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Algebra 1B Section 5-3
 

Algebra 1B Section 5-3

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Solving Multi-Step Inequalities

Solving Multi-Step Inequalities

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    Algebra 1B Section 5-3 Algebra 1B Section 5-3 Presentation Transcript

    • SECTION 5-3 Solving Multi-Step Inequalities Monday, September 30, 13
    • ESSENTIAL QUESTION • How do we solve multi-step inequalities? Monday, September 30, 13
    • WARM UP 2x − 7 = 9 + 4xa. 1 2 a = a − 3 4 b. Monday, September 30, 13
    • WARM UP 2x − 7 = 9 + 4xa. 1 2 a = a − 3 4 b. −2x Monday, September 30, 13
    • WARM UP 2x − 7 = 9 + 4xa. 1 2 a = a − 3 4 b. −2x −2x Monday, September 30, 13
    • WARM UP 2x − 7 = 9 + 4xa. 1 2 a = a − 3 4 b. −2x −2x−9 Monday, September 30, 13
    • WARM UP 2x − 7 = 9 + 4xa. 1 2 a = a − 3 4 b. −2x −2x−9−9 Monday, September 30, 13
    • WARM UP 2x − 7 = 9 + 4xa. 1 2 a = a − 3 4 b. −2x −2x−9−9 Monday, September 30, 13
    • WARM UP 2x − 7 = 9 + 4xa. 1 2 a = a − 3 4 b. −2x −2x−9−9 −16 = 2x Monday, September 30, 13
    • WARM UP 2x − 7 = 9 + 4xa. 1 2 a = a − 3 4 b. −2x −2x−9−9 −16 = 2x 22 Monday, September 30, 13
    • WARM UP 2x − 7 = 9 + 4xa. 1 2 a = a − 3 4 b. −2x −2x−9−9 −16 = 2x 22 −8 = x Monday, September 30, 13
    • WARM UP 2x − 7 = 9 + 4xa. 1 2 a = a − 3 4 b. −2x −2x−9−9 −16 = 2x 22 −8 = x x = −8 Monday, September 30, 13
    • WARM UP 2x − 7 = 9 + 4xa. 1 2 a = a − 3 4 b. −2x −2x−9−9 −16 = 2x 22 −8 = x x = −8 4 • 1 2 a = a − 3 4 • 4 Monday, September 30, 13
    • WARM UP 2x − 7 = 9 + 4xa. 1 2 a = a − 3 4 b. −2x −2x−9−9 −16 = 2x 22 −8 = x x = −8 4 • 1 2 a = a − 3 4 • 4 2a = a − 3 Monday, September 30, 13
    • WARM UP 2x − 7 = 9 + 4xa. 1 2 a = a − 3 4 b. −2x −2x−9−9 −16 = 2x 22 −8 = x x = −8 4 • 1 2 a = a − 3 4 • 4 2a = a − 3 −a−a Monday, September 30, 13
    • WARM UP 2x − 7 = 9 + 4xa. 1 2 a = a − 3 4 b. −2x −2x−9−9 −16 = 2x 22 −8 = x x = −8 4 • 1 2 a = a − 3 4 • 4 2a = a − 3 −a−a a = −3 Monday, September 30, 13
    • EXAMPLE 1 Maggie Brann has a budget of $115 for faxes.The fax service she uses charges $25 to activate an account and $0.08 per page to send faxes. How many pages can she fax and stay within her budget? Identify a variable, then set up and solve an inequality. Interpret your solution. Monday, September 30, 13
    • EXAMPLE 1 Maggie Brann has a budget of $115 for faxes.The fax service she uses charges $25 to activate an account and $0.08 per page to send faxes. How many pages can she fax and stay within her budget? Identify a variable, then set up and solve an inequality. Interpret your solution. p = pages she can fax Monday, September 30, 13
    • EXAMPLE 1 Maggie Brann has a budget of $115 for faxes.The fax service she uses charges $25 to activate an account and $0.08 per page to send faxes. How many pages can she fax and stay within her budget? Identify a variable, then set up and solve an inequality. Interpret your solution. p = pages she can fax 25+ .08p ≤115 Monday, September 30, 13
