ADDING AND SUBTRACTING RATIONAL EXPRESSIONS

Subscribe to our ▶️ YouTube channel 🔴 for the latest videos, updates, and tips.

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

Subscribe to our ▶️ YouTube channel 🔴 for the latest videos, updates, and tips.

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.

About Us  |  Contact Us  |  Privacy Policy

©All rights reserved. onlinemath4all.com

onlinemath4all_official_badge1.png

Recent Articles

  1. How to Solve Challenging Math Problems in SAT

    Feb 17, 26 06:42 PM

    How to Solve Challenging Math Problems in SAT

    Read More

  2. How to Solve the Hardest SAT Math Problems

    Feb 17, 26 10:06 AM

    digitalsatmath409.png
    How to Solve the Hardest SAT Math Problems

    Read More

  3. The Hidden Patterns in Hard SAT Math Questions

    Feb 17, 26 09:46 AM

    digitalsatmath422.png
    The Hidden Patterns in Hard SAT Math Questions

    Read More