8 Inequalities Concepts: Equivalent Inequalities Linear and Nonlinear Inequalities Absolute Value Inequalities (Sections.6 and.) 8. Equivalent Inequalities Definition 8. Two inequalities are equivalent if they have the same solution set. Operations that Produce Equivalent Inequalities. Add or Subtract the same value on both sides of the inequality. Multiply or Divide by the same positive value on both sides of the inequality. Multiply or Divide by the same negative value on both sides of the inequality AND change the direction of the inequality. Eample 8. For each pair of inequalities, determine if the two inequalities are equivalent. + 5 and + 5. Yes. (Try subtracting 5 from both sides of the first inequality.) 6 < 5 and > Yes. + > 3 and 3 > 0 Yes. + > 3 and > 3(+) No.(Notice =.5 is a solution to the first inequality, but + not the second.). You should NEVER multiply both sides of an inequality by an epression involving that you don t know the sign of. Why?
8. Solving a Linear or Nonlinear Inequality Eample 8.3 (Linear Inequality) Solve the inequality 5+3 6 7. Write the solution set in interval notation. By adding 7 and subtracting 6 from both sides of the inequality, we obtain 3. First, we will solve the equality 3 = 0 = 3 = Net, we make a sign chart using the solution(s) to the equality as follows: 3 is: - + = 0 = = Since we want those values of where 3, then our solution set is (, ] Note 8. There are many ways to solve this inequality algebraically. We will begin by using addition and subtraction to move all the nonzero quantities to one side. This is not necessary for this inequality, but it will help us to understand the process needed for solving more complicated inequalities. Graph the equations y = 5+3 and y = 6 7. How can you approimate the solutions of an inequality graphically? y 9 8 7 6 5 3 0-9 -8-7 -6-5 - -3 - - 0 3 5 6 7 8 9 - -3 - -5-6 -7-8 -9 5+3 0 3 8 3 - - - -7 6 7 0 6 - -8-3 - 0 The graphs intersect when =. The graph of y = 5 + 3 is below the graph of y = 6 7 when is in the interval (, ]. Thus, the y-value of y = 5+3 is less than the y value of y = 6 7. solving 5+3 6 7.
Thinking graphically can help us understand the algebraic procedure for solving Nonlinear Inequalities. Given an epression, such as (+3) ( )( 5), the epression is positive when the graph of y = ( + 3) ( )( 5) is above the -ais. The epression is negative when the graph of y = (+3) ( )( 5) is below the -ais. Eample 8.5 (Polynomial Inequality) Thegraphofy = (+3) ( )( 5) isshownbelow. Theviewingwindowis[ 0,0] [ 00,00]. Use the graph to help you approimate the solutions of the inequality. (+3) ( )( 5) > 0 (, 3) ( 3,) (5, ) The algebraic procedure for solving an inequality is based on the intuition we gain from the graphical solution. Algebraic Procedure for Solving Linear and Nonlinear Inequalities. Use addition and subtraction to move all nonzero quantities to one side. If fractional epressions are involved, simplify so the nonzero side is a single fractional epression.. Find the zeros of the epression AND the zeros of all denominators. (If you can factor the epression, this can help. Finding a zero means find when the epression equals zero.) 3. Make a sign chart to determine if the values between the zeros from step lead to positive or negative values of the polynomial.. Answer the question. Eample 8.6 Use the algebraic approach to solve ( + 3) ( )( 5) > 0. Be sure to write you answer in interval notation. (+3) ( )( 5) is: + + - + = = 0 = 3 = 6-3 5 Since we want the values of such that ( 3) ( )( 5) > 0, then are solution set contains the positive intervals. So, the solution set is (, 3) ( 3,) (5, ). 3
Eample 8.7 (Quadratic Inequality) + > 8 Rewriteas + 8> 0, then solvethe equality (+)( ) is: + - + + 8 = 0 (+)( ) = 0 + = 0 or = 0 = + The solution set is (, 5) (, ) Eample 8.8 (Rational Inequality) 3 > = 5 = 0 = 3 = = 3 > 3 > 0 (3 ) > 0 3 5 3 > 0 5 = 0 3 = 0 =.5 3 = 5 3 is: - + - = 0 =.75 = 6.5 3 So the solution set is (.5,3). Eample 8.9 (Rational Inequality) +3 ( ) (+3)( ) +3 +3 (+3) (+3)( ) (+3)( ) (+3)( ) = 0 (+3)( ) = 0 = +3 = 0 = 0 = 7 = 3 = + - + - = 8 = = 0 = -7-3 The solution set is [ 7, 3) (, ). Note, the points = 3 and = are ecluded because division by zero is undefined.
8.3 Absolute Value Inequalities Number lines can be really insightful when working with absolute value equations and inequalities. Recall that we think of the absolute value as a distance. Eample 8.0 (A Distance Eample) Solve 5 < 6 geometrically. Be sure to write your answer in interval notation. The distance from 5 to is less than 6. (-,) - -5 Eample 8. (Another Distance Eample) Solve +3 > 5 geometrically. Be sure to write your answer in interval notation. The distance from to -3 is greater than 5. (, 8) (, ) -8-3 Eample 8. (Another Distance Eample) -0-5 0 5 0 (a) Write a distance sentence that corresponds to this number line. The distance from to - is greater than or equal to 6. (b) Write an absolute value equation or inequality that corresponds to this number line. + 6 Eample 8.3 (The Algebraic Approach to Absolute Values) Solve each inequality algebraically. (a) +3 7 +3 7 or +3 7 0 +0 - + - + The solution set is (, 5] [, ). -5 (b) + + < 3 The solution set is (,0). + + < 3 + < + < or + > < 0 > 5