Ws13 Solving Inequalities

Date:

______/______/ 2009

Marks

Worksheet 13: Solving Inequalities 1

Basic Given that −4 ≤ x ≤ 3 and 1 ≤ y ≤ 8 , calculate (a) the largest value of xy , (b) the smallest value of x − y , x (c) the smallest value of , y (d) the largest value of x 2 + y 2 . Solution: (a) (b)

Largest value of xy = 3 × 8 = 24 Smallest value of x − y = −4 − 8

(c)

= − 12 x −4 Smallest value of = y 1 = −4

(d)

Largest value of x 2 + y 2 = ( −4 ) + 82

2

= 16 + 64 = 80 [Catholic High School / 2007] Ans: (a) 24 (b) –12 (c) – 4 (d) 80

2

Basic (a) Find the largest integer p such that 5 p < 23. (b) Find the smallest prime number q such that 3q > 69. Solution: 5 p < 23 (a) 23 p< 5 p < 4.6 Largest integer p = 4 (b)

3q > 69 q > 23

Smallest prime number q = 29 [Anglican High School / 2007] Ans: (a) p = 4 (b) q = 29

© Victoria School Math Department

3

Basic Solve the inequality 2 − 2 x > 9 . Hence write down the greatest integer value of x which satisfies 2 − 2 x > 9 . Solution: 2 − 2x > 9 −2 x > 7 x < −3.5 The greatest integer value of x = − 4

[GCE O Level / 2008] Ans: x = – 4

4

Basic (a)

Illustrate, on the number line given in the answer space below, the solution of the inequality 4 x − 10 ≤ x + 6 .

0

(b)

1

4

3

2

5

6

7

Hence, list all the possible natural numbers of the solution.

Solution: (a) 4 x − 10 ≤ x + 6 4 x − x ≤ 6 + 10 3 x ≤ 16

x≤5

0

(b)

1 3

1

2

3

5 51 3

4

6

7 x

All the possible natural numbers are 1, 2, 3, 4, and 5.

[Cedar Girl’s Sec School/2007] 1 Ans: (a) x ≤ 5 (b) 1, 2, 3, 4, 5 3

© Victoria School Math Department

5

Basic Solve the inequality 3 − 2 x <

1 (3x − 4) . 2

Solution:

1 ( 3x − 4 ) 2 3 − 2 x < 1.5 x − 2 3 + 2 < 1.5 x + 2 x 5 < 3.5 x 3 − 2x <

3.5 x > 5 x >1

3 7

[Cedar Girl’s Sec School/2008] 3 Ans: x > 1 7

6

Intermediate Solve the following inequality 5

1 1 ≤ 1 − 8k . 2 2

Solution:

1 1 ≤ 1 − 8k 2 2 1 1 8k ≤ 1 − 5 2 2 8 k ≤ −4

5

4 8 1 k≤− 2 k≤−

[Anglican High School / 2007] 1 Ans: k ≤ − 2

© Victoria School Math Department

7

Advance (a)

Solve −5 ≤ 2 x + 3 ≤ 10 .

(b)

Hence, write down the greatest and least integer values of x which satisfy

−5 ≤ 2 x + 3 ≤ 10 . Solution: (a) − 5 ≤ 2 x + 3 ≤ 10 − 5 ≤ 2 x + 3 and

2 x + 3 ≤ 10

− 8 ≤ 2x −4≤ x ∴ − 4 ≤ x ≤ 3.5 (b)

2x ≤ 7 x ≤ 3.5

Greatest integer of x = 3 Least integer of x = − 4

[Catholic High School / 2007] Ans: (a) − 4 ≤ x ≤ 3.5 (b) 3 , − 4

© Victoria School Math Department