This document provides an overview of adding and subtracting rational expressions. It begins with instructions on finding the least common denominator (LCD) and provides examples of adding fractions with unlike denominators. It then demonstrates subtracting fractions with unlike denominators by first making the denominators the same. Several examples of adding and subtracting rational expressions are worked through step-by-step. Special cases involving factoring denominators are also discussed.

1. Adding & Subtracting Rational
Expressions
By L.D.

2. Table of Contents
 Slide 3: Instructions: Find the LCD
 Slide 4: Least Common Denominator
 Slide 5: Back to Basics: LCD & Adding Fractions
 Slide 8: 1/12r , (r + 7)/4r2
 Slide 10: 4/(g2 – g -6), 5g2/(g + 2)2
 Slide 12: Instructions: Find the sum
 Slide 13: 12/5b + 18/5b
 Slide 15: 2h/(h + 2) + (h + 1)/(h + 2)
 Slide 17: Instructions: Find the difference
 Slide 18: Back to Basics: Subtracting Fractions
 Slide 19: (4x + 1)/3x – (x + 3)/3x
 Slide 21: (5x + 1)/3x – (3x + 3)/3x
 Slide 24:
 1. 6/18m + 4/6m2
 2. 5/2n – 3n/(n-1)
 Slide 25: Mini Lesson: Fraction Denominators
 Slide 27: Special Problem 1
 Slide 29: 3/(y + 1) + 2/(y2 – 1)
 Slide 32: 2x – 1/x2 + 2x – 8 – x + 1/ x2 – 4
 Slide 34: 2x – 1/x2 + 2x – 8 – x + 1/x2 - 4

3. Instructions
 Find the LCD

4. Least Common Denominator
 Product of the factors of both denominators with
each common factor only used once.

5. Back to Basics: LCD & Adding
Fractions
Example Problem 4/36 + 7/126
 When adding fractions the bottom always has to be the same, to make the
bottom the same you find the LCD, you do this by finding all the factors.
 Our two denominators are 126 and 36.
 126: 7, 3, 3, 2
 36: 3, 3, 2, 2
 What we now need to do is to choose the pairs of numbers that are the same and
place the number that they represent under this.
 3, 3, 2
 Now you need to take the remaining numbers and multiply those with the
representatives of the pairs of numbers.
 3 x 3 x 2 x 2 x 7 = 252

6. Back to Basics: Adding Fractions
 3 x 3 x 2 x 2 x 7 = 252
 That number is what we want and to get it with the different
denominators we multiply them by the “remaining factor” of the
other number.
 To get 36 to 252 we will multiply it by the “remaining factor” of
126 which is 7. We will also multiply the numerator by 7 to get
28/252.
 To get 126 to 252 we will multiply it by the “remaining factor” of
36 which is 2. We will also multiply the numerator by 2 to get
14/252.

7. Back to Basics: Adding Fractions
 Now we add the numerators of the two numbers (DON’T ADD
THE DENOMINATORS WE JUST MADE) to get the answer
which is below.
 28/252 + 14/252 = 42/252

8. Problem 1
1/12r , (r + 7)/4r2

9. Problem 1
1/12r , (r + 7)/4r2
The first thing to do here is to find the GCD as I explained in
the mini lesson.
Factors
12r: 3, 2, 2, r
4r2: 2, 2, r, r
2 x r x 2 = 4r
Now adding the remainders
4r x 3 x r
The LCD is 12r2

10. Problem 2
4/(g2 – g -6), 5g2/(g + 2)2

11. Problem 2
4/(g2 – g -6), 5g2/(g + 2)2
The first thing we do is place denominators in their
factored forms
(g + 2)2: (g + 2)(g + 2)
(g2 – g - 6): (g - 3 )(g + 2)
The LCD is (g – 3)(g + 2)2 since I won’t multiply the
things out since I find it not useful in Algebra 1.

12. Instructions
 Find the sum

13. Problem 3
12/5b + 18/5b

14. Problem 3
12/5b + 18/5b
In this problem, the denominators are already the same
so the numerators are the only ones that need to be
added and they make 30.
The answer 30/5b can be simplified by dividing the top
and bottom by 5 to get 6/b.

15. Problem 4
2h/(h + 2) + (h + 1)/(h + 2)

16. Problem 4
2h/(h + 2) + (h + 1)/(h + 2)
Since the denominators are the same, The numerators
can be added, we WILL NOT add the denominators.
2h + h + 1= 3h + 1 or (3h + 1)/(x + 2) as the final
answer.

17. Instructions
 Find the difference

18. Back to Basics: Subtracting Fractions
Example Problem 7/12 – 12/24
In this problem, we will not have to use LCD as 12/24 can be
simplified to 6/12 to match the other number. Now that the
two bottoms are the same, the two tops can be subtracted for
the answer.
7
-6
1
The final answer is 1/12.

