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7.6: Complex Fractions

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In this section we learn how to simplify what are called complex fractions, an example of which follows.

12+1314+23

Note that both the numerator and denominator are fraction problems in their own right, lending credence to why we refer to such a structure as a “complex fraction.”

There are two very different techniques we can use to simplify the complex fraction (1). The first technique is a “natural” choice.

Simplifying Complex Fractions — First Technique

To simplify a complex fraction, proceed as follows:

  1. Simplify the numerator.
  2. Simplify the denominator.
  3. Simplify the division problem that remains.

Let’s follow this outline to simplify the complex fraction (1). First, add the fractions in the numerator as follows.

12+13=36+26=56

Secondly, add the fractions in the denominator as follows.

14+23=312+812=1112

Substitute the results from (2) and (3) into the numerator and denominator of (1), respectively.

12+1314+23=561112

The right-hand side of (4) is equivalent to

56÷1112

This is a division problem, so invert and multiply, factor, then cancel common factors.

12+1314+23=561211=52322311=52322311=1011

Here is an arrangement of the work, from start to finish, presented without comment. This is a good template to emulate when doing your homework.

12+1314+23=36+26312+812=561112=561211=52322311=52322311=1011

Now, let’s look at a second approach to the problem. We saw that simplifying the numerator in (2) required a common denominator of 6. Simplifying the denominator in (3) required a common denominator of 12. So, let’s choose another common denominator, this one a common denominator for both numerator and denominator, namely, 12. Now, multiply top and bottom (numerator and denominator) of the complex fraction (1) by 12, as follows.

12+1314+23=(12+13)12(14+23)12

Distribute the 12 in both numerator and denominator and simplify.

(12+13)12(14+23)12=(12)12+(13)12(14)12+(23)12=6+43+8=1011

Let’s summarize this second technique.

Simplifying Complex Fractions — Second Technique

To simplify a complex fraction, proceed as follows:

  1. Find a common denominator for both numerator and denominator.
  2. Clear fractions from the numerator and denominator by multiplying each by the common denominator found in the first step.

Note that for this particular problem, the second method is much more efficient. It saves both space and time and is more aesthetically pleasing. It is the technique that we will favor in the rest of this section.

Let’s look at another example.

Example 7.6.1

Use both the First and Second Techniques to simplify the expression 1x111x2 State all restrictions.

Solution

Let’s use the first technique, simplifying numerator and denominator separately before dividing. First, make equivalent fractions with a common denominator for the subtraction problem in the numerator of (7) and simplify. Do the same for the denominator.

1x111x2=1xxxx2x21x2=1xxx21x2

Next, invert and multiply, then factor.

1x111x2=1xxx2x21=1xxx2(x+1)(x1)

Let’s invoke the sign change rule and negate two parts of the fraction (1 − x)/x, numerator and fraction bar, then cancel the common factors.

\[\dfrac{\dfrac{1}{x}-1}{1-\dfrac{1}{x^{2}}}=-\dfrac{x-1}{x} \cdot \dfrac{x^{2}}{(x+1)(x-1)}=-\dfrac{x-1}{\not{x}} \cdot \dfrac{x \not{x}}{(x+1)(x-1)} \nonumber \]

Hence,

1x111x2=xx+1

Now, let’s try the problem a second time, multiplying numerator and denominator by x2 to clear fractions from both the numerator and denominator.

1x111x2=(1x1)x2(11x2)x2=(1x)x2(1)x2(1)x2(1x2)x2=xx2x21

The order in the numerator of the last fraction intimates that a sign change would be helpful. Negate the numerator and fraction bar, factor, then cancel common factors.

\[\dfrac{\dfrac{1}{x}-1}{1-\dfrac{1}{x^{2}}}=-\dfrac{x^{2}-x}{x^{2}-1}=-\dfrac{x(x-1)}{(x+1)(x-1)}=-\dfrac{x(x-1)}{(x+1)(x-1)}=-\dfrac{x}{x+1} \nonumber \]

This is precisely the same answer found with the first technique. To list the restrictions, we must make sure that no values of x make any denominator equal to zero, at the beginning of the problem, in the body of our work, or in the final answer.

In the original problem, if x = 0, then both 1/x and 1/x2 are undefined, so x = 0 is a restriction. In the body of our work, the factors x + 1 and x − 1 found in various denominators make x = −1 and x = 1 restrictions. No other denominators supply restrictions that have not already been listed. Hence, for all x other than −1, 0, and 1, the left-hand side of

1x111x2=xx+1

is identical to the right-hand side. Again, the calculator’s table utility provides ample evidence of this fact in the screenshots shown in Figure 7.6.1.

