This is “Rational Functions: Addition and Subtraction”, section 4.6 from the book Advanced Algebra (v. 1.0). For details on it (including licensing), click here.

For more information on the source of this book, or why it is available for free, please see the project's home page. You can browse or download additional books there. You may also download a PDF copy of this book (41 MB) or just this chapter (3 MB), suitable for printing or most e-readers, or a .zip file containing this book's HTML files (for use in a web browser offline).

Has this book helped you? Consider passing it on:
Creative Commons supports free culture from music to education. Their licenses helped make this book available to you.
DonorsChoose.org helps people like you help teachers fund their classroom projects, from art supplies to books to calculators.

4.6 Rational Functions: Addition and Subtraction

Learning Objectives

  1. Add and subtract rational functions.
  2. Simplify complex rational expressions.

Adding and Subtracting Rational Functions

Adding and subtracting rational expressions is similar to adding and subtracting fractions. Recall that if the denominators are the same, we can add or subtract the numerators and write the result over the common denominator. When working with rational expressions, the common denominator will be a polynomial. In general, given polynomials P, Q, and R, where Q0, we have the following:

PQ±RQ=P±RQ

The set of restrictions to the domain of a sum or difference of rational expressions consists of the restrictions to the domains of each expression.

Example 1

Subtract: 4xx2643x+8x264.

Solution:

The denominators are the same. Hence we can subtract the numerators and write the result over the common denominator. Take care to distribute the negative 1.

4xx2643x+8x264=4x(3x+8)x264       Subtractthenumerators.=4x3x8x264          Simplify.=x81(x+8)(x8)Cancel.=1x+8Restrictionsx±8

Answer: 1x+8, where x±8

To add rational expressions with unlike denominators, first find equivalent expressions with common denominators. Do this just as you have with fractions. If the denominators of fractions are relatively prime, then the least common denominator (LCD) is their product. For example,

1x+1yLCD=xy=xy

Multiply each fraction by the appropriate form of 1 to obtain equivalent fractions with a common denominator.

1x+1y=1yxy+1xyx=yxy+xxyEquivalentfractionswithacommondenominator=y+xxy

In general, given polynomials P, Q, R, and S, where Q0 and S0, we have the following:

PQ±RS=PS±QRQS

Example 2

Given f(x)=5x3x+1 and g(x)=2x+1, find f+g and state the restrictions.

Solution:

Here the LCD is the product of the denominators (3x+1)(x+1). Multiply by the appropriate factors to obtain rational expressions with a common denominator before adding.

(f+g)(x)=f(x)+g(x)=5x3x+1+2x+1=5x(3x+1)(x+1)(x+1)+2(x+1)(3x+1)(3x+1)=5x(x+1)(3x+1)(x+1)+2(3x+1)(x+1)(3x+1)=5x(x+1)+2(3x+1)(3x+1)(x+1)=5x2+5x+6x+2(3x+1)(x+1)=5x2+11x+2(3x+1)(x+1)=(5x+1)(x+2)(3x+1)(x+1)

The domain of f consists all real numbers except 13, and the domain of g consists of all real numbers except −1. Therefore, the domain of f + g consists of all real numbers except −1 and 13.

Answer: (f+g)(x)=(5x+1)(x+2)(3x+1)(x+1), where x1,13

It is not always the case that the LCD is the product of the given denominators. Typically, the denominators are not relatively prime; thus determining the LCD requires some thought. Begin by factoring all denominators. The LCD is the product of all factors with the highest power.

Example 3

Given f(x)=3x3x1 and g(x)=414x3x24x+1, find fg and state the restrictions to the domain.

Solution:

To determine the LCD, factor the denominator of g.

(fg)(x)=f(x)g(x)=3x3x1414x3x24x+1=3x(3x1)414x(3x1)(x1)

In this case the LCD=(3x1)(x1). Multiply f by 1 in the form of (x1)(x1) to obtain equivalent algebraic fractions with a common denominator and then subtract.

