ADDITION AND SUBTRACTION OF RATIONAL EXPRESSIONS

Question 1 :

Simplify

(i) [x (x + 1)/(x - 2)] + [x(1 - x)/(x - 2)]

Solution :

Since the denominator of both fractions are same, we have to put only one denominator and add the fractions.

(ii)  (x + 2)/(x + 3) + (x - 1)/(x - 2)

Solution :

(iii)  [x³/(x - y)] + [y³/(y - x)]

Solution :

  =  [x³/(x - y)] + [y³/(y - x)]

  =  [x³/(x - y)] - [y³/(x - y)]

  =  (x³ - y³)/(x - y)

  =  (x - y) (x² + xy + y²)/(x - y)

  =  (x² + xy + y²)

Hence the value of the given rational expression is (x²+xy+y²).

Question 2 :

Simplify 

(i)  [(2x + 1)(x - 2)/(x - 4)] - [(2x² - 5x + 2)/(x - 4)]

Solution :

  =  [(2x + 1)(x - 2)/(x - 4)] - [(2x² - 5x + 2)/(x - 4)]

  =  [(2x² - 4x + x - 2) - (2x² - 5x + 2)]/(x - 4)

  =  [(2x² - 3x - 2 - 2x² + 5x - 2)]/(x - 4)

  =  (2x - 4)/(x - 4)

  =  2(x - 2)/(x - 4)

(ii)  4x/(x² - 1) - (x + 1)/(x - 1)

Solution :

  =  4x/(x² - 1) - (x + 1)/(x - 1)

  =  4x/(x + 1)(x - 1) - (x + 1)/(x - 1)

Taking L.C.M, we get

  =  4x/(x + 1)(x - 1) - (x + 1)(x + 1)/(x - 1)(x + 1)

  =  [4x - (x + 1)²]/(x + 1)(x - 1)

  =  [4x - (x² + 2x + 1)]/(x + 1)(x - 1)

  =  [4x - x² - 2x - 1]/(x + 1)(x - 1)

  =  (-x² + 2x - 1)/(x + 1)(x - 1)

  =  -(x² - 2x + 1)/(x + 1)(x - 1)

  =  -(x - 1)(x - 1)/(x + 1)(x - 1)

  =  -(x - 1)/(x + 1)

  =  (1 - x)/(1 + x)

Question 3 :

Subtract 1/(x² + 2) from (2x³ + x² + 3)/(x² + 2)²

Solution :

  =  [(2x³ + x² + 3)/(x² + 2)²] - [1/(x² + 2)]

By taking LCM, we get

  =  [(2x³ + x² + 3)/(x² + 2)²] - [(x² + 2)/(x² + 2)²]

  =  [(2x³ + x² + 3) - (x² + 2)/(x² + 2)²]

=  (2x³ + x² + 3 - x² - 2)/(x² + 2)²

=  (2x³+ 1)/(x² + 2)²

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