- 1. Addition and Subtraction II
- 2. Addition and Subtraction II Often we multiply the numerator and the denominator by –1 to change the denominator to it’s opposite,
- 3. a b Addition and Subtraction II Often we multiply the numerator and the denominator by –1 to change the denominator to it’s opposite, i.e. –a –b
- 4. Example A. Switch the following fractions to the opposite denominator. a. 2 3 = a b b. a – b = d. x + y 2x – y = Addition and Subtraction II Often we multiply the numerator and the denominator by –1 to change the denominator to it’s opposite, i.e. –3 –a –b
- 5. Example A. Switch the following fractions to the opposite denominator. a. 2 3 = a b –2 b. a – b = d. x + y 2x – y = Addition and Subtraction II Often we multiply the numerator and the denominator by –1 to change the denominator to it’s opposite, i.e. –3 –3 –a –b
- 6. Example A. Switch the following fractions to the opposite denominator. a. 2 3 = a b –2 b. a – b = –(b – a) 3 d. x + y 2x – y = Addition and Subtraction II Often we multiply the numerator and the denominator by –1 to change the denominator to it’s opposite, i.e. –3 –3 –a –b = b – a 3
- 7. Example A. Switch the following fractions to the opposite denominator. a. 2 3 = a b –2 b. a – b = –(b – a) 3 d. x + y 2x – y = –x – y y – 2x Addition and Subtraction II Often we multiply the numerator and the denominator by –1 to change the denominator to it’s opposite, i.e. –3 –3 –a –b = b – a 3
- 8. Example A. Switch the following fractions to the opposite denominator. a. 2 3 = a b –2 b. a – b = –(b – a) 3 d. x + y 2x – y = –x – y y – 2x Addition and Subtraction II Often we multiply the numerator and the denominator by –1 to change the denominator to it’s opposite, i.e. –3 –3 –a –b = b – a 3 When combining two fractions with opposite denominators, we may switch one of them to make their denominators the same.
- 9. Example B. Switch one of the denominators then combine. x + y 2x – y + x – 3y y – 2x Addition and Subtraction II
- 10. Example B. Switch one of the denominators then combine. x + y 2x – y + x – 3y y – 2x Opposites denominators Addition and Subtraction II
- 11. Example B. Switch one of the denominators then combine. x + y 2x – y + x – 3y y – 2x Opposites denominators = x + y 2x – y + 3y – x 2x – y switched Addition and Subtraction II
- 12. Example B. Switch one of the denominators then combine. x + y 2x – y + x – 3y y – 2x Opposites denominators = x + y 2x – y + 3y – x 2x – y switched x + y + 3y – x 2x – y = Addition and Subtraction II
- 13. Example B. Switch one of the denominators then combine. x + y 2x – y + x – 3y y – 2x Opposites denominators = x + y 2x – y + 3y – x 2x – y switched x + y + 3y – x 2x – y = 4y 2x – y= Addition and Subtraction II
- 14. Example B. Switch one of the denominators then combine. x + y 2x – y + x – 3y y – 2x Opposites denominators = x + y 2x – y + 3y – x 2x – y switched x + y + 3y – x 2x – y = 4y 2x – y= Addition and Subtraction II Another way to switch to a denominator to its opposite is to pull out a “–” and pass it to the numerator.
- 15. Example B. Switch one of the denominators then combine. x + y 2x – y + x – 3y y – 2x Opposites denominators = x + y 2x – y + 3y – x 2x – y switched x + y + 3y – x 2x – y = 4y 2x – y= Addition and Subtraction II Another way to switch to a denominator to its opposite is to pull out a “–” and pass it to the numerator. Specifically, for polynomials in x, make sure the leading term is positive.
- 16. Example B. Switch one of the denominators then combine. x + y 2x – y + x – 3y y – 2x Opposites denominators = x + y 2x – y + 3y – x 2x – y switched x + y + 3y – x 2x – y = 4y 2x – y= Addition and Subtraction II Example C. combine x + 3 (4 – x2) + 2x – 1 (x2 – 3x + 2) Another way to switch to a denominator to its opposite is to pull out a “–” and pass it to the numerator. Specifically, for polynomials in x, make sure the leading term is positive.
