ADDING AND SUBTRACTING RATIONAL EXPRESSIONS

How to add or subtract rational expressions with like denominators : 

1. To add or subtract rational expressions with like denominators, take the denominator once and combine the numerators. 

2. Simplify, if needed.

How to add or subtract rational expressions with unlike denominators : 

1. To add or subtract rational expressions with unlike denominators, find the least common multiple, or LCM of the denominators.   

2. Write each rational expression using LCM and make sure that each rational expression has the LCM as denominator. 

3. Take the denominator once and combine the numerators. 

4. Simplify, if needed. 

Example 1 :

Simplify :

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

Solution :

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

Since the denominators are same, take the denominator once and combine the numerators. 

=  [(x + 2) + (x - 1)]/(x + 3)]

=  (x + 2 + x - 1)/(x + 3)

=  (2x + 1)/(x + 3)

Example 2 :

Simplify : 

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

Solution :

=  [(x + 1)/(x - 1)2] + [1/(x + 1)]

Since the denominators are not the same, find the LCM.

The LCM of (x - 1)2 and (x + 1) is (x - 1)2(x + 1).

Rewrite the expressions using the LCM.

First expression : 

=  (x + 1)/(x - 1)2

=  [(x + 1)(x + 1)]/[(x - 1)2(x + 1)]

=  (x + 1)2/(x - 1)2(x + 1)

Second expression :

=  1/(x + 1)

=  1(x - 1)2/[(x + 1)(x - 1)2]

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

Add :

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

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

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

=  (2x2 + 2) / [(x - 1)2(x + 1)]

Example 3 :

Simplify :

[(4x + 3)/(x2 + 3x + 2)] + [(x + 2)/(x2 + 3x + 2)]

Solution :

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

Since the denominators are same, take the denominator once and combine the numerators. 

[(4x + 3) + (x + 2)]/(x2 + 3x + 2)

=  (4x + 3 + x + 2)/(x2 + 3x + 2)

=  (5x + 5)/(x2 + 3x + 2)

Factor the numerator and denominator. 

=  5(x + 1)/[(x + 1)(x + 2)]

Cancel common factors. 

=  5/(x + 2)

Example 4 :

Simplify :

1/(x2 + 5x + 6) - 1/(x2 + 6x + 8)

Solution :

=  1/(x2 + 5x + 6) - 1/(x2 + 6x + 8)

The denominators are not same. To find LCM of the denominators, factor both the denominators. 

=  1/[(x + 2)(x + 3)] + 1/[(x + 2)(x + 4)]

The LCM of denominators is (x + 2)(x + 3)(x + 4). 

Rewrite the expressions using the LCM.

First expression : 

=  1/[(x + 2)(x + 3)]

(x + 4)]/[(x + 2)(x + 3)(x + 4)]

Second expression :

=  1/(x + 2)(x + 4)

(x + 3)/[(x + 2)(x + 3)(x + 4)]

Now, the denominators are same. Take the denominator once and subtract the numerators. 

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

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

=  1/[(x + 2)(x + 3)(x + 4)]

Example 5 :

Simplify :

[x3/(x - 2)] + [8/(2 - x)]

Solution :

=  [x3/(x - 2)] + [8/(2 - x)]

Make the denominators same. 

=  [x3/(x - 2)] + [-8/(x - 2)]

Since the denominators are same, take the denominator once and combine the numerators. 

=  [x+ (-8)]/(x - 2)

=  (x- 8)/(x - 2)

=  (x- 23)/(x - 2)

The numerator is in the form of (a3 - b3), factor (x- 23) using the identity shown below. 

a3 - b3 = (a - b)(a2 + ab + b2)

Then, 

=  [(x - 2)(x2 + 2x + 22)]/(x - 2)

=  [(x - 2)(x2 + 2x + 4)]/(x - 2)

Cancel common factors. 

=  (x2 + 2x + 4)

Example 6 :

Simplify :

a3/(a - b) - b3/(a - b)

Solution :

=  a3/(a - b) - b3/(a - b)

Since the denominators are same, take the denominator once and combine the numerators. 

=  (a3 - b3)/(a - b)

Factor the numerator.

=  (a - b)(a2 + ab + b2)/(a - b)

Cancel common factors.

=  a2 + ab + b2

Example 7 :

Simplify :

x2/(x4 - y4) - y2/(x4 - y4)

Solution :

=  x2/(x4 - y4) - y2/(x4 - y4)

The denominators are not same. To get LCM of denominators, factor both the denominators. 

=  (x2 - y2)/(x4 - y4)

=  (x2 - y2)/[(x2)2 - (y2)2]

Let a = x2 and b = y2.

=  (a - b)/(a2 - b2)

Factor denominator. 

=  (a - b)/[(a + b)(a - b)]

Cancel common factors.

=  1/(a + b)

Replace a by x2 and by y2.

=  1/(x2 + y2)

Example 8 :

If P = x/(x + y) and Q = y/(x + y) find

[1/(P - Q)] - [2Q/(P- Q2)]

Solution :

P - Q :

=  x/(x + y) - y(x + y)

=  (x - y)/(x + y)

1/(P - Q) : 

=  (x + y)/(x - y)

P2 - Q2 :

=  [x/(x + y)]2 - [y(x + y)]2

=  [x2/(x + y)2] - [y2(x + y)2]

=  (x- y2)/(x + y)2

=  [(x + y)(x - y)]/(x + y)2

=  (x - y)/(x + y)

2Q :

= 2y/(x + y)

2Q/(P- Q2) :

=  [2y/(x + y)] ÷ [(x - y)/(x + y)]

=  [2y/(x + y)]  [(x + y)/(x - y)]

=  [2y(x + y)] / [(x + y)(x - y)]

=  2y(x - y)

[1/(P - Q)] - [2Q/(P- Q2)] :

=  [(x + y)/(x - y)] - [2y(x - y)]

=  [(x + y) - 2y] / (x - y)

=  (x + y - 2y) / (x - y)

=  (x - y)/(x - y)

=  1

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