Maths Made Easy

Easy Mathematics AFTERSCHOOOL www.afterschool.tk www.afterschoool.tk www.afterschoool.tk

Example—Method 1

3 5  Simplify.4 6 5 1  8 6

Solution

Write the numerator fractions as equivalent fractions with 3(3)12, 5(2) 3 5LCD, their and write 9 10 the 19    4(3) 6(2) 4 6 12 denominator fractions LCD,  12 12with their  5 1 15 4 11 5(3) 1(4)   24.  8

6

8(3)

6(4)

24

24

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24



continued

19  12 11 24 19 11   12 24 19 24   12 11

2

1

38  11 www.afterschoool.tk

Example—Method 2

3 5  Simplify.4 6 5 1  8 6

Solution Multiply the numerator 6 4 3 5  3  5  24 3 24 5 24       denominator by 24. 4 6   4 6 1 4 1 3 1 6 1 4 5 1 5 24 1 24  5 1      24  8 6 1 16 1 1 8  8 6 18  20 38   15  4 11 www.afterschoool.tk

and

Example



1 2 y Simplify.2 3 y

Solution

Write the numerator and denominator 1 2( y ) 1(1)as equivalent 2y 1 2 y  1fractions. 2   (Method y 1( y1) ) y (1) y y y 2 3 y



3( y ) 2(1)  1( y ) y (1)



3y 2  y y



3y  2 y

2 y 1 3y  2 2 y 1 y     y y y 3y  2 www.afterschoool.tk

2 y 1  3y  2

Example



x 3  2 x x 1 Simplify. 1 x

Solution

Multiply the numerator and byx the LCD of all  x 3 x 3denominator 2 x 3 2 x    2 x   (Method  the rational expressions. 2) 2 x  2 x   2 1 x 1  x 1 1 x

x 1   1 2 x x   

x2  6  2 x  2( x  1)

1 2x x 1 2x    1 1 x 1

x2  6 x2  6 x2  6   or 2x  2x  2 4x  2 2(2 x  1) www.afterschoool.tk

Example x y 2



2

Simplify. x 3  y 3

Solution 1

 1 using Rewrite 1  3 3 only 1  2  2  2 x y 2 2 2 x y  x y x y positive exponents.    x 3  y 3

xy 3  x 3 y  3 3 y x

1 1  3 3 x y

 1 1  3 3  x 3  y 3 x y  

xy ( y 2  x 2 )  ( y  x)( y 2  xy  x 2 ) www.afterschoool.tk

Simplify.

a) b) c) d)

a b c d

ab cd ac bd ad bc ad 1 bc www.afterschoool.tk

Simplify.

a) b) c) d)

a b c d

ab cd ac bd ad bc ad 1 bc www.afterschoool.tk

Simplify.

a)

a a 1

b)

a 1 a

1 a 1 1 a a

c) a + 1 d) a – 1

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Simplify.

a)

a a 1

b)

a 1 a

1 a 1 1 a a

c) a + 1 d) a – 1

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Example  Find

the

5 2 and 2 LCD. 12 y 8 y3

Solution We first factor the denominators 12y2 and 8y3 by writing their prime factorizations. 12 y 2  22 g3 gy 2 8 y 3  23 gy 3

The unique factors are 2, 3, and y. To generate the LCD, include 2, 3, and y the greatest number of times each appears in any of the factorizations. LCD = 23 g3 gy 3  24 y 3 www.afterschoool.tk

Example  Find

the

8 3 and LCD. 2 x  25 2 x  10

Solution Factor the denominators x2 – 25 and 2x – 10. x 2  25 

 x  5  x  5

2 x  10  2  x  5 

The unique factors are 2, (x + 5), and (x – 5). The greatest number of times that 2 appears is once. The greatest number of times that (x + 5) appears is once. The greatest number of times that (x – 5) appears is once. LCD = 2  x  5   x  5  www.afterschoool.tk

Adding or Subtracting Rational Expressions with Different Denominators 1. Find the LCD. 2. Write each rational expression as an equivalent expression with the LCD. 3. Add or subtract the numerators and keep the LCD as the denominator. 4. Simplify.

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Example x  3 5x 1 +  Add. 8x 6x2

Solution The LCD is 24x2.

x  3 5 x  1  x  3  3 x  5 x  1  4   + + 2  8x 6x  8 x   3x   6 x2   4 3x 2  9 x 20 x  4  + 2 24 x 24 x 2 3x 2  9 x  20 x  4  24 x 2 3x 2  29 x  4  2 24 x www.afterschoool.tk

Write equivalent rational expressions with the LCD, 24x2.

Add numerators. Note: Remember that to add polynomials, we combine like terms.

Example 7 3x +  Add. x6 6 x

Solution 3 x  1 7 7 3x  + + 6  x  1 x6 6 x x6 7 3x  + x6 x6 3x  7  x6 www.afterschoool.tk

Since x – 6 and 6 – x are additive inverses, we obtain the LCD by multiplying the numerator and denominator of one of the rational expressions by –1. We chose the second rational expression.

