algebra 1b section 5-3
DESCRIPTION
Solving Multi-Step InequalitiesTRANSCRIPT
SECTION 5-3Solving 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. 12
a = a − 34
b.
Monday, September 30, 13
WARM UP
2x − 7 = 9+ 4xa. 12
a = a − 34
b. −2x
Monday, September 30, 13
WARM UP
2x − 7 = 9+ 4xa. 12
a = a − 34
b. −2x −2x
Monday, September 30, 13
WARM UP
2x − 7 = 9+ 4xa. 12
a = a − 34
b. −2x −2x −9
Monday, September 30, 13
WARM UP
2x − 7 = 9+ 4xa. 12
a = a − 34
b. −2x −2x −9 −9
Monday, September 30, 13
WARM UP
2x − 7 = 9+ 4xa. 12
a = a − 34
b. −2x −2x −9 −9
Monday, September 30, 13
WARM UP
2x − 7 = 9+ 4xa. 12
a = a − 34
b. −2x −2x −9 −9
−16 = 2x
Monday, September 30, 13
WARM UP
2x − 7 = 9+ 4xa. 12
a = a − 34
b. −2x −2x −9 −9
−16 = 2x 2 2
Monday, September 30, 13
WARM UP
2x − 7 = 9+ 4xa. 12
a = a − 34
b. −2x −2x −9 −9
−16 = 2x 2 2
−8 = x
Monday, September 30, 13
WARM UP
2x − 7 = 9+ 4xa. 12
a = a − 34
b. −2x −2x −9 −9
−16 = 2x 2 2
−8 = x
x = −8
Monday, September 30, 13
WARM UP
2x − 7 = 9+ 4xa. 12
a = a − 34
b. −2x −2x −9 −9
−16 = 2x 2 2
−8 = x
x = −8
4 • 1
2a = a − 3
4• 4
Monday, September 30, 13
WARM UP
2x − 7 = 9+ 4xa. 12
a = a − 34
b. −2x −2x −9 −9
−16 = 2x 2 2
−8 = x
x = −8
4 • 1
2a = a − 3
4• 4
2a = a − 3
Monday, September 30, 13
WARM UP
2x − 7 = 9+ 4xa. 12
a = a − 34
b. −2x −2x −9 −9
−16 = 2x 2 2
−8 = x
x = −8
4 • 1
2a = a − 3
4• 4
2a = a − 3 −a −a
Monday, September 30, 13
WARM UP
2x − 7 = 9+ 4xa. 12
a = a − 34
b. −2x −2x −9 −9
−16 = 2x 2 2
−8 = x
x = −8
4 • 1
2a = a − 3
4• 4
2a = a − 3 −a −a
a = −3
Monday, September 30, 13
EXAMPLE 1Maggie 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 1Maggie 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 1Maggie 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 1Maggie 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 1Maggie 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 1Maggie 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 1Maggie 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 1Maggie 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 1Maggie 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 2Solve the inequality.
13−11d ≥ 79
Monday, September 30, 13
EXAMPLE 2Solve the inequality.
13−11d ≥ 79 −13 −13
Monday, September 30, 13
EXAMPLE 2Solve the inequality.
13−11d ≥ 79 −13 −13
−11d ≥ 66
Monday, September 30, 13
EXAMPLE 2Solve the inequality.
13−11d ≥ 79 −13 −13
−11d ≥ 66 −11 −11
Monday, September 30, 13
EXAMPLE 2Solve the inequality.
13−11d ≥ 79 −13 −13
−11d ≥ 66 −11 −11
d ≤ −6
Monday, September 30, 13
EXAMPLE 2Solve the inequality.
13−11d ≥ 79 −13 −13
−11d ≥ 66 −11 −11
d ≤ −6
d | d ≤ −6{ }
Monday, September 30, 13
EXAMPLE 3Define 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 3Define 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 3Define 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 3Define 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 3Define 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 3Define 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 3Define 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 3Define 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 3Define 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 3Define 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 3Define 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 ≥ −1This 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 ≥ −1This 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 ≤ 6This 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 ≤ 6This 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 ≤ 6This 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 ≤ 6This 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 BeaconsfieldMonday, September 30, 13