Note the ERR (error) messages at each of the restricted values of x, but also note the perfect agreement of Y1 and Y2 at all other values of x.

Let’s look at another example, an important example involving function notation.

Screen Shot 2019-07-16 at 9.39.29 PM.png
Figure 7.6.1. Using the table feature of the graphing calculator to check the identity in (8).
Example 7.6.2

Given that f(x)=1x, simplify the expression f(x)f(2)x2. List all restrictions.

Solution

Remember, f(2) means substitute 2 for x. Because f(x) = 1/x, we know that f(2) = 1/2, so

f(x)f(2)x2=1x12x2

To clear the fractions from the numerator, we’d use a common denominator of 2x. There are no fractions in the denominator that need clearing, so the common denominator for numerator and denominator is 2x. Multiply numerator and denominator by 2x.

f(x)f(2)x2=(1x12)2x(x2)2x=(1x)2x(12)2x(x2)2x=2x2x(x2)

Negate the numerator and fraction bar, then cancel common factors.

\[\dfrac{f(x)-f(2)}{x-2}=-\dfrac{x-2}{2 x(x-2)}=-\dfrac{x-2}{2 x(x-2)}=-\dfrac{1}{2 x} \nonumber \]

In the original problem, we have a denominator of x − 2, so x = 2 is a restriction. If the body of our work, there is a fraction 1/x, which is undefined when x = 0, so x = 0 is also a restriction. The remaining denominators provide no other restrictions. Hence, for all values of x except 0 and 2, the left-hand side of

f(x)f(2)x2=12x

is identical to the right-hand side.

Let’s look at another example involving function notation.

Example 7.6.3

Given f(x)=1x2, simplify the expression f(x+h)f(x)h List all restrictions.

Solution

The function notation f(x + h) is asking us to replace each instance of x in the formula 1/x2 with x + h. Thus, f(x+h)=1/(x+h)2.

Here is another way to think of this substitution. Suppose that we remove the x from

f(x)=1x2

so that it reads

f( )=1( )2

Now, if you want to compute f(2), simply insert a 2 in the blank area between parentheses. In our case, we want to compute f(x + h), so we insert an x + h in the blank space between parentheses in (12) to get

f(x+h)=1(x+h)2

With these preliminary remarks in mind, let’s return to the problem. First, we interpret the function notation as in our preliminary remarks and write

f(x+h)f(x)h=(f(x+h)f(x)h=x2(x2+2xh+h2)hx2(x+h)2=x2x22xhh2hx2(x+h)2=2xhh2hx2(x+h)2\)frac1(x+h)21x2h

The common denominator for the numerator is found by listing each factor to the highest power that it occurs. Hence, the common denominator is x2(x+h)2. The denominator has no fractions to be cleared, so it suffices to multiply both numerator and denominator by x2(x+h)2.

f(x+h)f(x)h=(1(x+h)21x2)x2(x+h)2hx2(x+h)2=(1(x+h)2)x2(x+h)2(1x2)x2(x+h)2hx2(x+h)2=x2(x+h)2hx2(x+h)2

We will now expand the numerator. Don’t forget to use parentheses and distribute that minus sign.

f(x+h)f(x)h=x2(x2+2xh+h2)hx2(x+h)2=x2x22xhh2hx2(x+h)2=2xhh2hx2(x+h)2

Finally, factor a −h out of the numerator in hopes of finding a common factor to cancel.

f(x+h)f(x)h=h(2x+h)hx2(x+h)2=h(2x+h)hx2(x+h)2=(2x+h)x2(x+h)2

We must now discuss the restrictions. In the original question (11), the h in the denominator must not equal zero. Hence, h = 0 is a restriction. In the final simplified form, the factor of x2 in the denominator is undefined if x = 0. Hence, x = 0 is a restriction. Finally, the factor of (x+h)2 in the final denominator is undefined if x+h = 0, so x = −h is a restriction. The remaining denominators provide no additional restrictions. Hence, provided h0,x0, and xh, for all other combinations of x and h, the left-hand side of

f(x+h)f(x)h=(2x+h)x2(x+h)2

is identical to the right-hand side.

Let’s look at one final example using function notation.

Example 7.6.4

If f(x)=xx+1 simplify f(f(x)).