- 17. Example B. Switch one of the denominators then combine. x + y 2x – y + x – 3y y – 2x Opposites denominators = x + y 2x – y + 3y – x 2x – y switched x + y + 3y – x 2x – y = 4y 2x – y= Addition and Subtraction II Example C. combine x + 3 (4 – x2) + 2x – 1 (x2 – 3x + 2) Another way to switch to a denominator to its opposite is to pull out a “–” and pass it to the numerator. Specifically, for polynomials in x, make sure the leading term is positive. We write the denominator 4 – x2 as –(x2 – 4)
- 18. Example B. Switch one of the denominators then combine. x + y 2x – y + x – 3y y – 2x Opposites denominators = x + y 2x – y + 3y – x 2x – y switched x + y + 3y – x 2x – y = 4y 2x – y= Addition and Subtraction II Example C. combine x + 3 (4 – x2) + 2x – 1 (x2 – 3x + 2) as Another way to switch to a denominator to its opposite is to pull out a “–” and pass it to the numerator. Specifically, for polynomials in x, make sure the leading term is positive. x + 3 (4 – x2) We write the denominator 4 – x2 as –(x2 – 4) then write x + 3 –(x2 – 4)
- 19. Example B. Switch one of the denominators then combine. x + y 2x – y + x – 3y y – 2x Opposites denominators = x + y 2x – y + 3y – x 2x – y switched x + y + 3y – x 2x – y = 4y 2x – y= Addition and Subtraction II Example C. combine x + 3 (4 – x2) + 2x – 1 (x2 – 3x + 2) as Another way to switch to a denominator to its opposite is to pull out a “–” and pass it to the numerator. Specifically, for polynomials in x, make sure the leading term is positive. x + 3 (4 – x2) We write the denominator 4 – x2 as –(x2 – 4) then write x + 3 –(x2 – 4) –(x + 3) x2 – 4
- 20. Example B. Switch one of the denominators then combine. x + y 2x – y + x – 3y y – 2x Opposites denominators = x + y 2x – y + 3y – x 2x – y switched x + y + 3y – x 2x – y = 4y 2x – y= Addition and Subtraction II Example C. combine x + 3 (4 – x2) + 2x – 1 (x2 – 3x + 2) as Another way to switch to a denominator to its opposite is to pull out a “–” and pass it to the numerator. Specifically, for polynomials in x, make sure the leading term is positive. x + 3 (4 – x2) We write the denominator 4 – x2 as –(x2 – 4) then write x + 3 –(x2 – 4) –(x + 3) x2 – 4 = –x – 3 x2 – 4
- 21. x + 3 (4 – x2) + 2x – 1 (x2 – 3x + 2) Switch to opposite denominator Addition and Subtraction II
- 22. x + 3 (4 – x2) + 2x – 1 (x2 – 3x + 2) Switch to opposite denominator and pass the “–” to the top. = –x – 3 (x2 – 4) + 2x – 1 (x2 – 3x + 2) Addition and Subtraction II
- 23. x + 3 (4 – x2) + 2x – 1 (x2 – 3x + 2) = –x – 3 (x2 – 4) + 2x – 1 (x2 – 3x + 2) = Addition and Subtraction II (x – 2)(x + 2)+ 2x – 1 (x – 2)(x – 1) –x – 3 Switch to opposite denominator and pass the “–” to the top.
- 24. x + 3 (4 – x2) + 2x – 1 (x2 – 3x + 2) = –x – 3 (x2 – 4) + 2x – 1 (x2 – 3x + 2) = The LCD is (x – 2)(x + 2)(x – 1), Addition and Subtraction II (x – 2)(x + 2)+ 2x – 1 (x – 2)(x – 1) –x – 3 Switch to opposite denominator and pass the “–” to the top.
