objective the student will be able to: solve inequalities. mfcr lesson 1-7 9-25-14

ObjectiveThe student will be able to:

solve inequalities.

MFCR Lesson 1-7

9-25-14

9-25-14 Bellwork

Section 1.7—Linear Inequalities in One Variable

• Copy Key concepts from p. 94

• Read examples 1 & 2

1.7—Linear Inequalities in One Variable

• A linear inequality in one variable can be written in the form

ax + b < 0,

ax + b> 0,

ax + b ≤ 0, or

ax + b ≥ 0

Properties of Inequalities

1. If a < b, then a + c < b + c.2. If a < b, then a – c < b – c.3. If c is positive and a < b, then ac < bc and a/c

< b/c.4. If c is negative and a < b, then ac > bc and a/c

> b/c. *Properties 3 and 4 indicate that if we multiply or divide

an inequality by a negative value, the direction of the inequality sign must be reversed.

1) Solve 5m - 8 > 12

+ 8 + 8

5m > 20

5 5

m > 4

5(4) – 8 = 12

1. Draw “the river” 2. Add 8 to both sides3. Simplify4. Divide both sides by 55. Simplify6. Check your answer7. Graph the solution

o4 53

2) Solve 12 - 3a > 18 - 12 - 12

-3a > 6

-3 -3

a < -2

12 - 3(-2) = 18

1. Draw “the river” 2. Subtract 12 from both

sides3. Simplify4. Divide both sides by -3 5. Simplify

(Switch the inequality!)6. Check your answer7. Graph the solution

o-2 -1-3

Which graph shows the solution to 2x - 10 ≥ 4?

1. .

2.

3.

4.

Answer NowAnswer Now

3) Solve 5m - 4 < 2m + 11-2m -2m

3m - 4 < 11

+ 4 + 4

3m < 15

3 3

m < 5

5(5) – 4 = 2(5) + 11

1. Draw “the river” 2. Subtract 2m from both

sides3. Simplify4. Add 4 to both sides5. Simplify6. Divide both sides by 3 7. Simplify8. Check your answer9. Graph the solution

o5 64

4) Solve 2r - 18 ≤ 5r + 3-2r -2r

-18 ≤ 3r + 3

- 3 - 3

-21 ≤ 3r

3 3

-7 ≤ r or r ≥ -7

2(-7) – 18 = 5(-7) + 3

1. Draw “the river” 2. Subtract 2r from both sides3. Simplify4. Subtract 3 from both sides5. Simplify6. Divide both sides by 3 7. Simplify8. Check your answer9. Graph the solution

●-7 -6-8

6) Solve -2x + 6 ≥ 3x - 41. x ≥ -2

2. x ≤ -2

3. x ≥ 2

4. x ≤ 2

Answer NowAnswer Now

5) Solve 26p - 20 > 14p + 64-14p -14p

12p – 20 > 64

+ 20 + 20

12p > 84

12 12

p > 7

26(7) – 20 = 14(7) + 64

1. Draw “the river” 2. Subtract 14p from both

sides3. Simplify4. Add 20 to both sides5. Simplify6. Divide both sides by 12 7. Simplify8. Check your answer9. Graph the solution

o7 86

What are the values of x if 3(x + 4) - 5(x - 1) < 5?

Answer NowAnswer Now

1. x < -6

2. x > -6

3. x < 6

4. x > 6

ObjectivesThe student will be able to:

1. solve compound inequalities.

2. graph the solution sets of compound inequalities.

Compound Inequalities

To solve a compound inequality, isolate the variable x in the “middle.” The operations performed on the middle portion of the inequality must also be performed on the left-hand side and right-hand side.

Ex. 1: -2 ≤ 3x + 1 < 5

Ex. 2: -8 < 5x – 3 ≤ 12

What is the difference between and and or?

AND means intersection-what do the two items

have in common?

OR means union-if it is in one item, it is in

the solution

A

A B

B

1) Graph x < 4 and x ≥ 2

3 42●

a) Graph x < 4

b) Graph x ≥ 23 42

o

c) Combine the graphs

●

3 42o

d) Where do they intersect?●

3 42o

2) Graph x < 2 or x ≥ 4

3 42●

a) Graph x < 2

b) Graph x ≥ 43 42

o

c) Combine the graphs

3 42o

3 42●

3) Which inequalities describe the following graph?

-2 -1-3oo

Answer NowAnswer Now

1. y > -3 or y < -1

2. y > -3 and y < -1

3. y ≤ -3 or y ≥ -1

4. y ≥ -3 and y ≤ -1

When written this way, it is the same thing as

6 < m AND m < 8

It can be rewritten as m > 6 and m < 8 and graphed as previously shown, however,

it is easier to graph everything

between 6 and 8!

4) Graph the compound inequality 6 < m < 8

7 86oo

5) Which is equivalent to-3 < y < 5?

1. y > -3 or y < 5

2. y > -3 and y < 5

3. y < -3 or y > 5

4. y < -3 and y > 5

Answer NowAnswer Now

6) Which is equivalent to x > -5 and x ≤ 1?

1. -5 < x ≤ 1

2. -5 > x ≥ 1

3. -5 > x ≤ 1

4. -5 < x ≥ 1

Answer NowAnswer Now

7) 2x < -6 and 3x ≥ 12

1. Solve each inequality for x

2. Graph each inequality3. Combine the graphs4. Where do they

intersect?5. They do not! x cannot

be greater than or equal to 4 and less than -3 No Solution!!

2 6

2 2 3

x

x

3x 12

3 3 x 4

-3 0-6o-3 0-6o

4 71o●4 71o●

8) Graph 3 < 2m – 1 < 9

Remember, when written like this, it is an AND problem!

3 < 2m – 1 AND 2m – 1 < 9

Solve each inequality.

Graph the intersection of 2 < m and m < 5.

0 5-5

9) Graph x < 2 or x ≥ 4

0 5-5

10) Graph x ≥ -1 or x ≤ 3

The whole line is shaded!!

0 5-5

Practice Problems

Solving a Compound Inequality Application

• Beth received grades of 87%, 82%, 96%, and 79% on her last four algebra tests. To graduate with honors, she needs at least a B in the course.a) What grade does she need to make on the 5th test to

get a B in the course? Assume that the tests are weighted equally and that to earn a B the average of the test grades must be at least 80% but less than 90%.

b) Is it possible for Beth to earn an A in the course if an A requires an average of 90% or more?

Solving a Linear Inequality Application

The number of registered passenger cars, N (in millions), in the U.S. has risen since 1960 according to the equation N = 2.5t + 64.4, where t represents the number of years after 1960 (t = 0 corresponds to 1960, t = 1 corresponds to 1961, etc.)a) For what years was the number of registered passenger

cars less than 89.4 million?

b) For what years was the number of registered passenger cars between 94.4 million and 101.9 million?

Exit Ticket – Hand in before leaving class.

a. What are the solutions to

-2 < x + 2 ≤ 5

b. Graph the solutions to the above compound inequality