HKDSE Mathematics Ronald Hui Tak Sun Secondary School

Missing Homework  Summer Holiday Homework 1  1, 10, 14  Summer Holiday Homework 2  1, 9, 14  SHW1-R1  9, October 2015 Ronald HUI

Missing Homework  SHW2-A1  10, 14  SHW2-B1  9, 14  SHW2-C1  8, 9, 10, 12, 13, 14, 20  RE2  9 22 October 2015 Ronald HUI

Missing Homework  SHW2-R1  I got 5 only!  SHW2-P1  I got 3 only!!!  SHW3-01  Coming! 22 October 2015 Ronald HUI

Summary on “AND” 22 October 2015 Ronald HUI

Book 5A Chapter 3 Compound Linear Inequalities in One Unknown

In summary, we have: Compound inequalitySolutions x > a and x > b x > b x > a and x < b a < x < b x b no solutions Note:The solving steps are similar when the inequality signs ‘ ’ are replaced by ‘  ’ and ‘  ’ respectively.

Compound Linear Inequalities Connected by ‘or’ e.g.(i)x  3 or x  1 (ii)x + 3 > 0 or 3x – 1 < 2 We have to find all values of x satisfying at least one of the linear inequalities. How can we solve this kind of compound linear inequalities?

Compound Linear Inequalities Connected by ‘or’ e.g.(i)x  3 or x  1 (ii)x + 3 > 0 or 3x – 1 < 2 How can we solve this kind of compound linear inequalities? Let me show you the steps of solving ‘x  3 or x  1’.

Graphical representation of x  1 Graphical representation of x  3 Solve ‘x  3 or x  1’. Step 1 Draw the graphical representations of the two linear inequalities on the same number line.

All the values of x in the combined region satisfy at least one of the inequalities. Step 2 Find the combined region of the two graphical representations. Solve ‘x  3 or x  1’. Combined region Hence, it represents the solutions of the compound inequality.

∴ The solutions can be represented graphically by: Solve ‘x  3 or x  1’. The solutions of ‘x  3 or x  1’ are x  1. Step 2 Find the combined region of the two graphical representations. Combined region

∴ The solutions of ‘x  –2 or x < 3’ are all real values of x. Similarly, we can solve the following compound linear inequalities connected by ‘or’. (a) Solve ‘ or ’. x  –2x < 3 Combined region Graphical representation:

(b) Solve ‘ or ’. x < –4x  4 Combined region ∴ The solutions of ‘x < –4 or x  4’ are x < –4 or x  4. Graphical representation:

In summary, we have: Compound inequalitySolutions Note:The solving steps are similar when the inequality signs ‘ ’ are replaced by ‘  ’ and ‘  ’ respectively. x < a or x < b x < b x > a or x < b all real values of x x b

When solving this compound inequality, we should solve each inequality separately first. Solve ‘x + 3 < 1 or 3x  1 < 2’.

x + 3 < 1 or 3x  1 < 2 x < – (1) or 3x < 3 x < (2) ∵ x must satisfy (1) or (2). ∴ The solutions of the compound inequality are x < 1. Graphical representation:

Follow-up question Solve ‘2x + 3  1 or 2x – 1 > x + 1’, and represent the solutions graphically. 2x + 3  1 2x  –2 x  – (1) or 2x – 1 > x + 1 x > (2) ∵ x must satisfy (1) or (2). ∴ The solutions of the compound inequality are x  –1. Graphical representation:

Problems Leading to Compound Linear Inequalities It is given that n is a positive integer. If 3n + 1 is greater than 5n  7, or 3n + 7 is less than 2n + 5, find the possible value(s) of n. The compound inequality is ‘3n + 1 > 5n  7 Step 1 Set up a compound inequality in n. or3n + 7 < 2n + 5’.

3n + 1 > 5n  7  2n >  8 n < 4 ……(1) or 3n + 7 < 2n + 5 n <  2 ……(2) ∵ n must satisfy (1) or (2). ∴ The solutions of the compound inequality are n < 4. Step 2 Solve the compound inequality obtained in step 1. n ‘3n + 1 > 5n  7 or 3n + 7 < 2n + 5’.

∵ n is a positive integer. ∴ The possible values of n are 1, 2 and 3. Step 3 Check whether there are restrictions on the unknown. ∴ The solutions of the compound inequality are n < 4.

No. of coinsTotal value $(30  2x) $2 coins 15  x $5 coins x $5x $2(15  x) Follow-up question Jenny has 15 coins. x of them are $5 coins and the rest are $2 coins. The total value of the coins is greater than $45 but less than $54. Find the possible value(s) of x. ∴ The total value of the coins = $5x + $(30  2x) = $(3x + 30)

∵ The total value of the coins is greater than $45 but less than $54. ‘3x + 30 > 45 and 3x + 30 < 54’. ∴ The compound inequality is 3x + 30 > 45 3x > 15 x > 5 ……(1) 3x + 30 < 54 3x < 24 x < 8 ……(2) ∵ x must satisfy both (1) and (2). ∴ The solutions of the compound inequality are 5 < x < 8. ∵ x must be an integer. ∴ The possible values of x are 6 and 7. and

Summary on “OR” 22 October 2015 Ronald HUI