Compound Inequalities

Blue Day: ___________________Gold Day: ___________________

Bell Ringer

1.Solve the inequality 4(k − 5) + 12k ≥ −4.

2.Solve the inequality, 3(x + 4) − x > 4?

3. How do you know that −5r + 6 ≤ −5r − 10 has no solution? Explain.

Agenda

How do we feel about multistep Inequalities? Solving compound inequalities Looking forward:

Work on solving word problems for the next class/maybe 2 classes

Review and Test the following 2 classes Tentative test days –

▪ 2nd Period – 10/9,8th Period- 10/12

Monday, October 12, 2015 – No school; teacher workday

Solve it!

https://media.pearsoncmg.com/curriculum/math/ma_hs16tx_a1te/scos/A0262886/player.html

2 types of compound inequalities“AND”

Statements

Can be written separately using the word “AND”

Can be written together without using the word “AND”

Graphs must cross to show what each answer shares.

Shading stops at the circles

“OR” Statements

Are only written separately using the word “OR”

Graphs can go in opposite directions, they do not have to cross.

Shading can continue past the circles.

1. Graph the solution set ofX > -5 and x < 1.

<---|---|---|---|---|---|---|---|---|--->

FIRST, write a Compound Inequality !!

-6 -5 -4 -3 -2 -1 0 1 2W

hat kin

d o

fC

ircle?

What kind of

Circle?

Now shadeBetween the

circles !!

When graphing COMPOUND INEQUALITIES ( the AND statements ), shading stops at the endpoints.

5 1x

2. Graph the solution set ofm > -2 and m < 3.

<---|---|---|---|---|---|---|---|---|--->

FIRST, write a Compound Inequality !!

-3 -2 -1 0 1 2

Wh

at kind

of

Circle?

What kind of

Circle?

Now shadeBetween the

circles !!

When graphing COMPOUND INEQUALITIES ( the AND statements ), shading stops at the endpoints.

3 4 5

2 3m

3. Solve and graph: x – 2 > - 4 and x + 2 <

5FIRST, Solve each for x.

NEXT, Write a Compound Inequality.

<---|---|---|---|---|---|---|---|---|---> -3 -2 -1 1 2 3 4 5

NOW, graph it!!

0

2 2x 2

2 2x 3

2 3x

4. Solve and graphx + 1 < 2 and -5x < 15

FIRST, Solve each for x.

NEXT, Write a Compound Inequality.

Start with the SMALLER number

NOW, graph it!!

<---|---|---|---|---|---|---|---|---|---> -3 -2 -1 0 1 2 3 4 5

1 1x 1

5 5x 3

3 1x

x + 1 < 2 and -5x < 15

5. Solve and graph:

8 2 4x FIRST, Solve for x.

Divide each term by 2.

22 2

4 x 2

<---|---|---|---|---|---|---|---|---|---> -5 -4 -3 -2 -1 0 1 2 3

NOW, graph it!!

6. Solve and graph:

9 3 12 9x Start in the middle and solve for x.

1212 123x3 21Now

divide each term by 3.

33 3x1 7

<---|---|---|---|---|---|---|---|---|---> 0 1 2 3 4 5 6 7 8

Now, let’s look at “OR” statements

7. Graph x < -2 OR x > 5

<---|---|---|---|---|---|---|---|---|---> -3 -2 -1 1 2 3 4 5 0

And that’s it!! The graph shows that the answers are true for one statement OR the other.

The arrows show that the shading does not stop, and that the answers continue on in that direction.

8. Graph: x ≤ 1 OR x < 3

<---|---|---|---|---|---|---|---|---|---> -3 -2 -1 1 2 3 4 5 0

<---|---|---|---|---|---|---|---|---|---> -3 -2 -1 1 2 3 4 5 0

Since these “or” solutions happen to cross, you can simplify the graph to look like this.

…because x < 3 crosses over or “covers up” x < 1

9. Solve and graph:

3 4m m 2 6m m OR

First, Solve for m.

mm2m 42 2m 2

mm3m 63 3m 2OR

<---|---|---|---|---|---|---|---|---|---> -3 -2 -1 1 2 3 4 5 0

10. Solve and graph

8 3 2x 12 11 1x

First, Solve for x.

OR8 83x 63 3x 2 OR

1111 11x

11111 x

1x Turn your answer around

<---|---|---|---|---|---|---|---|---|---> -3 -2 -1 1 2 3 4 5 0