    • EXAMPLE 1 Maggie Brann has a budget of $115 for faxes.The fax service she uses charges $25 to activate an account and $0.08 per page to send faxes. How many pages can she fax and stay within her budget? Identify a variable, then set up and solve an inequality. Interpret your solution. p = pages she can fax 25+ .08p ≤115 −25 −25 Monday, September 30, 13
    • EXAMPLE 1 Maggie Brann has a budget of $115 for faxes.The fax service she uses charges $25 to activate an account and $0.08 per page to send faxes. How many pages can she fax and stay within her budget? Identify a variable, then set up and solve an inequality. Interpret your solution. p = pages she can fax 25+ .08p ≤115 −25 −25 .08p ≤ 90 Monday, September 30, 13
    • EXAMPLE 1 Maggie Brann has a budget of $115 for faxes.The fax service she uses charges $25 to activate an account and $0.08 per page to send faxes. How many pages can she fax and stay within her budget? Identify a variable, then set up and solve an inequality. Interpret your solution. p = pages she can fax 25+ .08p ≤115 −25 −25 .08p ≤ 90 .08 .08 Monday, September 30, 13
    • EXAMPLE 1 Maggie Brann has a budget of $115 for faxes.The fax service she uses charges $25 to activate an account and $0.08 per page to send faxes. How many pages can she fax and stay within her budget? Identify a variable, then set up and solve an inequality. Interpret your solution. p = pages she can fax 25+ .08p ≤115 −25 −25 .08p ≤ 90 .08 .08 p ≤1125 Monday, September 30, 13
    • EXAMPLE 1 Maggie Brann has a budget of $115 for faxes.The fax service she uses charges $25 to activate an account and $0.08 per page to send faxes. How many pages can she fax and stay within her budget? Identify a variable, then set up and solve an inequality. Interpret your solution. p = pages she can fax 25+ .08p ≤115 −25 −25 .08p ≤ 90 .08 .08 p ≤1125 p | p ≤1125{ } Monday, September 30, 13
    • EXAMPLE 1 Maggie Brann has a budget of $115 for faxes.The fax service she uses charges $25 to activate an account and $0.08 per page to send faxes. How many pages can she fax and stay within her budget? Identify a variable, then set up and solve an inequality. Interpret your solution. p = pages she can fax 25+ .08p ≤115 −25 −25 .08p ≤ 90 .08 .08 p ≤1125 p | p ≤1125{ } Maggie can send up to 1125 faxes. Monday, September 30, 13
    • EXAMPLE 2 Solve the inequality. 13−11d ≥ 79 Monday, September 30, 13
    • EXAMPLE 2 Solve the inequality. 13−11d ≥ 79 −13 −13 Monday, September 30, 13
    • EXAMPLE 2 Solve the inequality. 13−11d ≥ 79 −13 −13 −11d ≥ 66 Monday, September 30, 13
    • EXAMPLE 2 Solve the inequality. 13−11d ≥ 79 −13 −13 −11d ≥ 66 −11 −11 Monday, September 30, 13
    • EXAMPLE 2 Solve the inequality. 13−11d ≥ 79 −13 −13 −11d ≥ 66 −11 −11 d ≤ −6 Monday, September 30, 13
    • EXAMPLE 2 Solve the inequality. 13−11d ≥ 79 −13 −13 −11d ≥ 66 −11 −11 d ≤ −6 d | d ≤ −6{ } Monday, September 30, 13
    • EXAMPLE 3 Define a variable, write an inequality and solve the problem.Then check your solution. Four times a number increased by twelve is less than the difference of that number and three. Monday, September 30, 13
    • EXAMPLE 3 Define a variable, write an inequality and solve the problem.Then check your solution. Four times a number increased by twelve is less than the difference of that number and three. n = the number Monday, September 30, 13