- 25. x + 3 (4 – x2) + 2x – 1 (x2 – 3x + 2) = –x – 3 (x2 – 4) + 2x – 1 (x2 – 3x + 2) = (x – 2)(x + 2)+ 2x – 1 (x – 2)(x – 1) The LCD is (x – 2)(x + 2)(x – 1), hence Addition and Subtraction II –x – 3 (x – 2)(x + 2)+ 2x – 1 (x – 2)(x – 1) –x – 3 Switch to opposite denominator and pass the “–” to the top.
- 26. x + 3 (4 – x2) + 2x – 1 (x2 – 3x + 2) = –x – 3 (x2 – 4) + 2x – 1 (x2 – 3x + 2) = (x – 2)(x + 2)+ 2x – 1 (x – 2)(x – 1) The LCD is (x – 2)(x + 2)(x – 1), hence Addition and Subtraction II –x – 3 (x – 2)(x + 2)+ 2x – 1 (x – 2)(x – 1) –x – 3 [ ] * (x – 2)(x + 2)(x – 1) LCD Switch to opposite denominator and pass the “–” to the top.
- 27. x + 3 (4 – x2) + 2x – 1 (x2 – 3x + 2) = –x – 3 (x2 – 4) + 2x – 1 (x2 – 3x + 2) = (x – 2)(x + 2)+ 2x – 1 (x – 2)(x – 1) The LCD is (x – 2)(x + 2)(x – 1), hence Addition and Subtraction II –x – 3 (x – 2)(x + 2)+ 2x – 1 (x – 2)(x – 1) –x – 3 [ ] * (x – 2)(x + 2)(x – 1) LCD Switch to opposite denominator and pass the “–” to the top. Distribute
- 28. x + 3 (4 – x2) + 2x – 1 (x2 – 3x + 2) = –x – 3 (x2 – 4) + 2x – 1 (x2 – 3x + 2) = (x – 2)(x + 2)+ 2x – 1 (x – 2)(x – 1) The LCD is (x – 2)(x + 2)(x – 1), hence Addition and Subtraction II –x – 3 (x – 2)(x + 2)+ 2x – 1 (x – 2)(x – 1) –x – 3 [ ]* (x – 2)(x + 2)(x – 1) LCD (x – 1) Switch to opposite denominator and pass the “–” to the top.
- 29. x + 3 (4 – x2) + 2x – 1 (x2 – 3x + 2) = –x – 3 (x2 – 4) + 2x – 1 (x2 – 3x + 2) = (x – 2)(x + 2)+ 2x – 1 (x – 2)(x – 1) The LCD is (x – 2)(x + 2)(x – 1), hence Addition and Subtraction II –x – 3 (x – 2)(x + 2)+ 2x – 1 (x – 2)(x – 1) –x – 3 [ ]* (x – 2)(x + 2)(x – 1) LCD (x – 1) (x + 2) Switch to opposite denominator and pass the “–” to the top.
- 30. x + 3 (4 – x2) + 2x – 1 (x2 – 3x + 2) = –x – 3 (x2 – 4) + 2x – 1 (x2 – 3x + 2) = (x – 2)(x + 2)+ 2x – 1 (x – 2)(x – 1) The LCD is (x – 2)(x + 2)(x – 1), hence Addition and Subtraction II –x – 3 (x – 2)(x + 2)+ 2x – 1 (x – 2)(x – 1) –x – 3 [ ]* (x – 2)(x + 2)(x – 1) LCD (x – 1) (x + 2) = [(–x – 3)(x – 1) + (2x – 1)(x + 2)] LCD Switch to opposite denominator and pass the “–” to the top.
- 31. x + 3 (4 – x2) + 2x – 1 (x2 – 3x + 2) = –x – 3 (x2 – 4) + 2x – 1 (x2 – 3x + 2) = (x – 2)(x + 2)+ 2x – 1 (x – 2)(x – 1) The LCD is (x – 2)(x + 2)(x – 1), hence Addition and Subtraction II –x – 3 (x – 2)(x + 2)+ 2x – 1 (x – 2)(x – 1) –x – 3 [ ]* (x – 2)(x + 2)(x – 1) LCD (x – 1) (x + 2) = [(–x – 3)(x – 1) + (2x – 1)(x + 2)] LCD = –x2 – 2x + 3 + 2x2 + 3x – 2[ ] LCD Switch to opposite denominator and pass the “–” to the top.