=3x(3x1)(x1)(x1)414x(3x1)(x1)=3x(x1)4+14x(3x1)(x1)=3x2+11x4(3x1)(x1)=(3x1)(x+4)(3x1)(x1)=(x+4)(x1)

The domain of f consists of all real numbers except 13, and the domain of g consists of all real numbers except 1 and 13. Therefore, the domain of fg consists of all real numbers except 1 and 13.

Answer: (fg)(x)=x+4x1, where x13,1

Example 4

Simplify and state the restrictions: 2xx+63x6x18(x2)x236.

Solution:

Begin by applying the opposite binomial property 6x=(x6).

2xx+63x6x18(x2)x236=2x(x+6)3x1(x6)18(x2)(x+6)(x6)=2x(x+6)+3x(x6)18(x2)(x+6)(x6)

Next, find equivalent fractions with the LCD=(x+6)(x6) and then simplify.

=2x(x+6)(x6)(x6)+3x(x6)(x+6)(x+6)18(x2)(x+6)(x6)=2x(x6)+3x(x+6)18(x2)(x+6)(x6)=2x2+12x+3x2+18x18x+36(x+6)(x6)=x2+12x+36(x+6)(x6)=(x+6)(x+6)(x+6)(x6)=x+6x6

Answer: x+6x6, where x±6

Try this! Simplify and state the restrictions: x+1(x1)22x214(x+1)(x1)2

Answer: 1x1, where x±1

Rational expressions are sometimes expressed using negative exponents. In this case, apply the rules for negative exponents before simplifying the expression.

Example 5

Simplify and state the restrictions: 5a2+(2a+5)1.

Solution:

Recall that xn=1xn. Begin by rewriting the rational expressions with negative exponents as fractions.

5a2+(2a+5)1=5a2+1(2a+5)1

Then find the LCD and add.

5a2+1(2a+5)1=5a2(2a+5)(2a+5)+1(2a+5)a2a2=5(2a+5)a2(2a+5)+a2a2(2a+5)Equivalentexpressionswithacommondenominator=10a+25+a2a2(2a+5)  Add.=a2+10a+25a2(2a+5)Simplfiy.=(a+5)(a+5)a2(2a+5)

Answer: (a+5)2a2(2a+5), where a52,0

Simplifying Complex Rational Expressions

A complex rational expressionA rational expression that contains one or more rational expressions in the numerator or denominator or both. is defined as a rational expression that contains one or more rational expressions in the numerator or denominator or both. For example, 412x+9x225x+3x2 is a complex rational expression. We simplify a complex rational expression by finding an equivalent fraction where the numerator and denominator are polynomials. There are two methods for simplifying complex rational expressions, and we will outline the steps for both methods. For the sake of clarity, assume that variable expressions used as denominators are nonzero.

Method 1: Simplify Using Division

We begin our discussion on simplifying complex rational expressions using division. Before we can multiply by the reciprocal of the denominator, we must simplify the numerator and denominator separately. The goal is to first obtain single algebraic fractions in the numerator and the denominator. The steps for simplifying a complex algebraic fraction are illustrated in the following example.

Example 6

Simplify: 412x+9x225x+3x2.

Solution:

Step 1: Simplify the numerator and denominator to obtain a single algebraic fraction divided by another single algebraic fraction. In this example, find equivalent terms with a common denominator in both the numerator and denominator before adding and subtracting.

412x+9x225x+3x2=41x2x212xxx+9x221x2x25xxx+3x2=4x2x212xx2+9x22x2x25xx2+3x2Equivalentfractionswithcommondenominators.=4x212x+9x22x25x+3x2Addthefractionsinthenumeratoranddenominator.

At this point we have a single algebraic fraction divided by another single algebraic fraction.

Step 2: Multiply the numerator by the reciprocal of the denominator.

4x212x+9x22x25x+3x2=4x212x+9x2x22x25x+3

Step 3: Factor all numerators and denominators completely.

=(2x3)(2x3)x2x2(2x3)(x1)

Step 4: Cancel all common factors.

=(2x3)(2x3)x2x2(2x3)(x1)=2x3x1

Answer: 2x3x1

Example 7

Simplify: 2xx1+7x+32xx15x3.

