Ws10 Solving Simple Algebraic Equations Ii

Date:

______/______/ 2009

Marks

Worksheet 10: Solving Simple Algebraic Equations II 1

Intermediate (a)

Solve the equation

(b)

If h =

2( x + 1) x − 2 x − = 3 4 6

h 2 + fk , find the value of k when h = 6 and f = − 4 . hk + fh

Solution:

2( x + 1) x − 2 x − = 3 4 6 8 ( x + 1) − 3 ( x − 2 ) 2 x = 12 12 8 x + 8 − 3x + 6 = 2 x 5 x + 14 = 2 x 3 x = −14

(a)

x = −4

2 3

(b) h = 6=

h 2 + fk hk + fh 6 2 + ( −4 ) k 6k + ( −4 )( 6 )

36 − 4k 6k − 24 36k − 144 = 36 − 4k 6=

36k + 4k = 36 + 144 40k = 180 1 k =4 2

[Bukit Panjang Government High School / 2007] 2 1 Ans: (a) −4 (b) 4 3 2

© Victoria School Math Department

2

Intermediate Solve the equation Solution:

5(2 y − 9) −8 = 2y . 3

5(2 y − 9) −8 = 2y 3 5(2 y − 9) − 24 = 6 y 10 y − 45 − 24 − 6 y = 0 4 y − 69 = 0 4 y = 69 y = 17

1 4

[Anglican High School / 2007] 1 Ans: y = 17 4

3

Intermediate Solve the equation

5−

2 x − 1 3x + 2 = 4 3

Solution: 2 x − 1 3x + 2 = 4 3 20 − ( 2 x − 1) 3 x + 2 = 4 3 3 ( 20 − 2 x + 1) = 4 ( 3 x + 2 ) 5−

60 − 6 x + 3 = 12 x + 8 63 − 8 = 12 x + 6 x 55 = 18 x 1 x=3 18

[Catholic High School / 2007] 1 Ans: x = 3 18

© Victoria School Math Department

4

Intermediate Given that

a + 2b 3 6b = , find the value of . a + 3b 4 7a

Solution: a + 2b 3 = a + 3b 4 4a + 8b = 3a + 9b

a=b 6b 6b = 7 a 7b 6 = 7 [Cedar Girls’ Sec School / 2008] Ans:

5

6 7

Advance (a) (b)

5 − 2x 2 − x 1 − = . 3 5 7 5x + y 6 2x If = , find the value of . 3x − 2 y 7 3y Solve the equation

Solution:

5 − 2x 2 − x 1 − = 3 5 7 5 (5 − 2x ) − 3( 2 − x ) 1 = 15 7 25 − 10 x − 6 + 3 x 1 = 15 7 7 (19 − 7 x ) = 15 (a)

133 − 49 x = 15 −49 x = 15 − 133 −49 x = −118 −118 x= −49 20 x=2 49

(b)

5x + y 6 = 3x − 2 y 7 35 x + 7 y = 18 x − 12 y 17 x = −19 y 19 x=− y 17  19  2 − y  2x 17  =  3y 3y 38 1 =− × 17 3 2x 38 =− 3y 51

[Cedar Girls’ Sec School / 2007] 20 38 (b) x = − Ans: (a) x = 2 49 51 © Victoria School Math Department

6

Advance

If x 3 + x 2 + x + 1 = 0 , what is the value of x 97 + x 98 + x 99 + x100 + x 101 + x102 + x 103 + x 104 ?

Solution:

x97 + x98 + x99 + x100 + x101 + x102 + x 103 + x104 = x 97 (1 + x + x 2 + x 3 + x 4 + x5 + x 6 + x 7 ) = x 97  0 + x 4 + x5 + x 6 + x 7  = x 97  0 + x 4 (1 + x + x 2 + x 3 )  = x 97  x 4 ( 0 )  = x 97 ( 0 ) =0 [Bukit Panjang Government High School / 2007] Ans: 0

© Victoria School Math Department