Solving Inequalities Using Addition or Subtraction Section 3-2

Goals Goal To use addition or subtraction to solve inequalities. Rubric Level 1 – Know the goals. Level 2 – Fully understand the goals. Level 3 – Use the goals to solve simple problems. Level 4 – Use the goals to solve more advanced problems. Level 5 – Adapts and applies the goals to different and more complex problems.

Vocabulary Equivalent Inequalities

Solving Inequalities Solving one-step inequalities is much like solving one-step equations. To solve an inequality, you need to isolate the variable using the properties of inequality and inverse operations. At each step, you will create an inequality that is equivalent to the original inequality. Equivalent inequalities have the same solution set.

Addition & Subtraction Properties of Inequalities

When you add or subtract the same number on both sides of an inequality, the resulting statement will still be true. –2 < < 12 You can find solution sets of inequalities the same way you find solutions of equations, by isolating the variable. Addition & Subtraction Properties of Inequalities

In Lesson 3-1, you saw that one way to show the solution set of an inequality is by using a graph. Another way is to use set-builder notation. The set of all numbers x such that x has the given property. {x : x < 6}{x : x < 6} Read the above as “the set of all numbers x such that x is less than 6.” Inequality Solutions

Solve and graph the inequality. x + 3 > –5 –3 x > –8 Since 3 is added x, subtract 3 from both sides. –9  8 –7  6  5  4  3  2  Example:

Check x + 3 > –5 –4 + 3 > –5  –1 > –5 Substitute –4 for x. According to the graph –4 should be a solution and – 9 should not be a solution. So –4 is a solution. x + 3 > –5 –9 + 3 > –5  –6 > –5 Substitute –9 for x. So –9 is not a solution.  Example: Continued

Since there can be an infinite number of solutions to an inequality, it is not possible to check all the solutions. You can check the endpoint and the direction of the inequality symbol. The solutions of x + 9 < 15 are given by x < 6. Checking Inequality Solutions

Helpful Hint Use an inverse operation to “undo” the operation in an inequality. If the inequality contains addition, use subtraction to undo the addition.

Solve the inequality and graph the solutions. x + 12 < 20 –12 –12 x + 0 < 8 x < 8 Since 12 is added to x, subtract 12 from both sides to undo the addition. –10 –8 –6–4 – The solution set is {x: x < 8}. Example:

d – 5 > –7 Since 5 is subtracted from d, add 5 to both sides to undo the subtraction. The solution set is {d: d > –2}. +5 d + 0 > –2 d > –2 d – 5 > –7 Solve the inequality and graph the solutions. –10 –8 –6–4 – Example:

Solve the inequality and graph the solutions. 0.9 ≥ n – 0.3 Since 0.3 is subtracted from n, add 0.3 to both sides to undo the subtraction. The solution set is {n: n ≤ 1.2} ≥ n – ≥ n or n ≤ ≥ n –  1.2 Example:

a. s + 1 ≤ 10 –1 s + 0 ≤ 9 s ≤ 9 Since 1 is added to s, subtract 1 from both sides to undo the addition. Solve each inequality and graph the solutions. s + 1 ≤ 10 9 –10 –8 –6–4 – The solution set is {s: s ≤ 9}. Your Turn:

b. > –3 + t Since –3 is added to t, add 3 to both sides. > –3 + t +3 > 0 + t t < –10 –8 –6–4 – Solve each inequality and graph the solutions. Your Turn:

c. q – 3.5 < q – 0 < 11 q < 11 Since 3.5 is subtracted from q, add 3.5 to both sides to undo the subtraction. q – 3.5 < 7.5 –7–5–3– Solve each inequality and graph the solutions.

While training for a race, Ann’s goal is to run at least 3.5 miles each day. She has already run 1.8 miles today. Write and solve an inequality to find out how many more miles she must run today. Let m = the number of additional miles m ≥ 3.5 –1.8 m ≥ 1.7 Ann should run at least 1.7 more miles. Since 1.8 is added to m, subtract 1.8 from both sides. 1.8 miles plus additional miles is at least 3.5 miles m ≥ 3.5 Example: Application

Tim’s company produces recycled paper. They produce 60.5 lb of paper each day. They have already produced at least 20.2 lb today. Write and solve an inequality to find out how many more pounds Tim’s company must produce. Let p = the number of additional pounds of paper p ≥ 60.5 –20.2 –  20.2 p ≥ 40.3 Tim’s company should produce at least 40.3 lb more of paper. Since 20.2 is added to p, subtract 20.2 from both sides lbs plus additional pounds is at least 60.5 lb p ≥ 60.5 Your Turn:

Joke Time What flower grows between your nose and your chin? Tulips How many sides are there to a circle? 2 – inside and outside. What do you get when you cross an elephant and Darth Vader? An elevader.

Assignment 3.2 Exercises Pg. 187 – 190: #10 – 66 even