HKDSE MATHEMATICS Ronald Hui Tak Sun Secondary School

MISSING HOMEWORK  Summer Holiday Homework 1  10  SHW1-R1  9 22 October 2015 Ronald HUI

MISSING HOMEWORK  SHW2-A1  14  SHW2-B1  9  SHW2-C1  10, 12, 13, 14, 20  RE2  9 22 October 2015 Ronald HUI

MISSING HOMEWORK  SHW2-R1  5J07, 3, 5, 8, 9, 10, 12, 13, 14, 15, 18, 19, 20, 21, 4 (RD)  SHW2-P1  5J07, 5P24, 1, 3, 5, 8, 9, 10, 11, 12, 13, 14, 15, 17, 18, 19, 20, 21, 23, 24, 4 (RD), 7 (RD) 22 October 2015 Ronald HUI

MISSING HOMEWORK  SHW3-01  5P24, 1, 3, 8, 9, 10, 12, 13, 14, 15, 20, 23  SHW3-A1  5P24, 1, 3, 8, 9, 10, 12, 13, 15, 19, 20, 23, 24  SHW3-B1  5J07, 1, 3, 9, 12, 13, 15, 20, 23, 24  SHW3-C1, D1, E1, R1, P1  Where are they? 22 October 2015 Ronald HUI

HOMEWORK CHECKING  Collection  This week  Checking  Next week  Return  Next next week 22 October 2015 Ronald HUI

LINEAR PROGRAMMING Form 5 Mathematics Chapter 4

SUMMARY ON “AND” 22 October 2015Ronald HUI

SUMMARY ON “OR” 22 October 2015Ronald HUI

Book 5A Chapter 4 Linear Inequalities in Two Unknowns

3x – 2y – 2 < 0 x + y  4 2x > y Linear Inequalities in Two Unknowns A linear inequality is an inequality of which the highest degree of the terms is 1. If the linear inequality contains two unknowns, it is called a linear inequality in two unknowns. Example: (i) (ii) (iii) These inequalities contain two unknowns x and y, and the highest degree of the terms is 1. x y x x y y

In general, a linear inequality in two unknowns can be written in one of the following forms: ax + by > c ax + by  c ax + by  c ax + by < c where a, b and c are constants, a and b are not both zero. Linear Inequalities in Two Unknowns

An ordered pair (x, y) that satisfies an inequality in two unknowns is a solution of the inequality.

For example, consider the inequality x + y  4 and two ordered pairs (2, 2) and (1, 4). For (2, 2), For (1, 4), L.H.S. = x + y = 5 = 4 = = R.H.S. L.H.S. = x + y = i.e. when x = 2 and y = 2,i.e. when x = 1 and y = 4, > R.H.S. ∴ (2, 2) and (1, 4) are two solutions of x + y  4. > 4

Yes! (2, 2) and (1, 4) are indeed just two of an infinite number of solutions of the inequality. Does the inequality x + y  4 have other solutions?

In fact, all the solutions of the inequality can be represented graphically.

Consider the graph of x + y = 4. The straight line x + y = 4 divides the coordinate plane into two regions. Each region is called a half-plane.

Consider the graph of x + y = 4. The half-plane above the straight line is called the upper half-plane. The half-plane below the straight line is called the lower half-plane. upper half-plane lower half-plane The straight line x + y = 4 is called the boundary. boundary

upper half-plane lower half-plane boundary Consider the graph of x + y = 4. In fact, one of the half-planes represents the solutions of x + y > 4, and the other represents the solutions of x + y < 4.

upper half-plane boundary Consider the graph of x + y = 4. Let’s see which half-plane represents the solutions of x + y > 4. lower half-plane

Consider a point A(1, 4). We have proved that its corresponding ordered pair (1, 4) is the solution of x + y > 4. A From the graph, A lies above the line x + y = 4.