Solution

We first evaluate f at x, then evaluate f at the result of the first computation. Thus, we work the inner function first to obtain

f(f(x))=f(xx+1)

The notation f(x/(x + 1)) is asking us to replace each occurrence of x in the formula x/(x + 1) with the expression x/(x + 1). Confusing? Here is an easy way to think of this substitution. Suppose that we remove x from

f(x)=xx+1

replacing each occurrence of x with empty parentheses, which will produce the template

f( )=( )( )+1

Now, if asked to compute f(3), simply insert 3 into the blank areas between parentheses. In this case, we want to compute f(x/(x+ 1)), so we insert x/(x+ 1) in the blank space between each set of parentheses in (15) to obtain

f(xx+1)=xx+1xx+1+1

We now have a complex fraction. The common denominator for both top and bottom of this complex fraction is x + 1. Thus, we multiply both numerator and denominator of our complex fraction by x + 1 and use the distributive property as follows.

xx+1xx+1+1=(xx+1)(x+1)(xx+1+1)(x+1)=(xx+1)(x+1)(xx+1)(x+1)+(1)(x+1)

Cancel and simplify.

(xx+1)(x+1)(xx+1)(x+1)+(1)(x+1)=xx+(x+1)=x2x+1

In the final denominator, the value x = −1/2 makes the denominator 2x + 1 equal to zero. Hence, x = −1/2 is a restriction. In the body of our work, several fractions have denominators of x + 1 and are therefore undefined at x = −1. Thus, x = −1 is a restriction. No other denominators add additional restrictions.

Hence, for all values of x, except x = −1/2 and x = −1, the left-hand side of

f(f(x))=x2x+1

is identical to the right-hand side.

Exercise

In Exercises 1-6, evaluate the function at the given rational number. Then use the first or second technique for simplifying complex fractions explained in the narrative to simplify your answer.

Exercise 7.6.1

Given

f(x)=x+12x,

evaluate and simplify f(12).

Answer

1

Exercise 7.6.2

Given

f(x)=2xx+5,

evaluate and simplify f(32).

Exercise 7.6.3

Given

f(x)=2x+34x,

evaluate and simplify f(13).

Answer

1

Exercise 7.6.4

Given

f(x)=32xx+5

evaluate and simplify f(25)

Exercise 7.6.5

Given

f(x)=52xx+4,

evaluate and simplify f(35).

Answer

1923

Exercise 7.6.6

Given

f(x)=2x911x,

evaluate and simplify f(43).

In Exercises 7-46, simplify the given complex rational expression. State all restrictions.

Exercise 7.6.7

5+6x25x36x3

Answer

Provided x0,65,or65,

x25x6.

Exercise 7.6.8

7+9x49x81x3

Exercise 7.6.9

7x25x78x7+3x+8

Answer

Provided x2,7,8,or4311,

(2x39)(x+8)(11x+43)(x2)

Exercise 7.6.10

9x+47x99x9+5x4

Exercise 7.6.11

3+7x9x249x4

Answer

Provided x0,73,or73,

x33x7.

Exercise 7.6.12

25x4x225x4

Exercise 7.6.13

9x+4+7x+99x+9+2x8

Answer

Provided x4,9,8,or5411,

(16x+109)(x8)(11x54)(x+4)

Exercise 7.6.14

4x6+9x99x6+8x9

Exercise 7.6.15

5x74x410x45x+2

Answer

Provided x7,4,2,or8,

x+25(x7)

Exercise 7.6.16

3x+6+7x+99x+64x+9

Exercise 7.6.17

6x3+5x89x3+7x8

Answer

Provided x3,8,or9316

11x6316x93

Exercise 7.6.18

7x74x27x76x2

Exercise 7.6.19

4x2+7x75x2+2x6

Answer

Provided x2,7,or397,

11x427x39

Exercise 7.6.20

9x+27x+54x+2+3x+5

Exercise 7.6.21

5+4x25x16x3

Answer

Provided x0,45,or45,

x25x4.

Exercise 7.6.22

6x+5+5x+48x+53x+4

Exercise 7.6.23

9x5+8x+45x54x+4

Answer

Provided x5,4,or40,

17x4x+40.

Exercise 7.6.24

4x6+4x96x6+6x9

Exercise 7.6.25

6x+8+5x25x22x+2

Answer

Provided x8,2,2,or143,

(11x+28)(x+2)(3x+14)(x+8).

