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Mathematics LibreTexts

7.3: Adding and Subtracting Rational Expressions

  • Anonymous
  • LibreTexts

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Learning Objectives
  • Add and subtract rational expressions with common denominators.
  • Add and subtract rational expressions with unlike denominators.
  • Add and subtract rational functions.

Adding and Subtracting with Common Denominators

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.

313+713=3+713=1013

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

In this section, assume that all variable factors in the denominator are nonzero.

Example 7.3.1

Add:

3y+7y

Solution:

Add the numerators 3 and 7, and write the result over the common denominator, y.

3y+7y=3+7y=10y

Answer:

10y

Example 7.3.2

Subtract:

x52x112x1

Solution:

Subtract the numerators x5 and 1, and write the result over the common denominator, 2x1.

x52x112x1=x512x1Simplifythenumerator.=x62x1

Answer:

x62x1

Example 7.3.3

Subtract:

2x+7(x+5)(x3)x+10(x+5)(x3)

Solution:

We use parentheses to remind us to subtract the entire numerator of the second rational expression.

2x+7(x+5)(x3)x+10(x+5)(x3)=(2x+7)(x+10)(x+5)(x3)Simplifythenumerator.=2x+7x10(x+5)(x3)Leavethedenominatorfactored.=1x3(x+5)(x3)Cancelcommonfactors.=1x+5

Answer:

1x+5

Example 7.3.4

Simplify:

2x2+10x+3x236x2+6x+5x236+x4x236

Solution:

Subtract and add the numerators. Make use of parentheses and write the result over the common denominator, x236.

Answer:

x1x6

Exercise 7.3.1

Subtract:

Answer

1x4

Adding and Subtracting with Unlike Denominators

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,

13+15LCD=35=15

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

13+15=1535+1353=515+315Equivalentfractionswithacommondenominator=5+315=815

The process of adding and subtracting rational expressions is similar. In general, given polynomials P, Q, R, and S, where Q0 and S0, we have the following:

PQ±RS=PS±QRQS

In this section, assume that all variable factors in the denominator are nonzero.

Example 7.3.5

Add:

1x+1y

Solution:

In this example, the LCD=xy. To obtain equivalent terms with this common denominator, multiply the first term by yy and the second term by xx.

1x+1y=1xyy+1yxx=yxy+xxyEquivalenttermswithacommondenominator.=y+xxyAddthenumeratorsandplacetheresultoverthecommondenominator,xy.

Answer:

y+xxy

Example 7.3.6

Subtract:

1y1y3

Solution:

Since the LCD=y(y3), multiply the first term by 1 in the form of (y3)(y3) and the second term by yy.

1y1y3=1y(y3)(y3)1y3yy=(y3)y(y3)yy(y3)=y3yy(y3)=3y(y3) or =3y(y3)

Answer:

3y(y3)

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. For example, given

1x3(x+2)(x3) and 1x(x+2)2

there are three base factors in the denominator: x,(x+2), and (x3). The highest powers of these factors are x3,(x+2)2, and (x3)1. Therefore,

LCD=x3(x+2)2(x3)

The general steps for adding or subtracting rational expressions are illustrated in the following example.

Example 7.3.7

Subtract:

Solution:

Step 1: Factor all denominators to determine the LCD.

The LCD is (x+1)(x+3)(x5).

Step 2: Multiply by the appropriate factors to obtain equivalent terms with a common denominator. To do this, multiply the first term by (x5)(x5) and the second term by (x+3)(x+3).

=x(x+1)(x+3)(x5)(x5)3(x+1)(x5)(x+3)(x+3)=x(x5)(x+1)(x+3)(x5)3(x+3)(x+1)(x+3)(x5)

Step 3: Add or subtract the numerators and place the result over the common denominator.

=x(x5)3(x+3)(x+1)(x+3)(x5)

Step 4: Simplify the resulting algebraic fraction.

Answer:

(x9)(x+3)(x5)

Example 7.3.8

Subtract:

Solution:

It is best not to factor the numerator, x29x+18, because we will most likely need to simplify after we subtract.

Answer:

18(x4)(x9)

Example 7.3.9

Subtract:

1x2412x

Solution:

First, factor the denominators and determine the LCD. Notice how the opposite binomial property is applied to obtain a more workable denominator.

1x2412x=1(x+2)(x2)11(x2)=1(x+2)(x2)+1(x2)

The LCD is (x+2)(x2). Multiply the second term by 1 in the form of (x+2)(x+2).

=1(x+2)(x2)+1(x2)(x+2)(x+2)=1(x+2)(x2)+x+2(x2)(x+2)

Now that we have equivalent terms with a common denominator, add the numerators and write the result over the common denominator.