Solution:

Obtain a single algebraic fraction in the numerator and in the denominator.

2xx1+7x+32xx15x3=2xx1(x+3)(x+3)+7x+3(x1)(x1)2xx1(x3)(x3)5x3(x1)(x1)=2x(x+3)+7(x1)(x1)(x+3)2x(x3)5(x1)(x1)(x3)=2x2+6x+7x7(x1)(x+3)2x26x5x+5(x1)(x3)=2x2+13x7(x1)(x+3)2x211x+5(x1)(x3)

Next, multiply the numerator by the reciprocal of the denominator, factor, and then cancel.

=2x2+13x7(x1)(x+3)(x1)(x3)2x211x+5=(2x1)(x+7)(x1)(x+3)(x1)(x3)(2x1)(x5)=(x+7)(x3)(x+3)(x5)

Answer: (x+7)(x3)(x+3)(x5)

Try this! Simplify using division: 1y21x21y+1x.

Answer: xyxy

Sometimes complex rational expressions are expressed using negative exponents.

Example 8

Simplify: 2y1x1x24y2.

Solution:

We begin by rewriting the expression without negative exponents.

2y1x1x24y2=2y1x1x24y2

Obtain single algebraic fractions in the numerator and denominator and then multiply by the reciprocal of the denominator.

2y1x1x24y2=2xyxyy24x2x2y2=2xyxyx2y2y24x2=2xyxyx2y2(y2x)(y+2x)

Apply the opposite binomial property (y2x)=(2xy) and then cancel.

=(2xy)xyx2xy2y(2xy)(y+2x)=xyy+2x

Answer: xyy+2x

Method 2: Simplify Using the LCD

An alternative method for simplifying complex rational expressions involves clearing the fractions by multiplying the expression by a special form of 1. In this method, multiply the numerator and denominator by the least common denominator (LCD) of all given fractions.

Example 9

Simplify: 412x+9x225x+3x2.

Solution:

Step 1: Determine the LCD of all the fractions in the numerator and denominator. In this case, the denominators of the given fractions are 1, x, and x2. Therefore, the LCD is x2.

Step 2: Multiply the numerator and denominator by the LCD. This step should clear the fractions in both the numerator and denominator.

412x+9x225x+3x2=(412x+9x2)x2(25x+3x2)x2    MultiplynumeratoranddenominatorbytheLCD.=4x212xx2+9x2x22x25xx2+3x2x2    Distributeandthencancel.=4x212x+92x25x+3

This leaves us with a single algebraic fraction with a polynomial in the numerator and in the denominator.

Step 3: Factor the numerator and denominator completely.

=4x212x+92x25x+3=(2x3)(2x3)(x1)(2x3)  

Step 4: Cancel all common factors.

=(2x3)(2x3)(x1)(2x3)=2x3x1

Note: This was the same problem presented in Example 6 and the results here are the same. It is worth taking the time to compare the steps involved using both methods on the same problem.

Answer: 2x3x1

It is important to point out that multiplying the numerator and denominator by the same nonzero factor is equivalent to multiplying by 1 and does not change the problem.

Try this! Simplify using the LCD: 1y21x21y+1x.

Answer: xyxy

Key Takeaways

  • Adding and subtracting rational expressions is similar to adding and subtracting fractions. A common denominator is required. If the denominators are the same, then we can add or subtract the numerators and write the result over the common denominator.
  • The set of restrictions to the domain of a sum or difference of rational functions consists of the restrictions to the domains of each function.
  • Complex rational expressions can be simplified into equivalent expressions with a polynomial numerator and polynomial denominator. They are reduced to lowest terms if the numerator and denominator are polynomials that share no common factors other than 1.
  • One method of simplifying a complex rational expression requires us to first write the numerator and denominator as a single algebraic fraction. Then multiply the numerator by the reciprocal of the denominator and simplify the result.
  • Another method for simplifying a complex rational expression requires that we multiply it by a special form of 1. Multiply the numerator and denominator by the LCD of all the denominators as a means to clear the fractions. After doing this, simplify the remaining rational expression.

