Ws6 Simplify Expression

Date:

______/______/ 2009

Marks

Worksheet 6: Fundamental Algebra (Simplifying Algebraic Expressions) 1

Basic Question (a) Simplify 2e + 3 f − 2(e + f ) . Solution: 2e + 3 f − 2(e + f ) = 2e + 3 f − 2e − 2 f = f

[Xin Min Sec School/2007] Ans: f (b) Simplify (−5m − 8n + 8 p ) − (−5n + 5m − 4 p ) . Solution: (−5m − 8n + 8 p ) − (−5n + 5m − 4 p ) = −5m − 8n + 8 p + 5n − 5m + 4 p = −10m − 3n + 12 p

[Anglican High School/2007] Ans: −10m − 3n + 12 p (c) Simplify 2b(c − a ) − [3c(a − b ) − 3a(b + c )] . Solution: 2b ( c − a ) − 3c ( a − b ) − 3a ( b + c )  = 2bc − 2ab − [3ac − 3bc − 3ab − 3ac] = 2bc − 2ab − [−3bc − 3ab] = 2bc − 2ab + 3bc + 3ab = ab + 5bc

[Chung Cheng High School/2007] Ans: ab + 5bc

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2

Basic Question (a) Express

3(5a − 1) 3a + 7 − as a single fraction. 4 6

Solution:

3(5a − 1) 3a + 7 − 4 6 9(5a − 1) − 2(3a + 7) = 12 45a − 9 − 6a − 14 = 12 39a − 23 = 12

[Anglican High School/2007] 39a − 23 Ans: 12 (b) Express

x−5 2 x − 3 3( x + 1) − + as a single fraction in its simplest form. 3 6 5

Solution:

x−5 2 x − 3 3( x + 1) − + 3 6 5 10( x − 5) − 5( 2 x − 3) + 18( x + 1) = 30 10 x − 50 − 10 x + 15 + 18 x + 18 = 30 18 x − 17 = 30

[Chung Cheng High School/2007] 18 x − 17 Ans: 30

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3

Intermediate Question 2 3 Simplify the expression 6a ÷ 3a . 5b 3 10b

Solution: 6a 2 3a 3 6a 2 10b ÷ = × 5b3 10b 5b3 3a 3 4a 2b = 3 3 1a b 4 = 2 ab [Cat High Sec 1 P1/2007] 4 Ans: ab 2

4

Intermediate Question 2

3m 2 (− 6mn ) ÷ . 12 23 n (b) Subtract (3x2 – 4xy + 5y2) from the sum of (3x2 – 2xy + 6y) and (2x3 + xy – 2y2).

(a) Simplify

Solution: 3m 2 ( −6mn ) (a) ÷ 23 n 12 2 3m 12 = 3 × 2 n 36m 2 n 2 1m 2 = 8m 2 n 3 1 = 3 8n

2

(b) ( 3 x 2 − 2 xy + 6 y ) + ( 2 x 3 + xy − 2 y 2 ) − ( 3 x 2 − 4 xy + 5 y 2 )  = 3 x 2 − 2 xy + 6 y + 2 x3 + xy − 2 y 2 − 3 x 2 + 4 xy − 5 y 2 = 2 x 3 + 3 xy + 6 y − 7 y 2

[Cedar Girls’ Sec School/2007] 1 Ans: (a) 3 (b) 2 x 3 + 3xy + 6 y − 7 y 2 8n

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5

Advance Question Simplify 2 x 2 15 xz 3 8 x 4 yz (a) × ÷ 3 4 6  12  (b) ( 8 x 3 y 4 ) ÷  xy   5  (c)

3 xy 5 × ( −2 x 3 )

2

2 3

( − xy )

Solution:

(a)

2 x 2 15 xz 3 8 x 4 yz 2 x 2 15 xz 3 6 × ÷ = × × 4 3 4 6 3 4 8 x yz 15 x3 z 3 = 4 8 x yz =

15 z 2 8 xyz

12 xy 8 x 3 y 4 5 = × (b)8 x y ÷ 5 1 12 xy 3

4

=

(c)

3 xy 5 × ( −2 x3 ) 2 3

( − xy )

10 x 2 y 3 3

2

=

3 xy 5 × 4 x 6 − x3 y 6

=

−12 x 4 y

Ans: (a)

10 x 2 y 3 15 z 2 −12x 4 (b) (c) 3 8 xyz y

© Victoria School Math Department