=1(x+2)(x2)+x+2(x2)(x+2)=1+x+2(x+2)(x2)=x+3(x+2)(x2)

Answer:

x+3(x+2)(x2)

Example 7.3.10

Simplify:

y1y+1y+1y1+y25y21

Solution:

Begin by factoring the denominator.

y1y+1y+1y1+y25y21=y1y+1y+1y1+y25(y+1)(y1)

We can see that the LCD is (y+1)(y1). Find equivalent fractions with this denominator.

Next, subtract and add the numerators and place the result over the common denominator.

Finish by simplifying the resulting rational expression.

Answer:

y5y1

Exercise 7.3.2

Simplify:

2x21+x1+x51x

Answer

x+3x1

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

Example 7.3.11

Simplify:

y2+(y1)1

Solution:

Recall that xn=1xn. We begin by rewriting the negative exponents as rational expressions.

y2+(y1)1=1y2+1(y1)1Replacenegativeexponents.=1y2(y1)(y1)+1(y1)y2y2Multiplybyfactorstoobtainequivalentexpressionswithacommondenominator.=(y1)y2(y1)+y2y2(y1)=(y1)+y2y2(y1)Addandsimplify.=y2+y1y2(y1)Thetrinomialdoesnotfactor.

Answer:

y2+y1y2(y1)

Adding and Subtracting Rational Functions

We can simplify sums or differences of rational functions using the techniques learned in this section. The restrictions of the result consist of the restrictions to the domains of each function.

Example 7.3.12

Calculate (f+g)(x), given f(x)=1x+3 and g(x)=1x2, and state the restrictions.

Solution:

(f+g)(x)=f(x)+g(x)=1x+3+1x2=1x+3(x2)(x2)+1x2(x+3)(x+3)=x2(x+3)(x2)+x+3(x2)(x+3)=x2+x+3(x+3)(x2)=2x+1(x+3)(x2)

Here the domain of f consists of all real numbers except 3, and the domain of g consists of all real numbers except 2. Therefore, the domain of f + g consists of all real numbers except 3 and 2.

Answer:

2x+1(x+3)(x2), where x3,2

Example 7.3.13

Calculate (fg)(x), given f(x)=x(x1)x225 and g(x)=x3x5, and state the restrictions to the domain.

Solution:

(fg)(x)=f(x)g(x)=x(x1)x225x3x5=x(x1)(x+5)(x5)(x3)(x5)(x+5)(x+5)=x(x1)(x3)(x+5)(x+5)(x5)=x2x(x2+5x3x15)(x+5)(x5)=x2x(x2+2x15)(x+5)(x5)=x2xx22x+15(x+5)(x5)=3x+15(x+5)(x5)=3(x5)(x+5)(x5)=3x+5

The domain of f consists of all real numbers except 5 and 5, and the domain of g consists of all real numbers except 5. Therefore, the domain of f − g consists of all real numbers except 5 and 5.

Answer:

3x+5, where x±5

Key Takeaways

  • When adding or subtracting rational expressions with a common denominator, add or subtract the expressions in the numerator and write the result over the common denominator.
  • To find equivalent rational expressions with a common denominator, first factor all denominators and determine the least common multiple. Then multiply numerator and denominator of each term by the appropriate factor to obtain a common denominator. Finally, add or subtract the expressions in the numerator and write the result over the common denominator.
  • The restrictions to the domain of a sum or difference of rational functions consist of the restrictions to the domains of each function.
Exercise 7.3.3 Adding and Subtracting with Common Denominators

Simplify. (Assume all denominators are nonzero.)

  1. 3x+7x
  2. 9x10x
  3. xy3y
  4. 4x3+6x3
  5. 72x1x2x1
  6. 83x83x3x8
  7. 2x9+x11x9
  8. y+22y+3y+32y+3
  9. 2x34x1x44x1
  10. 2xx13x+4x1+x2x1
  11. 13y2y93y135y3y
  12. 3y+25y10+y+75y103y+45y10
  13. x(x+1)(x3)3(x+1)(x3)
  14. 3x+5(2x1)(x6)x+6(2x1)(x6)
  15. xx236+6x236
  16. xx2819x281
  17. x2+2x2+3x28+x22x2+3x28
  18. x2x2x33x2x2x3
Answer

1. 10x

3. x3y

5. 7x2x1

7. 1

9. x+14x1

11. y1y

13. 1x+1

15. 1x6

17. x+5x+7

Exercise 7.3.4 Adding and Subtracting with Unlike Denominators

Simplify. (Assume all denominators are nonzero.)

