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HKDSE Mathematics Ronald Hui Tak Sun Secondary School.

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Presentation on theme: "HKDSE Mathematics Ronald Hui Tak Sun Secondary School."— Presentation transcript:

1 HKDSE Mathematics Ronald Hui Tak Sun Secondary School

2 Book 5A Chapter 3 Compound Linear Inequalities in One Unknown

3 (i) x > 1 x  3 and (ii) x + 1 < 0 2x  1 When two or more linear inequalities are connected together by logical operators (e.g. ‘and’, ‘or’), we call it a compound linear inequality. Example: or

4 Solve the compound linear inequality ‘x > 1 and x  3’. Solving Compound Linear Inequalities Connected by ‘and’ x > 1 x  3 The hollow circle ‘ ’ means that ‘1’ is NOT included. They are the solutions of ‘x > 1’ and ‘x  3’ respectively. The solid circle ‘ ’ means that ‘3’ is included. But what are the solutions for ‘x > 1 and x  3’?

5 Values of x that can satisfy both the linear inequalities are the solutions. Solving Compound Linear Inequalities Connected by ‘and’ Solve the compound linear inequality ‘x > 1 and x  3’. x > 1 x  3 Let me show you the steps!

6 Graphical representation of x  3 Solve the compound linear inequality ‘x > 1 and x  3’. Step 1 Draw the graphical representations of the two linear inequalities on the same number line. Graphical representation of x > 1

7 All the values of x in the overlapping region satisfy both inequalities. Hence, it represents the solutions of the compound inequality. Step 2 Find the overlapping region of the two graphical representations. Solve the compound linear inequality ‘x > 1 and x  3’. Overlapping region

8 Step 2 Find the overlapping region of the two graphical representations. Solve the compound linear inequality ‘x > 1 and x  3’. Overlapping region ∴ The solutions can be represented graphically by: The solutions of ‘x > 1 and x  3’ are x  3.

9 (a) Solve ‘ and ’. Similarly, we can solve the following compound linear inequalities connected by ‘and’. x  –2x < 3 Overlapping region ∴ The solutions of ‘x  –2 and x < 3’ are –2  x < 3. Graphical representation:

10 (b) Solve ‘ and ’. x < –4x  4 No overlapping region for these two inequalities In this case, the compound inequality has no solutions.

11 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.

12 Solve ‘x + 1  –2 and x  3  0’. When solving this compound inequality, we should solve each inequality separately first.

13 Solve ‘x + 1  –2 and x  3  0’.x + 1  –2 and x  3  0 x  –3......(1) x  3......(2) and ∵ x must satisfy both (1) and (2). ∴ The solutions of the compound inequality are x  –3. Graphical representation:

14 Follow-up question Solve, and represent the solutions graphically.    5 – x < 4 2x + 1  7 2x  6 –x < –1 5 – x < 4 x > 1......(1) x  3......(2) and2x + 1  7 ∵ x must satisfy both (1) and (2). ∴ The solutions of the compound inequality are 1 < x  3. Graphical representation:

15 Summary on “AND” 22 October 2015 Ronald HUI


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