ADDING AND SUBTRACTING RATIONAL EXPRESSIONS WITH UNLIKE DENOMINATOR

Problem 1 :

Solution :

The denominators are almost same, using the negative sign in the middle, we get

Problem 2 :

Solution :

(a-4) and (4-a) both are almost same.

By factoring the negative sign from (4-a), we get -(4-a).

= a²/(a - 4) - 4²/(a - 4)

= (a²- 4²)/(a - 4)

Expanding a²- 4² using the algebraic identity 

a²- b² = (a + b)(a - b)

= (a + 4)(a - 4)/(a - 4)

= a + 4

Problem 3 :

Solution :

Since the denominators are not the same, we are using cross multiplication.

= [(x - 1)(x - 2) + (x + 3)(x + 6)] / (x + 6)(x - 2)

= [(x² - 2x -1x + 2) + (x² + 6x + 3x + 18)] / (x + 6)(x - 2)

= [x² - 3x + 2 + (x² + 9x + 18] / (x + 6)(x - 2)

= (2x² + 6x + 20) / (x + 6)(x - 2)

= 2(x² + 3x + 10) / (x + 6)(x - 2)

= 2(x² + 3x + 10) / (x + 6)(x - 2)

Problem 4 :

Solution :

Factoring 3 from the denominator of the first fraction and factoring 2 from the denominator of the second fraction, we get

= (y - 5) / 3(y + 3) -  (y + 1) / 2(y + 3)

= [2(y - 5) - 3(y + 1)] / 6(y + 3)

= (2y - 10 - 3y - 3) / 6(y + 3)

= (-y - 13) / 6(y + 3)

= -(y + 13) / 6(y + 3)

Problem 5 :

Solution :

Since the denominators are not the same, we are taking the common factor of 2b + 6, we get

= [(b + 3) /2 (b + 3)] - 2/3b

= 1/2 - 2/3b

= (3b - 4)/6b

Problem 6 :

Solution :

Factoring negative from the denominator of the second fraction, we get

= (y + 7)/(y² - 49) + (3y + 1)/(y² - 49)

= [(y + 7) + (3y + 1)]/(y² - 49)

= (4y + 8)/(y² - 49)

Factoring 2 from the numerator and using algebraic identity (a² - b²) = (a + b)(a - b)

= 4(y + 2)/(y + 7)(y - 7)

Problem 7 :

Solution :

= (a - 5) / a(a - 5) + (a + 5) / (a² - 5²)

= [(a - 5) / a(a - 5)] + [(a + 5) / (a + 5)(a - 5)]

= [(a - 5)(a + 5) + a(a + 5)] / a(a - 5)(a + 5)

= [a² -5a + 5a - 25 + a² + 5a] / a(a - 5)(a + 5)

= [2a² + 5a - 25] / a(a - 5)(a + 5)

= [2a² + 5a - 25] / a(a - 5)(a + 5)

= (a + 5)(2a - 5) / a(a - 5)(a + 5)

= (2a - 5) / a(a - 5)

Problem 8 :

Solution :

= [(b - 3) / (b² - 3²)] + [(b + 3) / (b + 3)(b + 3)]

= [(b - 3) / (b + 3)(b - 3)] + [(b + 3) / (b + 3)(b + 3)]

= 1/(b + 3) + 1/(b + 3)

= (1 + 1)/(b + 3)

= 2/(b + 3)

Problem 9 :

Solution :

Since the denominators are not the same, we are using the least common multiple.

= [(x + 2)/x(x + 1)] - (1/x) + 3/(x + 1)

LCM = x(x + 1)

= [x + 2 - (x + 1) + 3x] / x(x + 1)

= [x + 2 - x + 1 + 3x] / x(x + 1)

= (3x + 3) / x(x + 1)

= 3(x + 1)/x(x + 1)

Cancelling the common factor, we get

= 3/x

Problem 10 :

Solution :

By factoring the denominators, we get

x² + 3x = x (x + 3)

x² - x - 12 = (x - 4)(x + 3)

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

LCM = x(x + 3)(x - 4)

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

= [5x - 20 - 4x]/x(x + 3)(x - 4)

= (x - 20)/x(x + 3)(x - 4)

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