9.2 Solving Single-Step Inequalities Math 9

9.2 We can solve inequalities using the exact same methods learned in Chapter 8. Our goal is to ISOLATE the variable. In order to do this, we need to explore how inequalities work in an equation.

9.2 Investigation: Let us investigate how an inequality symbol works in an inequality by filling out the following tables and answering the observation questions

Inequality Operation on both sides New 6 > -4 Add 2   8 > - 2 Subtract 2   4 > -6 Multiply 2   12 > -8 Divide 2   3 > -2

This isn’t true. We must flip the sign Inequality Operation on both sides New 6 > -4 Add (-2)   4 > -6 Subtract (-2)   8 > -2 Multiply (-2)   -12 > 8 Divide (-2)   -3 > 2 This isn’t true. We must flip the sign -12 < 8 -3 < 2 And again

Does the direction of the inequality stay the same when you do the following? add the same positive number to both sides? Yeah multiply both sides by the same positive number? add the same negative number to both sides? multiply both sides by the same negative number? No subtract the same positive number to both sides? divide both sides by the same positive number? divide both sides by the same negative number?

Conclusion ** If you multiply or divide both sides of an inequality equation by a negative number then the inequality sign must be flipped.

Example 1: Solving Inequalities We solve inequalities by isolating the variable using techniques learned from Chapter 8 a) x + 1.6 ≤ -5.6 -1.6 -1.6 x ≤ -7.2 b) 10 > -4x ÷(-4) ÷(-4) -2.5 < x Don’t forget the flip!

Less than boundary point Example 2 – Verifying Solutions to Inequalities We verify solutions of an inequality by substituting the value found for x into the inequality and see if it is true. Verify that x ≤ 32 is the solution to x −12 ≤ 20 Boundary point x −12 = 20 32 -12 = 20 20 = 20 Less than boundary point x −12 ≤ 20 30 -12 ≤ 20 18 ≤ 20

Less than boundary point Verify that x < −6 is the solution to −5x < −6 Boundary point −5x = −6 -5(-6) = -6 30 = -6 Not true Less than boundary point −5x < −6 -5(-7) < -6 35 < -6 Not true x < −6 is not the correct answer

Example 3 – Model and Solve a Problem Yvonne is planting trees as a summer job. She gets paid $0.10 per tree planted. She wants to earn at least $20/h. How many trees must she plant per hour in order to achieve her goal?

Let t = number of trees planted Example 3 – Model and Solve a Problem Yvonne is planting trees as a summer job. She gets paid $0.10 per tree planted. She wants to earn at least $20/h. How many trees must she plant per hour in order to achieve her goal? a) Write the inequality to model the number of trees Let t = number of trees planted 0.10t ≥ 20

Whole numbers = 0, 1, 2, 3… Integers = …-2, -1, 0, 1, 2… Example 3 – Model and Solve a Problem Yvonne is planting trees as a summer job. She gets paid $0.10 per tree planted. She wants to earn at least $20/h. How many trees must she plant per hour in order to achieve her goal? b) Will the solution be a set of whole numbers or integers? Explain. Whole numbers = 0, 1, 2, 3… Integers = …-2, -1, 0, 1, 2… This means that…

0.10t ≥ 20 ÷0.10 ÷0.10 t ≥ 200 Example 3 – Model and Solve a Problem Yvonne is planting trees as a summer job. She gets paid $0.10 per tree planted. She wants to earn at least $20/h. How many trees must she plant per hour in order to achieve her goal? c) Solve the inequality and interpret on a number line. 0.10t ≥ 20 ÷0.10 ÷0.10 t ≥ 200 100 200 300 400 500 600

April 23, 2012 9.2 Assignment Joe Schmoe 17, 19, 21, 22, *23, *27, *28