Exercise 7.6.26

7x+9+9x24x2+7x+1

Exercise 7.6.27

7x+75x+48x+73x+4

Answer

Provided x7,4,or115,

2x75x+11.

Exercise 7.6.28

2516x25+4x

Exercise 7.6.29

64x25x385x

Answer

Provided x0or58,

8x+5x2.

Exercise 7.6.30

4x+2+5x67x65x+7

Exercise 7.6.31

2x64x+93x66x+9

Answer

Provided x6,9,or21,

23.

Exercise 7.6.32

3x+64x+46x+68x+4

Exercise 7.6.33

9x264x438x

Answer

Provided x0or83,

3x+8x3.

Exercise 7.6.34

9x225x435x

Exercise 7.6.35

4x48x74x7+2x+2

Answer

Provided x4,7,2,or1,

2(x+2)3(x4).

Exercise 7.6.36

27x449x2

Exercise 7.6.37

3x2+8x9+3x2819x281+9x28x9

Answer

Provided x1,9,9,1,5,

(x5)(x+1)3(x+5)(x1)

Exercise 7.6.38

7x25x14+2x27x185x27x18+8x26x27

Exercise 7.6.39

2x2+8x+7+5x2+13x+427x2+13x+42+6x2+3x18

Answer

Provided x1,7,6,3,2113,

(7x+17)(x3)(13x+21)(x+1)

Exercise 7.6.40

3x2+5x14+3x27x983x27x98+3x215x+14

Exercise 7.6.41

\(\frac{\frac{6}{x^2+11x+24}−\frac{6}{x^2+13+40}}{\frac{9}{x^2+13x+40}−{\frac{9}{x^2−3x−40}}\)

Answer

Provided x3,8,5,8,

1(x8)12(x+3)

Exercise 7.6.42

7x2+19x+90+7x2+19x+909x2+19x+90+9x2+7x18

Exercise 7.6.43

7x26x+5+7x2+2x358x2+2x35+8x2+8x+7

Answer

Provided x1,5,7,1,2,

7(x+3)(x+1)8(x2)(x1)

Exercise 7.6.44

2x24x122x2x302x2x302x24x45

Exercise 7.6.45

4x2+6x74x2+2x34x2+2x34x2+5x+6

Answer

Provided x7,1,3,2,

4(x+2)3(x+7)

Exercise 7.6.46

9x2+3x4+8x27x+64x27x+6+9x210x+24

Exercise 7.6.47

Given f(x)=2x, simplify​​​​​​​

f(x)f(3)x3.

State all restrictions.

Answer

Provided x0,3,

23x

Exercise 7.6.48

Given f(x)=5x, simplify​​​​​​​

f(x)f(2)x2.

State all restrictions.

Exercise 7.6.49

Given f(x)=3x2, simplify​​​​​​​

f(x)f(1)x1.

State all restrictions.

Answer

Provided x0,1,

3(x+1)x2

Exercise 7.6.50

Given f(x)=5x2, simplify​​​​​​​

f(x)f(2)x2.

State all restrictions.

Exercise 7.6.51

Given f(x)=7x, simplify​​​​​​​

f(x+h)f(x)h.

State all restrictions.

Answer

Provided x0,h, and h0,

7h(x+h)

Exercise 7.6.52

Given f(x)=4x, simplify​​​​​​​

f(x+h)f(x)h.

State all restrictions.

Exercise 7.6.53

Given

f(x)=x+13x,

find and simplify f(1x). State all restrictions.

Answer

Provided x0,13,

x+13x1

Exercise 7.6.54

Given

f(x)=2x3x+4

find and simplify f(2x). State all restriction.​​​​​​​

Exercise 7.6.55

Given

f(x)=x+125x,

find and simplify f(5x). State all restrictions.

Answer

Provided x0,252,

x+52x25

Exercise 7.6.56

Given

f(x)=2x34+x,

find and simplify f(1x). State all restrictions.

Exercise 7.6.57

Given

f(x)=xx+2,

find and simplify f(f(x)). State all restrictions.

Answer

Provided x2,43,

x3x+4

Exercise 7.6.58

Given

f(x)=2xx+5

find and simplify f(f(x)). State all restrictions.

​​​​​​​


This page titled 7.6: Complex Fractions is shared under a CC BY-NC-SA 2.5 license and was authored, remixed, and/or curated by David Arnold via source content that was edited to the style and standards of the LibreTexts platform.

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