- 32. x + 3 (4 – x2) + 2x – 1 (x2 – 3x + 2) = –x – 3 (x2 – 4) + 2x – 1 (x2 – 3x + 2) = (x – 2)(x + 2)+ 2x – 1 (x – 2)(x – 1) The LCD is (x – 2)(x + 2)(x – 1), hence Addition and Subtraction II –x – 3 (x – 2)(x + 2)+ 2x – 1 (x – 2)(x – 1) –x – 3 [ ]* (x – 2)(x + 2)(x – 1) LCD (x – 1) (x + 2) = [(–x – 3)(x – 1) + (2x – 1)(x + 2)] LCD = –x2 – 2x + 3 + 2x2 + 3x – 2[ ] LCD = x2 + x + 1 (x + 2)(x – 2)(x – 1) Switch to opposite denominator and pass the “–” to the top.
- 33. x + 3 (4 – x2) + 2x – 1 (x2 – 3x + 2) = –x – 3 (x2 – 4) + 2x – 1 (x2 – 3x + 2) = (x – 2)(x + 2)+ 2x – 1 (x – 2)(x – 1) The LCD is (x – 2)(x + 2)(x – 1), hence Addition and Subtraction II –x – 3 (x – 2)(x + 2)+ 2x – 1 (x – 2)(x – 1) –x – 3 [ ]* (x – 2)(x + 2)(x – 1) LCD (x – 1) (x + 2) = [(–x – 3)(x – 1) + (2x – 1)(x + 2)] LCD = –x2 – 2x + 3 + 2x2 + 3x – 2[ ] LCD = x2 + x + 1 (x + 2)(x – 2)(x – 1) This is the simplified answer. Switch to opposite denominator and pass the “–” to the top.
- 34. Addition and Subtraction II We’re less likely to make mistakes in a subtraction problem if the problem is changed to an addition problem.
- 35. Addition and Subtraction II We’re less likely to make mistakes in a subtraction problem if the problem is changed to an addition problem. We do this by distributing the subtraction as a negative sign to the numerator.
- 36. Example D. Combine x + 3 (x2 – x – 2 ) x – 2 (x2 – 2x – 3 ) – Addition and Subtraction II We’re less likely to make mistakes in a subtraction problem if the problem is changed to an addition problem. We do this by distributing the subtraction as a negative sign to the numerator.
- 37. Example D. Combine x + 3 (x2 – x – 2 ) x – 2 (x2 – 2x – 3 ) = – Addition and Subtraction II x + 3 (x2 – x – 2 ) x – 2 (x2 – 2x – 3 ) – x + 3 (x2 – x – 2 ) –(x – 2) (x2 – 2x – 3 ) + distribute the subtraction to the numerator We’re less likely to make mistakes in a subtraction problem if the problem is changed to an addition problem. We do this by distributing the subtraction as a negative sign to the numerator.
- 38. Example D. Combine x + 3 (x2 – x – 2 ) x – 2 (x2 – 2x – 3 ) = – Addition and Subtraction II x + 3 (x2 – x – 2 ) x – 2 (x2 – 2x – 3 ) – x + 3 (x2 – x – 2 ) –(x – 2) (x2 – 2x – 3 ) + distribute the subtraction to the numerator = x + 3 (x + 1)(x – 2 ) –x + 2 (x + 1)(x – 3 ) + We’re less likely to make mistakes in a subtraction problem if the problem is changed to an addition problem. We do this by distributing the subtraction as a negative sign to the numerator.