Topic Exercises

    Part A: Adding and Subtracting Rational Functions

      State the restrictions and simplify.

    1. 3x3x+4+23x+4
    2. 3x2x12x+12x1
    3. x22x211x6+x+32x211x6
    4. 4x13x2+2x5x63x2+2x5
    5. 1x2x
    6. 4x31x
    7. 1x1+5
    8. 1x+71
    9. 1x213x+4
    10. 25x2+xx+3
    11. 1x2+1x2
    12. 2xx+2x2
    13. 3x7x(x7)+17x
    14. 28x+3x21x2(x8)
    15. x1x2252x210x+25
    16. x+12x2+5x3x4x21
    17. xx2+4x2x2+8x+16
    18. 2x14x2+8x534x2+20x+25
    19. 5x7x+x2x+249x2
    20. 2x4x2+xx+18x2+6x+1
    21. x12x27x4+2x1x25x+4
    22. 2(x+3)3x25x2+4x3x2+10x+3
    23. x24+2x22x4+2x2
    24. 3x4x4+6x32x26x3+9x2
    25. 3x212x48x2+16x2+24x2
    26. x22x2+1+6x2242x47x24

      Given f and g, simplify the sum f+g and difference fg. Also, state the domain using interval notation.

    1. f(x)=1x, g(x)=5x2
    2. f(x)=1x+2, g(x)=2x1
    3. f(x)=x2x+2, g(x)=x+2x2
    4. f(x)=x2x1, g(x)=2x2x+1
    5. f(x)=63x2+x, g(x)=189x2+6x+1
    6. f(x)=x1x28x+16, g(x)=x2x24x
    7. f(x)=xx225, g(x)=x1x24x5
    8. f(x)=2x3x24, g(x)=x2x2+3x2
    9. f(x)=13x2x2, g(x)=14x23x1
    10. f(x)=66x2+13x5, g(x)=22x2+x10

      State the restrictions and simplify.

    1. 1+3x5x1x2
    2. 4+2x6x1x2
    3. 2xx813x+12x+93x223x8
    4. 4xx2103x+119x+183x25x2
    5. 1x1+1(x1)21x21
    6. 1x21x24+1(x2)2
    7. 2x+1x13x2x23x+1+x+1x2x2
    8. 5x22x2+2xx24xx22x+4+2x24+2x2x2
    9. x+22x(3x2)+4x(x2)(3x2)3x+22x(x2)
    10. 10xx(x5)2x2(2x5)(x5)5xx(2x5)

      Simplify the given algebraic expressions. Assume all variable expressions in the denominator are nonzero.

    1. x2+y2

    2. x2+(2y)2

    3. 2x1+y2

    4. x24y1

    5. 16x2+y2

    6. xy1yx1

    7. 3(x+y)1+x2

    8. 2(xy)2(xy)1

    9. a2(a+b)1

    10. (ab)1(a+b)1

    11. xn+yn

    12. xyn+yxn

    Part B: Simplifying Complex Rational Expressions

      Simplify. Assume all variable expressions in the denominators are nonzero.

    1. 75x2(x3)225x3x3
    2. x+536x3(x+5)39x2
    3. x23632x5x64x3
    4. x856x2x2647x3
    5. 5x+12x2+x1025x2+10x+14x225
    6. 4x227x74x21x76x2x1
    7. x24x52x2+3x+1x210x+252x2+7x+3
    8. 5x2+9x2x2+4x+410x2+3x14x2+7x2
    9. x2153x
    10. 4x32x2
    11. 131x191x2
    12. 25+1x4251x2
    13. 1y23661y
    14. 151y1y2125
    15. 16x+8x235x2x2
    16. 2+13x7x23+1x10x2
    17. 912x+4x294x2
    18. 425x248x5x2
    19. 1x+53x123x11x
    20. 2x51x1x3x5
    21. 1x+1+2x22x31x2
    22. 4x+51x33x3+12x1
    23. x13x11x+1x1x+12x+1
    24. x+13x+51x+32x+3x+1x+3
    25. 2x+32x3+2x32x+32x+32x32x32x+3
    26. x1x+1x+1x1x+1x1x1x+1
    27. 12x+512x5+4x4x22512x+5+12x5+4x4x225
    28. 13x1+13x+13x3x113x+16x9x21
    29. 11+11+1x
    30. 1x111+1x
    31. 1y1x1y21x2
    32. 2y+1x4y21x2
    33. 125y21x21x15y
    34. 16y21x21x4y
    35. 1b+1a1b3+1a3
    36. 1a1b1b31a3
    37. xyyx1y22xy+1x2
    38. 2y5x4x25y2x
    39. x1+y1y2x2
    40. y225x25x1y1
    41. 1x1xx1
    42. 16x2x14
    43. 14x121x212x115x2
    44. x14(3x2)138x1+16(3x2)1
    45. (x3)1+2x1x13(x3)1
    46. (4x5)1+x2x2+(3x10)1
    47. Given f(x)=1x, simplify f(b)f(a)ba.