    • EXAMPLE 3 Define a variable, write an inequality and solve the problem.Then check your solution. Four times a number increased by twelve is less than the difference of that number and three. n = the number 4n +12 < n − 3 Monday, September 30, 13
    • EXAMPLE 3 Define a variable, write an inequality and solve the problem.Then check your solution. Four times a number increased by twelve is less than the difference of that number and three. n = the number 4n +12 < n − 3 −n −n Monday, September 30, 13
    • EXAMPLE 3 Define a variable, write an inequality and solve the problem.Then check your solution. Four times a number increased by twelve is less than the difference of that number and three. n = the number 4n +12 < n − 3 −n −n 3n +12 < −3 Monday, September 30, 13
    • EXAMPLE 3 Define a variable, write an inequality and solve the problem.Then check your solution. Four times a number increased by twelve is less than the difference of that number and three. n = the number 4n +12 < n − 3 −n −n 3n +12 < −3 −12 −12 Monday, September 30, 13
    • EXAMPLE 3 Define a variable, write an inequality and solve the problem.Then check your solution. Four times a number increased by twelve is less than the difference of that number and three. n = the number 4n +12 < n − 3 −n −n 3n +12 < −3 −12 −12 3n < −15 Monday, September 30, 13
    • EXAMPLE 3 Define a variable, write an inequality and solve the problem.Then check your solution. Four times a number increased by twelve is less than the difference of that number and three. n = the number 4n +12 < n − 3 −n −n 3n +12 < −3 −12 −12 3n < −15 3n < −15 Monday, September 30, 13
    • EXAMPLE 3 Define a variable, write an inequality and solve the problem.Then check your solution. Four times a number increased by twelve is less than the difference of that number and three. n = the number 4n +12 < n − 3 −n −n 3n +12 < −3 −12 −12 3n < −15 3n < −15 3 3 Monday, September 30, 13
    • EXAMPLE 3 Define a variable, write an inequality and solve the problem.Then check your solution. Four times a number increased by twelve is less than the difference of that number and three. n = the number 4n +12 < n − 3 −n −n 3n +12 < −3 −12 −12 3n < −15 3n < −15 3 3 n < −5 Monday, September 30, 13
    • EXAMPLE 3 Define a variable, write an inequality and solve the problem.Then check your solution. Four times a number increased by twelve is less than the difference of that number and three. n = the number 4n +12 < n − 3 −n −n 3n +12 < −3 −12 −12 3n < −15 3n < −15 3 3 n < −5 n | n < −5{ } Monday, September 30, 13
    • EXAMPLE 4 6c + 3(2 − c) > −2c +1a. Monday, September 30, 13
    • EXAMPLE 4 6c + 3(2 − c) > −2c +1a. 6c + 6 − 3c > −2c +1 Monday, September 30, 13
    • EXAMPLE 4 6c + 3(2 − c) > −2c +1a. 6c + 6 − 3c > −2c +1 3c + 6 > −2c +1 Monday, September 30, 13
    • EXAMPLE 4 6c + 3(2 − c) > −2c +1a. 6c + 6 − 3c > −2c +1 3c + 6 > −2c +1 +2c +2c Monday, September 30, 13
    • EXAMPLE 4 6c + 3(2 − c) > −2c +1a. 6c + 6 − 3c > −2c +1 3c + 6 > −2c +1 +2c +2c 5c + 6 > +1 Monday, September 30, 13