- 39. Example D. Combine x + 3 (x2 – x – 2 ) x – 2 (x2 – 2x – 3 ) = – LCD = (x + 1)(x – 2)(x – 3) Addition and Subtraction II We’re less likely to make mistakes in a subtraction problem if the problem is changed to an addition problem. We do this by distributing the subtraction as a negative sign to the numerator. x + 3 (x2 – x – 2 ) x – 2 (x2 – 2x – 3 ) – x + 3 (x2 – x – 2 ) –(x – 2) (x2 – 2x – 3 ) + distribute the subtraction to the numerator = x + 3 (x + 1)(x – 2 ) –x + 2 (x + 1)(x – 3 ) +
- 40. Example D. Combine x + 3 (x2 – x – 2 ) x – 2 (x2 – 2x – 3 ) = – LCD = (x + 1)(x – 2)(x – 3) Addition and Subtraction II x + 3 (x2 – x – 2 ) x – 2 (x2 – 2x – 3 ) – x + 3 (x2 – x – 2 ) –(x – 2) (x2 – 2x – 3 ) + distribute the subtraction to the numerator = x + 3 (x + 1)(x – 2 ) –x + 2 (x + 1)(x – 3 ) + = x + 3 (x + 1)(x – 2 ) –x + 2 (x + 1)(x – 3 ) +[ ]*(x + 1)(x – 2)(x – 3) LCD We’re less likely to make mistakes in a subtraction problem if the problem is changed to an addition problem. We do this by distributing the subtraction as a negative sign to the numerator.
- 41. Example D. Combine x + 3 (x2 – x – 2 ) x – 2 (x2 – 2x – 3 ) = – LCD = (x + 1)(x – 2)(x – 3) Addition and Subtraction II x + 3 (x2 – x – 2 ) x – 2 (x2 – 2x – 3 ) – x + 3 (x2 – x – 2 ) (x2 – 2x – 3 ) + distribute the subtraction to the numerator = x + 3 (x + 1)(x – 2 ) –x + 2 (x + 1)(x – 3 ) + = x + 3 (x + 1)(x – 2 ) –x + 2 (x + 1)(x – 3 ) +[ ]*(x + 1)(x – 2)(x – 3) LCD (x – 3) –(x – 2) We’re less likely to make mistakes in a subtraction problem if the problem is changed to an addition problem. We do this by distributing the subtraction as a negative sign to the numerator.
- 42. Example D. Combine x + 3 (x2 – x – 2 ) x – 2 (x2 – 2x – 3 ) = – LCD = (x + 1)(x – 2)(x – 3) Addition and Subtraction II x + 3 (x2 – x – 2 ) x – 2 (x2 – 2x – 3 ) – x + 3 (x2 – x – 2 ) (x2 – 2x – 3 ) + distribute the subtraction to the numerator = x + 3 (x + 1)(x – 2 ) –x + 2 (x + 1)(x – 3 ) + = x + 3 (x + 1)(x – 2 ) –x + 2 (x + 1)(x – 3 ) +[ ]*(x + 1)(x – 2)(x – 3) LCD (x – 3) (x – 2) –(x – 2) We’re less likely to make mistakes in a subtraction problem if the problem is changed to an addition problem. We do this by distributing the subtraction as a negative sign to the numerator.
- 43. Example D. Combine x + 3 (x2 – x – 2 ) x – 2 (x2 – 2x – 3 ) = = [(x + 3)(x – 3) + (–x + 2)(x – 2)] – LCD = (x + 1)(x – 2)(x – 3) Addition and Subtraction II x + 3 (x2 – x – 2 ) x – 2 (x2 – 2x – 3 ) – x + 3 (x2 – x – 2 ) (x2 – 2x – 3 ) + distribute the subtraction to the numerator = x + 3 (x + 1)(x – 2 ) –x + 2 (x + 1)(x – 3 ) + = x + 3 (x + 1)(x – 2 ) –x + 2 (x + 1)(x – 3 ) +[ ]*(x + 1)(x – 2)(x – 3) LCD (x – 3) (x – 2) LCD –(x – 2) We’re less likely to make mistakes in a subtraction problem if the problem is changed to an addition problem. We do this by distributing the subtraction as a negative sign to the numerator.