    48. Given f(x)=1x1, simplify f(b)f(a)ba.

    49. Given f(x)=1x, simplify the difference quotient f(x+h)f(x)h.

    50. Given f(x)=1x+1, simplify the difference quotient f(x+h)f(x)h.

    Part C: Discussion Board

    1. Explain why the domain of a sum of rational functions is the same as the domain of the difference of those functions.

    2. Two methods for simplifying complex rational expressions have been presented in this section. Which of the two methods do you feel is more efficient, and why?

Answers

  1. 3x+23x+4; x43

  2. 1x6; x12,6

  3. 12x2x; x0

  4. 5x4x1; x1

  5. 2(x+3)(x2)(3x+4); x43,2

  6. (x1)(x+2)x2(x2); x0,2

  7. 2x7x(x7); x0,7

  8. x28x5(x+5)(x5)2; x±5

  9. x+2(x+4)2; x0,4

  10. 7(52x)x(7+x)(7x); x7,0,7

  11. x(5x2)(x4)(x1)(2x+1); x12,1,4

  12. x222x2; x0

  13. x2+5(x+2)(x2); x±2

  14. (f+g)(x)=x+5x2; (fg)(x)=x5x2; Domain: (,0)(0,)

  15. (f+g)(x)=2(x2+4)(x+2)(x2); (fg)(x)=8x(x+2)(x2); Domain: (,2)(2,2)(2,)

  16. (f+g)(x)=6(6x+1)x(3x+1)2; (fg)(x)=6x(3x+1)2; Domain: (,13)(13,0)(0,)

  17. (f+g)(x)=2x2+5x5(x+1)(x+5)(x5); (fg)(x)=3x5(x+1)(x+5)(x5); Domain: (,5)(5,1)(1,5)(5,)

  18. (f+g)(x)=1(3x+2)(4x+1); (fg)(x)=7x+3(x1)(3x+2)(4x+1); Domain: (,23)(23,14)(14,1)(1,)

  19. (x1)2x2; x0

  20. 2x1x8; x13,8

  21. x2+1(x1)2(x+1); x±1

  22. 2x+1x; x0,12,1

  23. 0; x0,23,2

  24. y2+x2x2y2
  25. x+2y2xy2
  26. x2y2+16x2
  27. 3x2+x+yx2(x+y)
  28. a+ba2a2(a+b)
  29. xn+ynxnyn
  1. 3x(x3)
  2. x+68x2
  3. 2x5(x2)(5x+1)
  4. x+3x5
  5. 5x3x15
  6. 3xx+3
  7. 6y+1y
  8. x43x+1
  9. 3x23x+2
  10. 8x1x1
  11. 3x(x3)(x+1)(x1)
  12. x3x1
  13. 4x2+912x

  14. 2x54x
  15. x+12x+1
  16. xyx+y
  17. x+5y5xy
  18. a2b2a2ab+b2
  19. xy(x+y)xy
  20. xyxy
  21. 1x+1
  22. x7x5
  23. 3(x2)2x+3
  24. 1ab
  25. 1x(x+h)
  1. Answer may vary