Chapter 1 Section 2 Linear Inequalities Read pages 10 – 15 Study all Examples in the section

Rule for Solving an Inequality When solving for a variable (usually y), recall that you change the direction of the inequality sign only when you multiply, or divide, a negative_number to each side of the inequality sign!

General Form vs. Standard Form Take the following general form of inequalities and change them to standard form. 2 x + 2/3 y < 5 2/3 y < – 2 x + 5 – 5 x – 1/3 y < – 6 – 1/3 y < 5 x – 6 y < – 3 x + 15/2 y > – 15 x + 18

Graphing Linear Inequalities for Linear Programming Problems Blue-grey box on page 12 1.Draw the graph of y = m x + b 2.THROW AWAY / CROSS OUT the portion of the plane that does NOT satisfy the inequality Step (2) different from what you have learned.

Reason for change in directions in step 2 Later you will have to graph a system of inequalities similar to: y > – 3 x + 8 y > – x + 4 y < 7 x < 3 y > 0 x > 0

Which part of the plane to shade / cross- off for the linear inequality? 1.For y > m x + b –Consists of all points above and on the line –Shade BELOW the line –Note: Inequality sign pointing towards ‘b’ 2.For y < m x + b –Consists of all points below and on the line –Shade ABOVE the line

(CONTINUED) 3.For x > c –Consists of all the points to the right of c –Shade to the LEFT 4.For x < c –Consists of all the points to the left of c –Shade to the RIGHT

Exercise Graph: 4 x – 5 y + 25 > 0 Solution: 4 x – 5 y + 25 > 0 – 5 y > – 4 x – 25 y < 4/5 x + 5 y-intercept: ( 0, 5 ) x-intercept: ( – 25/4, 0 )

Exercise Graph (-25/4, 0) ( 0, 5 ) y = 4/5 x + 5 y-axis x-axis

Exercise 43 (page 17) Graph: x + 5 y < 10 x + y < 3 x > 0 y > 0 y < – 1/5 x + 2 y < – x + 3 x > 0 y > 0 x- and y- intercepts for: y < – 1/5 x +2( 10, 0 ) and ( 0, 2 ) y < – x + 3( 3, 0 ) and ( 0, 3 )

Exercise 41 Initial Graph y-axisx = 0 x-axisy = 0 ( 0, 3 ) ( 0, 2 ) ( 3, 0 ) ( 10, 0 ) y = – x + 3 y = – (1/5) x + 2

Exercise 41 Shaded Graph y-axis x-axis y = – x + 3 y = – (1/5) x + 2

Exercise 41 Solution Feasible Set y = – (1/5) x + 2 y = – x + 3 y = 0 x = 0

Using Exercise 41 Solution Question: Is ( 2, 7/4 ) in the feasible set (i.e. does ( 2, 7/4 ) satisfy all the inequalities)? Solution: y < – 1/5 x + 2(7/4) < (– 1/5)(2) + 27/4 < 8/5 y < – x + 3(7/4) < – (2) + 37/4 < 1 x > 0(2) > 02 > 0 y > 0(7/4) > 07/4 > 0 Since 7/4 < 8/5 and 7/4 < 1 are false statements, then ( 2, 7/4 ) is NOT in the feasible set (i.e. (2, 7/4 ) does not satisfy ALL of the inequalities) !