- 44. Example D. Combine x + 3 (x2 – x – 2 ) x – 2 (x2 – 2x – 3 ) = = [(x + 3)(x – 3) + (–x + 2)(x – 2)] = [x2 – 9 – x2 + 4x – 4] – LCD = (x + 1)(x – 2)(x – 3) Addition and Subtraction II x + 3 (x2 – x – 2 ) x – 2 (x2 – 2x – 3 ) – x + 3 (x2 – x – 2 ) (x2 – 2x – 3 ) + distribute the subtraction to the numerator = x + 3 (x + 1)(x – 2 ) –x + 2 (x + 1)(x – 3 ) + = x + 3 (x + 1)(x – 2 ) –x + 2 (x + 1)(x – 3 ) +[ ]*(x + 1)(x – 2)(x – 3) LCD (x – 3) (x – 2) LCD LCD –(x – 2) We’re less likely to make mistakes in a subtraction problem if the problem is changed to an addition problem. We do this by distributing the subtraction as a negative sign to the numerator.
- 45. Example D. Combine x + 3 (x2 – x – 2 ) x – 2 (x2 – 2x – 3 ) = = [(x + 3)(x – 3) + (–x + 2)(x – 2)] = [x2 – 9 – x2 + 4x – 4] – LCD = (x + 1)(x – 2)(x – 3) Addition and Subtraction II x + 3 (x2 – x – 2 ) x – 2 (x2 – 2x – 3 ) – x + 3 (x2 – x – 2 ) (x2 – 2x – 3 ) + distribute the subtraction to the numerator = x + 3 (x + 1)(x – 2 ) –x + 2 (x + 1)(x – 3 ) + = x + 3 (x + 1)(x – 2 ) –x + 2 (x + 1)(x – 3 ) +[ ]*(x + 1)(x – 2)(x – 3) LCD (x – 3) (x – 2) LCD LCD 4x – 13 = (x + 1)(x – 2)(x – 3) –(x – 2) We’re less likely to make mistakes in a subtraction problem if the problem is changed to an addition problem. We do this by distributing the subtraction as a negative sign to the numerator.
- 46. Addition and Subtraction II The special case of combining two “easy” fractions
- 47. Addition and Subtraction II The special case of combining two “easy” fractions When adding or subtracting two easy fractions, we may use the cross multiplication method.
- 48. Addition and Subtraction II The special case of combining two “easy” fractions When adding or subtracting two easy fractions, we may use the cross multiplication method. That is, a b c d±
- 49. Addition and Subtraction II The special case of combining two “easy” fractions When adding or subtracting two easy fractions, we may use the cross multiplication method. That is, a b c d± = ad ±bc
- 50. Addition and Subtraction II The special case of combining two “easy” fractions When adding or subtracting two easy fractions, we may use the cross multiplication method. That is, a b c d± = ad ±bc bd
- 51. Addition and Subtraction II The special case of combining two “easy” fractions When adding or subtracting two easy fractions, we may use the cross multiplication method. That is, a b c d± = ad ±bc bd Example E. Combine x + 1 x– 2 – 3 x + 4
- 52. Addition and Subtraction II The special case of combining two “easy” fractions When adding or subtracting two easy fractions, we may use the cross multiplication method. That is, a b c d± = ad ±bc bd Example E. Combine x + 1 x– 2 – 3 x + 4 x + 1 x – 2 – 3 x + 4±
- 53. Addition and Subtraction II The special case of combining two “easy” fractions When adding or subtracting two easy fractions, we may use the cross multiplication method. That is, a b c d± = ad ±bc bd Example E. Combine x + 1 x– 2 – 3 x + 4 x + 1 x – 2 – 3 x + 4 =± (x + 1)(x + 4) – 3(x – 2)
- 54. Addition and Subtraction II The special case of combining two “easy” fractions When adding or subtracting two easy fractions, we may use the cross multiplication method. That is, a b c d± = ad ±bc bd Example E. Combine x + 1 x– 2 – 3 x + 4 x + 1 x – 2 – 3 x + 4 =± (x + 1)(x + 4) – 3(x – 2) (x – 2)(x + 4)
- 55. Addition and Subtraction II The special case of combining two “easy” fractions When adding or subtracting two easy fractions, we may use the cross multiplication method. That is, a b c d± = ad ±bc bd Example E. Combine x + 1 x– 2 – 3 x + 4 x + 1 x – 2 – 3 x + 4 =± (x + 1)(x + 4) – 3(x – 2) (x – 2)(x + 4) No cancellation!
