Algebra 1B Section 5-3

SECTION 5-3Solving Multi-Step Inequalities

Monday, September 30, 13

ESSENTIAL QUESTION

• How do we solve multi-step inequalities?


WARM UP

2x − 7 = 9+ 4xa. 12

a = a − 34

b.


WARM UP

2x − 7 = 9+ 4xa. 12

a = a − 34

b. −2x


WARM UP

2x − 7 = 9+ 4xa. 12

a = a − 34

b. −2x −2x


WARM UP

2x − 7 = 9+ 4xa. 12

a = a − 34

b. −2x −2x −9


WARM UP

2x − 7 = 9+ 4xa. 12

a = a − 34

b. −2x −2x −9 −9


WARM UP

2x − 7 = 9+ 4xa. 12

a = a − 34

b. −2x −2x −9 −9

−16 = 2x


WARM UP

2x − 7 = 9+ 4xa. 12

a = a − 34

b. −2x −2x −9 −9

−16 = 2x 2 2


WARM UP

2x − 7 = 9+ 4xa. 12

a = a − 34

b. −2x −2x −9 −9

−16 = 2x 2 2

−8 = x


WARM UP

2x − 7 = 9+ 4xa. 12

a = a − 34

b. −2x −2x −9 −9

−16 = 2x 2 2

−8 = x

x = −8


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


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


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


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


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+ .08p ≤115 −25 −25






25+ .08p ≤115 −25 −25

.08p ≤ 90






25+ .08p ≤115 −25 −25

.08p ≤ 90 .08 .08






25+ .08p ≤115 −25 −25

.08p ≤ 90 .08 .08

p ≤1125






25+ .08p ≤115 −25 −25

.08p ≤ 90 .08 .08

p ≤1125

p | p ≤1125{ }






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



13−11d ≥ 79 −13 −13



13−11d ≥ 79 −13 −13

−11d ≥ 66



13−11d ≥ 79 −13 −13

−11d ≥ 66 −11 −11



13−11d ≥ 79 −13 −13

−11d ≥ 66 −11 −11

d ≤ −6



13−11d ≥ 79 −13 −13

−11d ≥ 66 −11 −11

d ≤ −6

d | d ≤ −6{ }


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





n = the number 4n +12 < n − 3





n = the number 4n +12 < n − 3 −n −n






3n +12 < −3






3n +12 < −3 −12 −12






3n +12 < −3 −12 −12 3n < −15






3n +12 < −3 −12 −12 3n < −15

3n < −15






3n +12 < −3 −12 −12 3n < −15

3n < −15 3 3






3n +12 < −3 −12 −12 3n < −15

3n < −15 3 3 n < −5






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.


EXAMPLE 4

6c + 3(2 − c) > −2c +1a. 6c + 6 − 3c > −2c +1


EXAMPLE 4

6c + 3(2 − c) > −2c +1a. 6c + 6 − 3c > −2c +1

3c + 6 > −2c +1


EXAMPLE 4

6c + 3(2 − c) > −2c +1a. 6c + 6 − 3c > −2c +1

3c + 6 > −2c +1 +2c +2c


EXAMPLE 4

6c + 3(2 − c) > −2c +1a. 6c + 6 − 3c > −2c +1

3c + 6 > −2c +1 +2c +2c

5c + 6 > +1


EXAMPLE 4

6c + 3(2 − c) > −2c +1a. 6c + 6 − 3c > −2c +1

3c + 6 > −2c +1 +2c +2c

5c + 6 > +1 −6 −6


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


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


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


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{ }


EXAMPLE 4

−7(k + 4)+11k ≥ 8k − 2(2k +1)b.


EXAMPLE 4

−7(k + 4)+11k ≥ 8k − 2(2k +1)b. −7k − 28 +11k ≥ 8k − 4k −1


EXAMPLE 4

−7(k + 4)+11k ≥ 8k − 2(2k +1)b. −7k − 28 +11k ≥ 8k − 4k −1

4k − 28 ≥ 4k −1


EXAMPLE 4

−7(k + 4)+11k ≥ 8k − 2(2k +1)b. −7k − 28 +11k ≥ 8k − 4k −1

4k − 28 ≥ 4k −1 −4k −4k


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


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!


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!

∅


EXAMPLE 4

2(4r + 3) ≤ 22 + 8(r − 2)c.


EXAMPLE 4

2(4r + 3) ≤ 22 + 8(r − 2)c. 8r + 6 ≤ 22 + 8r −16


EXAMPLE 4

2(4r + 3) ≤ 22 + 8(r − 2)c. 8r + 6 ≤ 22 + 8r −16

8r + 6 ≤ 8r + 6


EXAMPLE 4

2(4r + 3) ≤ 22 + 8(r − 2)c. 8r + 6 ≤ 22 + 8r −16

8r + 6 ≤ 8r + 6 −8r −8r


EXAMPLE 4

2(4r + 3) ≤ 22 + 8(r − 2)c. 8r + 6 ≤ 22 + 8r −16

8r + 6 ≤ 8r + 6 −8r −8r

6 ≤ 6


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!


EXAMPLE 4

2(4r + 3) ≤ 22 + 8(r − 2)c. 8r + 6 ≤ 22 + 8r −16

8r + 6 ≤ 8r + 6 −8r −8r


r | r is any real number{ }


EXAMPLE 4

2(4r + 3) ≤ 22 + 8(r − 2)c. 8r + 6 ≤ 22 + 8r −16

8r + 6 ≤ 8r + 6 −8r −8r


r | r is any real number{ }or


EXAMPLE 4

2(4r + 3) ≤ 22 + 8(r − 2)c. 8r + 6 ≤ 22 + 8r −16

8r + 6 ≤ 8r + 6 −8r −8r


r | r is any real number{ }

r | r = { }or


SUMMARIZER −3p + 7 > 2How could you solve without multiplying

or dividing each side by a negative number.


PROBLEM SETS


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

Algebra 1B Section 5-3

Education

Transcript of Algebra 1B Section 5-3