    • EXAMPLE 4 6c + 3(2 − c) > −2c +1a. 6c + 6 − 3c > −2c +1 3c + 6 > −2c +1 +2c +2c 5c + 6 > +1 −6 −6 Monday, September 30, 13
    • EXAMPLE 4 6c + 3(2 − c) > −2c +1a. 6c + 6 − 3c > −2c +1 3c + 6 > −2c +1 +2c +2c 5c + 6 > +1 −6 −6 5c > −5 Monday, September 30, 13
    • EXAMPLE 4 6c + 3(2 − c) > −2c +1a. 6c + 6 − 3c > −2c +1 3c + 6 > −2c +1 +2c +2c 5c + 6 > +1 −6 −6 5c > −5 5 5 Monday, September 30, 13
    • EXAMPLE 4 6c + 3(2 − c) > −2c +1a. 6c + 6 − 3c > −2c +1 3c + 6 > −2c +1 +2c +2c 5c + 6 > +1 −6 −6 5c > −5 5 5 c > −1 Monday, September 30, 13
    • EXAMPLE 4 6c + 3(2 − c) > −2c +1a. 6c + 6 − 3c > −2c +1 3c + 6 > −2c +1 +2c +2c 5c + 6 > +1 −6 −6 5c > −5 5 5 c > −1 c | c > −1{ } Monday, September 30, 13
    • EXAMPLE 4 −7(k + 4)+11k ≥ 8k − 2(2k +1)b. Monday, September 30, 13
    • EXAMPLE 4 −7(k + 4)+11k ≥ 8k − 2(2k +1)b. −7k − 28 +11k ≥ 8k − 4k −1 Monday, September 30, 13
    • EXAMPLE 4 −7(k + 4)+11k ≥ 8k − 2(2k +1)b. −7k − 28 +11k ≥ 8k − 4k −1 4k − 28 ≥ 4k −1 Monday, September 30, 13
    • EXAMPLE 4 −7(k + 4)+11k ≥ 8k − 2(2k +1)b. −7k − 28 +11k ≥ 8k − 4k −1 4k − 28 ≥ 4k −1 −4k −4k Monday, September 30, 13
    • EXAMPLE 4 −7(k + 4)+11k ≥ 8k − 2(2k +1)b. −7k − 28 +11k ≥ 8k − 4k −1 4k − 28 ≥ 4k −1 −4k −4k −28 ≥ −1 Monday, September 30, 13
    • EXAMPLE 4 −7(k + 4)+11k ≥ 8k − 2(2k +1)b. −7k − 28 +11k ≥ 8k − 4k −1 4k − 28 ≥ 4k −1 −4k −4k −28 ≥ −1 This is an untrue statement! Monday, September 30, 13
    • EXAMPLE 4 −7(k + 4)+11k ≥ 8k − 2(2k +1)b. −7k − 28 +11k ≥ 8k − 4k −1 4k − 28 ≥ 4k −1 −4k −4k −28 ≥ −1 This is an untrue statement! ∅ Monday, September 30, 13
    • EXAMPLE 4 2(4r + 3) ≤ 22 + 8(r − 2)c. Monday, September 30, 13
    • EXAMPLE 4 2(4r + 3) ≤ 22 + 8(r − 2)c. 8r + 6 ≤ 22 + 8r −16 Monday, September 30, 13
    • EXAMPLE 4 2(4r + 3) ≤ 22 + 8(r − 2)c. 8r + 6 ≤ 22 + 8r −16 8r + 6 ≤ 8r + 6 Monday, September 30, 13
    • EXAMPLE 4 2(4r + 3) ≤ 22 + 8(r − 2)c. 8r + 6 ≤ 22 + 8r −16 8r + 6 ≤ 8r + 6 −8r −8r Monday, September 30, 13
    • EXAMPLE 4 2(4r + 3) ≤ 22 + 8(r − 2)c. 8r + 6 ≤ 22 + 8r −16 8r + 6 ≤ 8r + 6 −8r −8r 6 ≤ 6 Monday, September 30, 13
    • EXAMPLE 4 2(4r + 3) ≤ 22 + 8(r − 2)c. 8r + 6 ≤ 22 + 8r −16 8r + 6 ≤ 8r + 6 −8r −8r 6 ≤ 6 This is a true statement! Monday, September 30, 13
    • EXAMPLE 4 2(4r + 3) ≤ 22 + 8(r − 2)c. 8r + 6 ≤ 22 + 8r −16 8r + 6 ≤ 8r + 6 −8r −8r 6 ≤ 6 This is a true statement! r | r is any real number{ } Monday, September 30, 13
    • EXAMPLE 4 2(4r + 3) ≤ 22 + 8(r − 2)c. 8r + 6 ≤ 22 + 8r −16 8r + 6 ≤ 8r + 6 −8r −8r 6 ≤ 6 This is a true statement! r | r is any real number{ } or Monday, September 30, 13
    • EXAMPLE 4 2(4r + 3) ≤ 22 + 8(r − 2)c. 8r + 6 ≤ 22 + 8r −16 8r + 6 ≤ 8r + 6 −8r −8r 6 ≤ 6 This is a true statement! r | r is any real number{ } r | r = { } or Monday, September 30, 13
    • SUMMARIZER −3p + 7 > 2How could you solve without multiplying or dividing each side by a negative number. Monday, September 30, 13
    • PROBLEM SETS Monday, September 30, 13
    • PROBLEM SETS Problem Set 1: p. 298 #1-11 Problem Set 2: p. 298 #13-39 odd (skip #37), "The secret of success in life is for a man to be ready for his opportunity when it comes." - Earl of Beaconsfield Monday, September 30, 13