis substitution. Suppose tat we remove from f( +, replacing eac occurrence of wit empty parenteses, wic will produce te template f( ( ( +. (5 Now, if asked to compute f(3, simply insert 3 into te blank areas between parenteses. In tis case, we want to compute f(/(+, so we insert /(+ in te blank space between eac set of parenteses in (5 to obtain f ( We now ave a comple fraction. Te common denominator for bot top and bottom of tis comple fraction is +. Tus, we multiply bot numerator and denominator of our comple fraction by + and use te distributive property as follows ( + ( + + ( + ( + ( + ( ( + + ( + + (( + Cancel and simplify. ( ( + + ( ( + + (( ( In te final denominator, te value /2 makes te denominator 2 + equal to zero. Hence, /2 is a restriction. In te body of our work, several fractions ave denominators of + and are terefore undefined at. Tus, is a restriction. No oter denominators add additional restrictions. Hence, for all values of, ecept /2 and, te left-and side of f(f( 2 + is identical to te rigt-and side.

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