Preview Warm Up California Standards Lesson Presentation

Warm Up Solve. 1. n + 42 > 27 2. r + 15 < 39 3. –17 < w – 52 4. 34 < m - 19 n > –15 r < 24 35 < w 53 < m

California Standards Preview of Grade 7 AF4.0 Students solve simple linear equations and inequalities over the rational numbers.

When you multiply or divide both sides of an inequality by the same positive number, the statement will still be true. However, when you multiply or divide both sides by the same negative number, you need to reverse the direction of the inequality symbol for the statement to be true.

2 < 4 (-1)2 (-1)4 –2 –4 –2 > –4 2 is less than 4. –4 –2 0 2 4 6 8 10 2 < 4 2 is less than 4. (-1)2 (-1)4 Multiply both sides by –1. Use the number line to determine the correct inequality symbol. –2 –4 Reverse the inequality symbol because –2 is greater than –4. –2 > –4

Additional Example 1: Solving Inequalities by Multiplying Solve. c 4 A. ≤ –4 c 4 ≤ –4 c 4 (4) ≤ (–4) (4) Multiply both sides by 4. c ≤ –16

Additional Example 1: Solving Inequalities by Multiplying Solve. t B. > 0.3 –4 t –4 > 0.3 t (–4) < (–4)0.3 Multiply both sides by –4 and reverse the inequality symbol. –4 t < –1.2

Additional Example 1B Continued Check t B. > 0.3 –4 -2 –2 is less than –1.2 Substitute –2 for t. > 0.3 ? –4 0.5 > 0.3

n 6 A. ≤ –5 n 6 ≤ –5 n 6 (6) ≤ (–5) (6) Multiply both sides by 6. Check It Out! Example 1 Solve. n 6 A. ≤ –5 n 6 ≤ –5 n 6 (6) ≤ (–5) (6) Multiply both sides by 6. n ≤ –30

Multiply both sides by –3 and reverse the inequality symbol. –3 Check It Out! Example 1 Solve. r B. > 0.9 –3 r –3 > 0.9 r (–3) < (–3)0.9 Multiply both sides by –3 and reverse the inequality symbol. –3 r < –2.7

Check It Out! 1B Continued r B. > 0.9 –3 -3 –3 is less than –2.7 Substitute –3 for r. > 0.9 ? –3 1 > 0.9

Additional Example 2: Solving Inequalities by Dividing Solve. Check your answer. A. 5a ≥ 23 5a ≥ 23 Divide both sides by 5. 5 5 23 5 3 5 a ≥ , or 4 Check 5a ≥ 23 3 5 ? 5 is greater than 4 . 5(5) 23 ≥ Substitute 5 for a. ? 25 23 ≥

Additional Example 2: Solving Inequalities by Dividing Solve. Check your answer. B. 192< –24b 192 > –24b Divide both sides by –24, and reverse the inequality symbol. –24 –24 –8 > b Check 192 < -24b 192< -24(–10) ? –10 is less than –8. Substitute –10 for b. ? 192 < 240

Check It Out! Example 2 Solve. Check your answer. A. 6b ≥ 25 6b ≥ 25 Divide both sides by 6. 6 6 25 6 1 6 , or 4 b ≥ Check 6b ≥ 25 1 6 ? 6 is greater than 4 . 6(6) 25 ≥ Substitute 6 for b. ? 36 25 ≥

B. 85 < –17b 85 > –17b –17 –17 –5 > b 85 < –17b Check It Out! Example 2 Solve. Check your answer. B. 85 < –17b 85 > –17b Divide both sides by –17, and reverse the inequality symbol. –17 –17 –5 > b Check 85 < –17b 85 < –17(–6) ? –6 is less than –5. Substitute –6 for b. ? 85 < 102

Additional Example 3: Application It cost Josh $85 to make candles for the craft fair. How many candles must he sell at $4.00 each to make a profit? Since profit is the amount earned minus the amount spent, Josh needs to earn more than $85. Let c represent the number of candles that must be sold. 4c > 85 Write an inequality. 4c > 85 Divide both sides by 4. 4 4 c > 21.25 Josh cannot sell 0.25 candle, so he needs to sell more than 21 candles to earn a profit.

Check It Out! Example 3 It cost the class $15 to make cookies for the bake sale. How many cookies must they sell at 10¢ each to make a profit? Since profit is the amount earned minus the amount spent, the class needs to earn more than $15. Let c represent the number of cookies that must be sold. 0.10c > 15 Write an inequality. 0.10c > 15 Divide both sides by 0.10. 0.10 0.10 c > 150 The class must sell more than 150 cookies to make a profit.

Solve. Check each answer. 3. 18w < 4 4. –4f > 36 Lesson Quiz Solve. 1. 2. Solve. Check each answer. 3. 18w < 4 4. –4f > 36 5. It costs a candle company $51 to make a dozen candles. How many candles must it sell at $7 apiece to make a profit? s 9 > 12 s > 108 b –14 > 6 b < –84 2 9 w < f < –9 more than 7 candles, or at least 8 candles