Try other points B(2, 3) and C(4, 2), whose corresponding ordered pairs are the solutions of x + y > 4.  For B(2, 3), x + y = = 5 > 4 For C(4, 2), x + y = = 6 > 4 C B From the graph, B and C lie above the line x + y = 4. A

Actually, any ordered pairs (x, y) that satisfy x + y > 4 lie above the line x + y = 4. Therefore, the solutions of x + y > 4 are represented by the upper half-plane. Solutions of x + y > 4 Try other points B(2, 3) and C(4, 2), whose corresponding ordered pairs are the solutions of x + y > 4.  For B(2, 3), x + y = = 5 > 4 For C(4, 2), x + y = = 6 > 4 C B A

Consider three points D(0, 0), E(2, 1) and F(4, –2), whose corresponding ordered pairs are the solutions of x + y < 4.  For D(0, 0), x + y = = 0 < 4 For E(2, 1), x + y = = 3 < 4 For F(4, –2), x + y = 4 + (–2) = 2 < 4 D F E From the graph, D, E and F lie below the line x + y = 4.

D F E Actually, any ordered pairs (x, y) that satisfy x + y < 4 lie below the line x + y = 4. Therefore, the solutions of x + y < 4 are represented by the lower half-plane. Solutions of x + y < 4 Consider three points D(0, 0), E(2, 1) and F(4, –2), whose corresponding ordered pairs are the solutions of x + y < 4.  For D(0, 0), x + y = = 0 < 4 For E(2, 1), x + y = = 3 < 4 For F(4, –2), x + y = 4 + (–2) = 2 < 4

We usually represent the solutions of a linear inequality in two unknowns graphically by shading the relevant half-plane. Solutions of x + y > 4 Solutions of x + y < 4 Note that when the boundary is not a part of the solutions, it is drawn as a dotted line.

Solutions of x + y  4 Solutions of x + y  4 Note that when the boundary is a part of the solutions, it is drawn as a solid line. We usually represent the solutions of a linear inequality in two unknowns graphically by shading the relevant half-plane.

To determine which half-plane represents the solutions of the inequality,

we should choose a point, which does not lie on the boundary, as a test point.

Choose (0, 0) as the test point. When x = 0 and y = 0, x + 2y = 0 + 2(0) = 0  4 ∴ The half-plane containing the test point (0, 0) together with the boundary represents the solutions of the inequality x + 2y  4. Consider the inequality x + 2y  4.

Let’s summarize the steps for solving a linear inequality in two unknowns.

Solve the inequality 2x + y  0 graphically. Step 1 Draw the boundary with a dotted/solid line according to the inequality sign. Inequality signBoundary ‘  ’ or ‘  ’ ‘>’ or ‘<‘ solid line dotted line

Solve the inequality 2x + y  0 graphically. Step 1 Draw the boundary with a dotted/solid line according to the inequality sign. Draw the solid line 2x + y = 0.

Step 2 Choose a test point and check which half-plane represents the solutions of the inequality. Solve the inequality 2x + y  0 graphically. Choose (1, 1) as the test point. We can choose any point that does not lie on the boundary as the test point. ∴ The half-plane containing the test point (1, 1) together with the boundary represents the solutions of the inequality. When x = 1 and y = 1, 2x + y = 2(1) + 1 = 3  0

Graphical representation of 2x + y  0 Step 3 Shade the half-plane found in step 2. Solve the inequality 2x + y  0 graphically.

Follow-up question Solve the inequality 3x – 2y > 2 graphically. Draw the dotted line 3x – 2y = 2. Choose (0, 0) as the test point. When x = 0 and y = 0, 3x – 2y = 3(0) – 2(0) = 0 ≯ 2  ‘ ≯ ’ means ‘not greater than’.

Follow-up question Solve the inequality 3x – 2y > 2 graphically. Draw the dotted line 3x – 2y = 2. ∴ The half-plane without the test point (0, 0) represents the solutions of the inequality.  The solutions of the inequality lie in the shaded region, excluding the boundary.