- 56. Addition and Subtraction II The special case of combining two “easy” fractions When adding or subtracting two easy fractions, we may use the cross multiplication method. That is, a b c d± = ad ±bc bd Example E. Combine x + 1 x– 2 – 3 x + 4 x + 1 x – 2 – 3 x + 4 =± (x + 1)(x + 4) – 3(x – 2) (x – 2)(x + 4) = x2 + 5x + 4 – 3x + 6 (x – 2)(x + 4) No cancellation! Expand
- 57. Addition and Subtraction II The special case of combining two “easy” fractions When adding or subtracting two easy fractions, we may use the cross multiplication method. That is, a b c d± = ad ±bc bd Example E. Combine x + 1 x– 2 – 3 x + 4 x + 1 x – 2 – 3 x + 4 =± (x + 1)(x + 4) – 3(x – 2) (x – 2)(x + 4) = x2 + 5x + 4 – 3x + 6 (x – 2)(x + 4) = x2 + 2x + 10 (x – 2)(x + 4) No cancellation! Expand
- 58. Addition and Subtraction II The special case of combining two “easy” fractions When adding or subtracting two easy fractions, we may use the cross multiplication method. That is, a b c d± = ad ±bc bd Example E. Combine x + 1 x– 2 – 3 x + 4 x + 1 x – 2 – 3 x + 4 =± (x + 1)(x + 4) – 3(x – 2) (x – 2)(x + 4) = x2 + 5x + 4 – 3x + 6 (x – 2)(x + 4) = x2 + 2x + 10 (x – 2)(x + 4) This method won’t work well with example D. Their cross– multiplication is messy. No cancellation! Expand
- 59. Addition and Subtraction II The special case of combining two “easy” fractions When adding or subtracting two easy fractions, we may use the cross multiplication method. That is, a b c d± = ad ±bc bd Example E. Combine x + 1 x– 2 – 3 x + 4 x + 1 x – 2 – 3 x + 4 =± (x + 1)(x + 4) – 3(x – 2) (x – 2)(x + 4) = x2 + 5x + 4 – 3x + 6 (x – 2)(x + 4) = x2 + 2x + 10 (x – 2)(x + 4) No cancellation! This method won’t work well with example D. Their cross– multiplication is messy. Hence this is for two ± “easy” fractions. Expand
- 60. Ex. Combine and simplify the answers. –3 x – 3 + 2x –6 – 2x 3. 2x – 3 x – 3 – 5x + 4 5 – 15x 4. 3x + 1 6x – 4 – 2x + 3 2 – 3x5. –5x + 7 3x – 12+ 4x – 3 –2x + 86. 3x + 1 + x + 3 4 – x211. x2 – 4x + 4 x – 4 + x + 5 –x2 + x + 2 12. x2 – x – 6 3x + 1 + 2x + 3 9 – x213. x2 – x – 6 3x – 4 – 2x + 5 x2 – x – 6 14. –x2 + 5x + 6 3x + 4 + 2x – 3 –x2 – 2x + 3 15. x2 – x 5x – 4 – 3x – 5 1 – x216. x2 + 2x – 3 –3 2x – 1 + 2x 2 – 4x 1. 2x – 3 x – 2 + 3x + 4 5 – 10x 2. 3x + 1 2x – 5 – 2x + 3 5 – 10x 9. –3x + 2 3x – 12 + 7x – 2 –2x + 8 10. 3x + 5 3x –2 – x + 3 2 – 3x7. –5x + 7 3x – 4 + 4x – 3 –6x + 88. Addition and Subtraction II
- 61. 3x + 1 – x – 2 x2 – 4x +4 17. 4 – x2 x + 1 – 2x – 1 x2 + 3x + 3 18. x2 + 2x – 4 2x – 3 – x – 3 x2 – 6x + 8 19. x2 – 4 4x – 1 – 2x +3 x2 + 3x – 4 20. 1 – x2 2x – x + 4 x2 – 5x + 4 21. 16 – x2 2x + 1 – x +3 x2 + 25 22. x2 – 3x – 10 Addition and Subtraction II