ADDING AND SUBTRACTING RATIONAL EXPRESSIONS WITH LIKE DENOMINATOR

Addition of Rational expression :

To add rational expressions with a like denominators, we will put only one denominator and add the numerators. If it is possible, reduce it into lowest terms and get the result.

If p, q, and r are polynomials where r ≠ 0, then

Subtraction of Rational expression :

To subtract rational expressions we will put only one denominator and subtract the numerators. If it is possible, reduce it into lowest terms and get the result.

If p, q, and r are polynomials where r ≠ 0, then

Problem 1 :

Solution :

Since the denominators are the same, add the numerators.

= (3b + 5b) / b²

= 8b / b²

= 8/b

Problem 2 :

Solution :

Since the denominators are the same, add the numerators.

= (5x + 50) / (x + 10)

Factoring 5 from the numerator, we get

= 5(x + 10) / (x + 10)

= 5

Problem 3 :

Solution :

Since the denominators are the same, subtract the numerators.

= [(5a + 2) - (2a - 4)]/(a² - 4)

= [5a + 2 - 2a + 4)]/(a² - 4)

= (3a + 6)/(a² - 4)

Factoring 3 from the numerator, we get

= 3(a + 2) / (a² - 2²)

= 3(a + 2) / (a + 2)(a - 2)

= 3 / (a - 2)

Problem 4 :

Solution :

Since the denominators are the same, subtract the numerators.

= [(3m - 6) - (-m + 2)] / (m² + m - 6)

= [3m - 6 + m - 2] / (m² + m - 6)

= [4m - 8] / (m² + m - 6)

Factoring 4 from the numerator and factoring the trinomial at the denominator, we get

= 4(m - 2) / (m + 3)(m - 2)

= 4/(m + 3)

Problem 5  :

Solution :

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

= (2x - 6 + 3x + 12)/5x

= (5x + 6)/5x

So, option B is correct.

Problem 6 :

Solution :

= (14q + 9 + 1 - 10q)/4q

= (4q + 10)/4q

Factoring 2 from the numerator, we get

= 2(2q + 5) / 4q

= (2q + 5) / 2q

Problem 7 :

Solution :

= [(x² + 3x) + (2x + 6)] / (x + 2)

= [x² + 3x + 2x + 6] / (x + 2)

= [x² + 5x + 6] / (x + 2)

Factoring the trinomial at the numerator, we get

= (x + 2)(x + 3) / (x + 2)

= x + 3

Problem 8 :

Solution :

= [(x - 1) + (x + 7)] / (x + 3)

= (x - 1 + x + 7) / (x + 3)

= (2x + 6) / (x + 3)

Factoring 2 from the numerator, we get

= 2(x + 3) / (x + 3)

= 2

Problem 9 :

If

(6x² - 5x + 4)/(-3x + 1) = -2x + 1 + [A/(-3x + 1)]

what is the value of A ?

Solution :

(6x² - 5x + 4)/(-3x + 1) = -2x + 1 + [A/(-3x + 1)]

Considering the right side,

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

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

= [(6x² - 5x + 1) + A] / (-3x + 1)

Combining the constants, 

= [6x² - 5x + (1 + A)] / (-3x + 1)

Comparing the like terms, we get

(6x² - 5x + 4)/(-3x + 1) = [6x² - 5x + (1 + A)] / (-3x + 1)

1 + A = 4

A = 4 - 1

A = 3

So, the value of A is 3.

Problem 10 :

(x² - x - a) / (x - 2) = (x + 1) + [8/(x - 2)]

In the equation above, what is the value of a ?

Solution :

(x² - x - a) / (x - 2) = (x + 1) + [8/(x - 2)]

Comparing the right side,

= (x + 1) + [8/(x - 2)]

= [(x + 1)(x - 2) + 8]/(x - 2)

= [(x² + x - 2x - 2) + 8]/(x - 2)

= [x² - x - 2 + 8]/(x - 2)

= [x² - x + 6]/(x - 2)

Comparing the corresponding terms, we get

-a = 6

a = 6

So, the value of a is 6.

Problem 11 :

(14x² + 9x - 20) / (ax - 1) = (7x + 8) + [-12/(ax - 1)]

In the equation above, a is a constant and ax - 1 ≠ 0, what is the value of a ?

Solution :

(14x² + 9x - 20) / (ax - 1) = (7x + 8) + [-12/(ax - 1)]

= [(7x + 8)(ax - 1) - 12]/(ax - 1)

= [(7ax² - 7x + 8ax - 8) - 12]/(ax - 1)

= [(7ax² - 7x + 8ax - 20]/(ax - 1)

Comparing the corresponding terms, we get

(14x² + 9x - 20) = [(7ax² - 7x + 8ax - 20]

7a = 14

a = 14/7

a = 2

So, the value of a is 2.

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