19. Problem 5
(4x + 1)/3x – (x + 3)/3x

20. Problem 5
(4x + 1)/3x – (x + 3)/3x
Since the denominators are the same, one has to simply
minus the numerators.
4x + 1
- x + 3
3x + -2
This makes the final answer (3x – 2)/3x

21. Problem 5
(5x + 1)/3x – (3x + 3)/3x

22. Problem 5
(5x + 1)/3x – (3x + 3)/3x
An alternate way to solve problems like this to take the
two denominators out (after they are the same) until the
problem looks like this
(5x + 1) – (3x + 3)
You then can solve that to get the answer that you will
place on top of the denominator (3x).

23. Problem 5
(5x + 1) – (3x + 3)
5x + 1 – 3x – 3
2x – 2
Now we will place that on the denominator underneath
2x – 2/3x

24. Example Problems
1. 6/18m + 4/6m2
2. 5/2n – 3n/(n-1)

25. Mini Lesson: Fraction Denominators
 If you get a denominator like (9 – t), it is best not to
try to mess with it when you are trying to find the
LCD between it and something like 6t, it is best to
leave the new denominator as 6t(9 – t)
 In Algebra 1, if you get a fraction denominator like
5u(6 –u), it is best to leave it like that and not
multiply it out.

26. Example Problems
1. 6/18m + 4/6m2
(6/18m x m/m) + (4/6m2 x 3/3) (I found the LCD and used it here)
6m/18m2 + 12/18m2
6m + 12/18m2 (Now Simplify)
m + 2/3m2 (Done)
2. 5/2n – 3n/(n-1) (The thing in parenthesis can’t be separated so the
LCD is 2n(n – 1))
5(n – 1)/2n(n – 1) – 3n(2n)/2n(n – 1)
5n – 5/2n(n – 1) – 6n2/2n(n – 1)
5n – 5 – 6n2/2n(n – 1) (Done)

27. Special Problem 1
y/(y + 1) + 3/(y + 2)

28. Special Problem 1
y/(y + 1) + 3/(y + 2)
The LCD here is (y + 1)(y + 2) so I will adapt that to the
problem below.
y(y + 2)/(y + 1)(y + 2) + 3(y + 1) /(y + 2)(y + 1)
y2 + 2y + 3y + 3/(y + 2)(y + 1)
y2 + 5y + 3/(y + 2)(y + 1)

29. Special Problem 2
3/(y + 1) + 2/(y2 – 1)

30. Special Problem 2
3/(y + 1) + 2/(y2 – 1)
Now you need to find the LCD, on a problem like (y2 –
1), the LCD would be (y + 1)(y – 1).
(y + 1): (y + 1)
(y2 – 1): (y + 1)(y – 1)
So the final LCD is (y + 1)(y – 1).

31. Special Problem 2
3/(y + 1) + 2/(y2 – 1)
Now that the LCD ((y + 1)(y – 1)) is known, it needs to be
applied.
3(y – 1)/(y + 1)(y – 1)+ 2/(y2 – 1)
3y – 3/(y2 – 1) + 2/(y2 – 1)
3y – 3 + 2 /(y2 – 1)
3y - 1/(y2 – 1)

32. Special Problem 3
2x – 1/x2 + 2x – 8 – x + 1/ x2 - 4

33. Special Problem 3
2x – 1/x2 + 2x – 8 – x + 1/ x2 – 4
In this one, I suggest you factor out the separate denominators.
2x – 1/(x + 4)(x – 2)– x + 1/(x + 2)(x – 2)
Now you need to use the LCD which is (x + 2)(x – 2)(x + 4)
2x – 1(x + 2)/(x + 4)(x – 2)(x + 2) – x + 1(x + 4)/(x + 2)(x – 2)(x + 4)
(2x -1)(x + 2) – (x + 1)(x + 4)/(x + 4)(x – 2)(x + 2)
x2 – 2x – 6/(x + 4)(x – 2)(x + 2)

34. Special Problem 4
2x – 1/x2 + 2x – 8 – x + 1/x2 - 4

35. Special Problem 4
2x – 1/x2 + 2x – 8 – x + 1/x2 – 4
(2x – 1)(x + 2)/(x + 4)(x – 2)(x + 2) – (x + 1)(x + 4)/(x + 2)(x – 2)
LCD: (x + 4)(x – 2)(x + 2)
(2x – 1)(x + 2) – (x + 1)(x + 4)/(x + 4)(x – 2)(x + 2)
X2 – 2x – 6/(x + 4)(x – 2)(x + 2)