  1. 12+13x
  2. 15x21x
  3. 112y2+310y3
  4. 1x12y
  5. 1y2
  6. 3y+24
  7. 2x+4+2
  8. 2y1y2
  9. 3x+1+1x
  10. 1x12x
  11. 1x3+1x+5
  12. 1x+21x3
  13. xx+12x2
  14. 2x3x+5xx3
  15. y+1y1+y1y+1
  16. 3y13yy+4y2
  17. 2x52x+52x+52x5
  18. 22x12x+112x
  19. 3x+4x828x
  20. 1y1+11y
  21. 2x2x29+x+159x2
  22. xx+3+1x315x(x+3)(x3)
  23. 2x3x113x+1+2(x1)(3x1)(3x+1)
  24. 4x2x+1xx5+16x3(2x+1)(x5)
  25. x3x+2x2+43x(x2)
  26. 2xx+63x6x18(x2)(x+6)(x6)
  27. xx+51x7257x(x+5)(x7)
  28. xx22x3+2x3
  29. 1x+5x2x225
  30. 5x2x242x2
  31. 1x+16x3x27x8
  32. 3x9x21613x+4
  33. 2xx21+1x2+x
  34. x(4x1)2x2+7x4x4+x
  35. 3x23x2+5x22x3x1
  36. 2xx411x+4x22x8
  37. x2x+1+6x242x27x4
  38. 1x2x6+1x23x10
  39. xx2+4x+33x24x5
  40. y+12y2+5y3y4y21
  41. y1y2252y210y+25
  42. 3x2+24x22x812x4
  43. 4x2+28x26x728x7
  44. a4a+a29a+18a213a+36
  45. 3a12a28a+16a+24a
  46. a2142a27a451+2a
  47. 1x+3xx26x+9+3x29
  48. 3xx+72xx2+23x10x2+5x14
  49. x+3x1+x1x+2x(x+11)x2+x2
  50. 2x3x+14x2+4(x+5)3x25x2
  51. x14x1x+32x+33(x+5)8x2+10x3
  52. 3x2x322x+36x25x94x29
  53. 1y+1+1y+2y21
  54. 1y1y+1+1y1
  55. 52+21
  56. 61+42
  57. x1+y1
  58. x2y1
  59. (2x1)1x2
  60. (x4)1(x+1)1
  61. 3x2(x1)12x
  62. 2(y1)2(y1)1
Answer

1. 3x+26x

3. 5y+1860y3

5. 12yy

7. 2(x+5)x+4

9. 4x+1x(x+1)

11. 2(x+1)(x3)(x+5)

13. x24x2(x2)(x+1)

15. 2(y2+1)(y+1)(y1)

17. 40x(2x+5)(2x5)

19. 3(x+2)x8

21. 2x+5x+3

23. 2x+13x+1

25. x2+4x+43x(x2)

27. x6x7

29. x5x2(x+5)(x5)

31. 5x8

33. 2x1x(x1)

35. x(x4)(x+2)(3x1)

37. x+62x+1

39. x9(x5)(x+3)

41. y28y5(y+5)(y5)2

43. 4xx+1

45. a+5a4

47. 6x(x+3)(x3)2

49. x7x+2

51. x54x1

53. 2y1y(y1)

55. 2750

57. x+yxy

59. (x1)2x2(2x1)

61. x(x+2)x1

Exercise 7.3.5 Adding and Subtracting Rational Functions

Calculate (f+g)(x) and (fg)(x) and state the restrictions to the domain.

  1. f(x)=13x and g(x)=1x2
  2. f(x)=1x1 and g(x)=1x+5
  3. f(x)=xx4 and g(x)=14x
  4. f(x)=xx5 and g(x)=12x3
  5. f(x)=x1x24 and g(x)=4x26x16
  6. f(x)=5x+2 and g(x)=3x+4
Answer

1. (f+g)(x)=2(2x1)3x(x2);(fg)(x)=2(x+1)3x(x2);x0,2

3. (f+g)(x)=x1x4;(fg)(x)=x+1x4;x4

5. (f+g)(x)=x(x5)(x+2)(x2)(x8);(fg)(x)=x213x+16(x+2)(x2)(x8);x2,2,8

Exercise 7.3.6 Adding and Subtracting Rational Functions

Calculate (f+f)(x) and state the restrictions to the domain.

  1. f(x)=1x
  2. f(x)=12x
  3. f(x)=x2x1
  4. f(x)=1x+2
Answer

1. (f+f)(x)=2x;x0

3. (f+f)(x)=2x2x1;x12

Exercise 7.3.7 Discussion Board
  1. Explain to a classmate why this is incorrect: 1x2+2x2=32x2.
  2. Explain to a classmate how to find the common denominator when adding algebraic expressions. Give an example.
Answer

1. Answer may vary


This page titled 7.3: Adding and Subtracting Rational Expressions is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by Anonymous via source content that was edited to the style and standards of the LibreTexts platform.

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