1. ALGEBRA READINESS LESSON 9-2 Warm Up Lesson 9-2 Warm-Up

2. ALGEBRA READINESS LESSON 9-2 Warm Up Lesson 9-2 Warm-Up

3. ALGEBRA READINESS “Solving Inequalities by Adding or Subtracting” (9-2) (3-1) What are “equivalent inequalities”? What is the “Addition Property of Inequality”? Equivalent Inequalities: two inequalities with the same solutions Example: x < 3 has the same solution as x + 4 < 7. As you can see in the picture below, if you take 4 blocks away from both sides of x + 4 7, you get x < 3. Rule: Addition Property of Inequality: If you add the same number to both sides of an inequality, the inequality is equivalent to the original inequality. Note: This property is also true for ≥ and ≤. if a > b, then a + c > b + c if a < b, then a + c < b + c Example: 3 > 1, so 3 + 1 > 1 + 1 Example: -5 < 4, so -5 + 2 < 4 + 2 4 > 2 -3 < 6 Notice the left side is stays 2 bigger than the right side in the example on the left and 9 less than the right side in the example on the right if you add the same amount to both sides of the inequality.

4. ALGEBRA READINESS “Solving Inequalities by Adding and Subtracting” (9-2) (3-1) What is the “Subtraction Property of Inequality”? Rule: Subtraction Property of Inequality: If you subtract the same number to both sides of an inequality, the inequality is equivalent to the original inequality. if a > b, then a - c > b - c if a < b, then a - c < b - c Example: 3 > 1, so 3 - 1 > 1 - 1 Example: -5 < 4, so -5 - 2 < 4 - 2 2 > 0 -7 < 2 Note: This property is also true for ≥ and ≤.

5. ALGEBRA READINESS How do you check to make sure an equality it written or graphed correctly? To check an inequality, replace the variable with one of the solutions of the inequality Example: Solve and graph x – 3 < 5. x – 3 + 3 < 5 + 3 Add 3 to both sides. x < 8Simplify Check (Step 1): Both sides of the inequality should be equal when you substitute the variable with the solution. 8 – 3 = 5 Substitute or “Plug In” 8 for x. 5 = 5  (True Statement)Simplify Check (Step 2): Check to make sure the inequality symbol in pointing the correct way by substituting the variable with an solution. 7 – 3 < 5 7 is one of the solutions for x < 8. 4 < 5  (True Statement) Simplify “Solving Inequalities by Adding or Subtracting” (9-2)

6. ALGEBRA READINESS Solve p –3 < –5. p – 3 < – 5 Add 3 to both sides to isolate the variable + 3 + 3 Step 1: Check to see if your answer makes both sides equal. Check: p – 3 = – 5 Write the related equation. – 5 = – 5 Substitute –2 for p. (–2) – 3 – 5 Solving Inequalities by Adding or Subtracting LESSON 9-2 Additional Examples p < – 2 Simplify. + 0 Step 2: Check the inequality symbol by substituting into the inequality. p – 3 < – 5 Substitute a number less than –2 for p. (–3) – 3 < – 5 – 6 < – 5

7. ALGEBRA READINESS Solve 8 d – 2. Graph and check your solution. > + 2 + 2Add 2 to each side. 10 d, or d 10Simplify (Switch the sign if you switch the sign around). > < Check: 8 = d – 2Check to see if d makes both sides equal. 8 (10) – 2Substitute 10 for d. 8 = 8 8 ≥ d – 2 Check the direction of the inequality. 8 ≥ (9) – 2 Substitute 9 for d. 8 ≥ 7 Solving Inequalities by Addition and Subtracting LESSON 9-2 Additional Examples 8 d – 2 > + 0

8. ALGEBRA READINESS The Drama Club can spend no more than $120 on costumes. They spent $79. How much more can they spend? ≤ Words amount plus amount left is no more $120 spent to spend than Let c = the amount left to spend on costumes. Inequality 79 + c 120 Subtract 79 from both sides to isolate the variable. – 79 79 + c ≤ 120 The Drama Club can spend at most $41 more on costumes. c ≤ 41 Simplify. Solving Inequalities by Adding or Subtracting LESSON 9-2 Additional Examples 0 +

9. ALGEBRA READINESS Solve each inequality. 1.q – 5 82.r + 10 < 4 3.6 + x 21 4.A lamp can use light bulbs of up to 75 watts. The lamp is using a 45-watt bulb. At most, how many watts are available for brighter light? > – < – r < – 6 30 watts x 15 < – q 13 > – Solving Inequalities by Adding or Subtracting LESSON 9-2 Lesson Quiz