2y  5 7y  2 . ExampleSubtract: 2  2 3y 1 1 3 y

Solution

2 y  5 (7 y  2)(1)  2  3 y  1 (1  3 y 2 )(1) 2 y  5 7 y  2  2  3y 1 3 y2 1 2 y  5  (7 y  2)  3 y2 1 2y  5 7y  2 9y  7   2 2 3y 1 3y 1 www.afterschoool.tk

a 1 5  3a Subtract. 2  2 a  2a  1 a  2a  1

a)

2a  6

 a  1 b) 2  a  3 2  a  1 c) d)

2

4 a 1 4

 a  1

2

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a 1 5  3a Subtract. 2  2 a  2a  1 a  2a  1

a)

2a  6

 a  1 b) 2  a  3 2  a  1 c) d)

2

4 a 1 4

 a  1

2

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5 8 Add.  n4 3n  12 23 a) 3n  12 13 b) 3n  12 13  n  4  c) 3n  12 5 d) 3n  12 www.afterschoool.tk

5 8 Add.  n4 3n  12 23 a) 3n  12 13 b) 3n  12

c) 13  n  4  3n  12

5 d) 3n  12 www.afterschoool.tk

5x 2x  3 Subtract.  x 5 5 x 3  x  1 a) x5 b) 7 x  3 x5 7x  3 c) x5 7x  3 d) x5 www.afterschoool.tk

5x 2x  3 Subtract.  x 5 5 x 3  x  1 a) x5 b) 7 x  3 x5 7x  3 c) x5 7x  3 d) x5 www.afterschoool.tk

Example 7y y   Add. 18 18

Solution 7y y 7 y  y Since the rational expressions have the   same denominator, we add numerators 18 18 18 and keep the same denominator. 8y  18 2 g2 g2 gy Factor.  3 g3 g2 4y Divide out the common  factor, 2. 9 www.afterschoool.tk

Example x2 16   Subtract. x4 x4

Solution

2

x 16  x4 x4

x 2  16  x4

Note: The numerator can be factored, so we may be able to simplify.

x  4  x  4   x4

 x4

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Divide out the common factor, x – 4.

Example b 2  7b b 2  4b  2  Add. 2 b  5b b  5b

Solution

b 2  7b b 2  4b  2 2 b  5b b  5b



2 2 b  7 b  b     4b 

b 2  5b 2b 2  3b Combine like terms in  2 the numerator. b  5b b  2b  3 Factor the numerator  b  b  5  and the denominator. 2b  3  b5

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Divide out the common factor, b.

Example x2  5x  6 x 2  3x  2   Add. 3 2x  4x 2 x3  4 x

Solution x  5x  6 x  3x  2   3 3 2x  4x 2x  4x 2

2

2 2 x  5 x  6  x     3x  2 

2 x3  4 x

Combine like terms in the numerator.

2 x2  8x  8  2 x3  4 x

Factor the numerator and the denominator.

2  x  2  x  2  2x  x  2  x  2

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continued 

2  x  2  x  2  2x  x  2  x  2 x  x  2 x2  2 x  or x2 x2

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Divide out the common factors, 2 and x + 2.

Example 3x 2  2 x  7 x2  x  8   Subtract. 3x  1 3x  1

Solution 2 2 3 x  2 x  7  x  x  8 3x 2  2 x  7 x2  x  8      3x  1 3x  1 3x  1

Note: To write an equivalent addition, change the operation symbol from a minus sign to a plus sign and change all the signs in the subtrahend (second) polynomial.



2 2 3 x  2 x  7   x     x  8

2x2  x  1  3x  1

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3x  1

6n  4n Simplify. 2 2n  2n 2

a) 5 2 3 n  2n b) n2  n 3n 2  2n c) n 1

3n  2 d) n 1 www.afterschoool.tk

6n  4n Simplify. 2 2n  2n 2

a) 5 2 3 n  2n b) n2  n 3n 2  2n c) n 1

3n  2 d) n 1 www.afterschoool.tk

x  10 x  25 Simplify. 2 x  25 2

a) –10x b) 10 x x5 x 5 c) x5 x5 d) x 5 www.afterschoool.tk

x  10 x  25 Simplify. 2 x  25 2

a) –10x b) 10 x x5 x 5 c) x5 x5 d) x 5 www.afterschoool.tk

m 2  6m  9 m  2 Multiply. g 2 m 4 m3

m3 a) m2 m3 b) m2 m3 c) m2 m3 d) m2 www.afterschoool.tk

m 2  6m  9 m  2 Multiply. g 2 m 4 m3

m3 a) m2 m3 b) m2 m3 c) m2 m3 d) m2 www.afterschoool.tk

n  5n  6 n2 Divide.  n6 n6 2

a) n  3 n3 b) n2

c)  n  3  n  2 

n2 n  3  n  2   d) n2 www.afterschoool.tk

n  5n  6 n2 Divide.  n6 n6 2

a) n  3 n3 b) n2

c)  n  3  n  2 

n2 n  3  n  2   d) n2 www.afterschoool.tk

Thanks 

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