Essential Question: How do you add and subtract two rational expressions?

Just like with fractions, terms need to have a common denominator in order for their numerators to be added or subtracted together.

After they’re added/subtracted, fractions can be simplified.

Sometimes, you will need to find the least common denominator (LCD). To do this, you find the least common multiple of the denominators.

4

3x2

3x6

3x2

x

To find the least common multiple1) Find the prime factors of each expression2) Write each prime factor the greatest number

of times it appears in either expression3) Simplify where possible

Find the LCM of 4x2 – 36 and 6x2 + 36x + 54 Step 1

4x2 – 36 6x2 + 36x + 54

Step 2

Step 3

= 4(x2 – 9)= (2)(2)= 6(x2 + 6x +

9)= (2)

(2) appears twice,

(3) appears once, (x – 3) appears

once,(x + 3) appears twice

(x – 3)(x + 3) (3)(x + 3)(x +

3)

(2)(2)(3)(x – 3)(x + 3)(x + 3) = 12(x – 3)(x + 3)2

Your Turn #1 Find the LCM of 3x2 – 9x - 30 and 6x +

30Step 1

3x2 – 9x – 30 6x + 30

Step 2

Step 3

= 3(x2 – 3x – 10)

= (3)= 6(x + 5)

= (2)

(2) appears once,

(3) appears once, (x - 5) appears

once,(x + 2) appears once

(x – 5)(x + 2)(3)(x +

5)

(2)(3)(x – 5)(x + 5)(x + 2) = 6(x – 5)(x + 5)(x + 2)

(x + 5) appears once,

Your Turn #2 Find the LCM of 5x2 + 15x + 10 and 2x2

– 8Step 1

5x2 + 15x + 10 2x2 – 8

Step 2

Step 3

= 5(x2 + 3x + 2)

= (5)= 2(x2 – 4)

= (2)

(2) appears once,

(5) appears once, (x + 2) appears

once,(x – 2) appears once

(x + 2)

(x + 1)(x +

2)(x – 2)

(2)(5)(x + 2)(x + 1)(x – 2) = 10(x + 2)(x + 1)(x – 2)

(x + 1) appears once,

Now that we understand how to find the least common multiple (which is also our least common denominator), let’s use that to add/subtract fractions

We multiply each term in the problem by what is missing from the LCD

Simplify

Find the LCM (x + 4)(x + 1) with (3)(x + 1) = (3)(x + 4)(x + 1) Multiply left side (top and bottom) by (3) Multiply right side (top and bottom) by (x + 4)

2

1 5

5 4 3 3

x

x x x

1 5

( 1)( 4) 3( 1)

3 ( 4)

3 ( 4)

x

xx x x

x

23 5 20

3( 1)( 4) 3( 1)( 4)

x x

x x x x

25 20 3

3( 1)( 4)

x x

x x

Simplify

Find the LCM Multiply left side (top and bottom) by Multiply right side (top and bottom) by

1 3

( 6)( 2) 4( 2)

4 ( 6)

4 ( 6)

x

xx x x

x

23 18 4

4( 6)( 2)

x x

x x

2

1 3

4 12 4 8

x

x x x

(x – 6)(x + 2) with (2)(2)(x + 2) = (4)(x – 6)(x + 2) (4)

(x - 6)

24 3 18

4( 6)( 2) 4( 6)( 2)

x x

x x x x

Simplify

Find the LCM Multiply left side (top and bottom) by Multiply right side (top and bottom) by

7 4

5( 5)( 5) 3( 5)

3 5( 5)

3 5( 5)

y

y

y

y y y

100

15( 5)( 5)

y

y y

5(y + 5)(y – 5) with (3)(y + 5) = (3)(5)(y + 5)(y – 5) (3)

5(y - 5)

21 20 100

15( 5)( 5) 15( 5)( 5)

y y

y y y y

2

7 4

5 125 3 15

y

y y

Page 517 – 518Problems 5 – 21 (odd)Show your work!!!

Essential Question: How do you add and subtract two rational expressions?

A complex fraction is a fraction that has a fraction in its numerator, denominator, or both. A few examples:

To simplify a complex fraction, multiply all terms by the LCD of all embedded fractions

1

xy

3

1 1

2y

x 2x

2

x 13

x 1 1

x 1

Simplify

This function has two embedded denominators, x and y, so the LCD of all embedded denominators is xy.

Multiply all terms by xy.

13

54

x

y

13

54

x

y

xy xy

xy xy

3

5 4

y xy

x xy

Simplify

Multiply all terms by:

1xy

1

xy

1xx

xy

x

Simplify

Multiply all terms by:

31

12y

6

2 1

y

y

2y

12

2

3

12

2

y

yy

y

Simplify

Multiply all terms by:x(x + 1)(x – 1)

2 21

3 11 1

xx x

x x

2 21

3

( 1)( 1) ( 1)( 1)

(1

11)( 1) ( 1)(

11)

xx x x x x x

x x x x

x

x xx x

x

( 1)( 1) ( 1)

( 1)

( 2) 2

3 1 ( 1)

x x x x

x

x

x x x

2 2

2 2

( 2) 2( 1) ( )

( ) ( )3 1

x x x

x x x x

x

3 2 2

2 2

2 2 2 2

3 3

x x x x x

x x x x

3 2

2

4 2

2 4

x x x

x x

Page 517 – 518Problems 22 – 30 (all)Show your work!!!