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Warm up, Tuesday 1/21 Make a number line for the following

inequali>es: X > 4 4 > x

x ! 4 x ! "4

5.2.1 How can I solve inequali>es? Solving Inequali>es with One or Two Variables Student pages for this lesson are 235 – 240. In Sec>on 5.1, you developed many strategies for solving equa>ons with one variable and systems of equa>ons with two variables. But what if you want to solve an inequality or system of inequali>es instead? Today you will explore how to use familiar strategies to find solu>ons for an inequality. As you work, the ques>ons below can help focus team discussions: •  What strategy should we use? •  How can we know if this solu>on is correct? •  How can we be sure we found all solu>ons?

5-‐54

Chapter 5: Solving and Intersections 429

5.2.1 How can I solve inequalities? • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •

Solving Inequalities with One or Two Variables

Student pages for this lesson are 235 – 240.

In Section 5.1, you developed many strategies for solving equations with one variable and systems of equations with two variables. But what if you want to solve an inequality or system of inequalities instead? Today you will explore how to use familiar strategies to find solutions for an inequality. As you work, the questions below can help focus team discussions:

What strategy should we use?

How can we know if this solution is correct?

How can we be sure we found all solutions?

5-54. In Lessons 5.1.1 and 5.1.2, you learned how to use the graph of a system to solve an

equation. How can the graphs of y = 2x2 + 5x ! 3 and y = x2 + 4x + 3 (shown at right) help you solve an inequality? Consider this as you answer the questions below. [ a: The x-coordinate for each point of intersection; x = !3 and x = 2 ; b: The graph of y = 2x

2+ 5x – 3 increases faster or substitute a number such as 0 or –2

for x in both equations, or the y-intercept of y = 2x2 + 5x ! 3 is –3 and the other is 3, etc. c: The solutions are the x-values of the points where the graph of y = 2x2 + 5x ! 3 intersects and dips below the graph of y = x2 + 4x + 3 ; there are infinite solutions; !3 " x " 2 . d and e: See number lines below right. e: x < !3!or!x > 2 ]

a. How are the solutions of 2x2 + 5x ! 3 = x2 + 4x + 3 represented on this graph? What are the solutions?

b. Obtain a Lesson 5.2.1 Resource Page from your teacher. On the resource page, label each graph with its equation and highlight each function with a different color. How did you decide which graph matches which function?

Problem continues on next page.!

x

y
















x

y

5-‐54
















x

y
















x

y

They are represented as points of intersec>on on the graph, in this case, those intersec>ons occur when x=-‐3 and x=2. The actual coordinates seem to be (-‐3,0) and (2,15)

.

.

5-‐54
















x

y
















x

y

They are represented as points of intersec>on on the graph, in this case, those intersec>ons occur when x=-‐3 and x=2. The actual coordinates seem to be (-‐3,0) and (2,15)

.

.

Remember, if all points on a func>on are a solu>on, then the point or points where mul>ple func>ons meet is the solu>on to that system!

5-‐54
















x

y

Obtain a Lesson 5.2.1 Resource Page from your teacher. On the resource page, label each graph with its equa>on and highlight each func>on with a different color. How did you decide which graph matches which func>on?

5-‐54
















x

y

Obtain a Lesson 5.2.1 Resource Page from your teacher. On the resource page, label each graph with its equa>on and highlight each func>on with a different color. How did you decide which graph matches which func>on?

One way to deal with this is by graphing each func>on separately:

Algebra 2 Connections 430

5-54. Problem continued from previous page.

c. On the graph, identify the x-values for which 2x2 + 5x ! 3 " x2 + 4x + 3 . How did you locate the solutions? How many solutions are there? Find a way to describe all of the solutions.

d. How can these solutions be represented on a number line? Locate the number line labeled with 2x2 + 5x ! 3 " x2 + 4x + 3 below the graph on your resource page. Use a colored marker to highlight the solutions to the inequality on the number line.

e. What about the inequality2x2 + 5x ! 3 > x2 + 4x + 3? What are the solutions to this inequality? Represent your solutions algebraically and on a number line.

5-55. Now consider the inequality 2x ! 5 > 1 . [ a: Answers vary, x > 3 ; b: 3; c: See

number line far below right. x > 3 ]

a. List at least three solutions of this inequality.

b. What is the smallest number that will make this inequality true? If you cannot find the smallest number, which number is the largest that makes it not true? This number is a boundary .

c. Draw a number line like the one at right on your paper and mark the boundary point for 2x ! 5 > 1 . Is the point part of the solution of the inequality? If yes, fill in a point on the line; if not draw an open circle. On which side of the boundary do the solutions lie? Indicate the solutions by “bolding” the appropriate portion of the number line and representing the graph algebraically.

5-56. Consider the inequality 4 x +1 ! 2 > 6 . [ a: There are two boundaries: x = 1 and

x = !3 . Both should be unfilled. b: There are three regions to test; the solutions are x > 1 or x < !3 . See number line below right. ]

a. How many boundary points are there? What are they? Should they be marked with filled or unfilled circles? Make the appropriate markings on a number line.

b. Which portions of the number line contain the solutions? How many regions do you need to test? Represent the solutions algebraically and on a number line.

–3 0 3

0 3 6 - 3 - 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–3 0 3















–3 0 3

0 3 6 - 3 - 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–3 0 3















–3 0 3

0 3 6 - 3 - 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–3 0 3















–3 0 3

0 3 6 - 3 - 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–3 0 3















–3 0 3

0 3 6 - 3 - 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–3 0 3

By comparing the two, you can see which one is increasing faster. One way to think about that is by seeing which one is ge]ng narrower faster. The one on the le^ seems a bit “sharper” than the one on the right















–3 0 3

0 3 6 - 3 - 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–3 0 3















–3 0 3

0 3 6 - 3 - 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–3 0 3

By comparing the two, you can see which one is increasing faster. One way to think about that is by seeing which one is ge]ng narrower faster. The one on the le^ seems a bit “sharper” than the one on the right

By comparing the two func>ons, the one with the larger coefficient should be increasing faster, just like they do with exponen>al func>ons from the first semester.

Speaking of equa>ons, you can also just plug in numbers for x and see what you get…















–3 0 3

0 3 6 - 3 - 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–3 0 3

=y

If we plug in 0 for example: 2(0)^2 + 5(0) – 3 = y 2 + 5 – 3 = y -‐3 = y

Speaking of equa>ons, you can also just plug in numbers for x and see what you get…















–3 0 3

0 3 6 - 3 - 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–3 0 3

=y

If we plug in 0 for example: 2(0)^2 + 5(0) – 3 = y 2 + 5 – 3 = y -‐3 = y

That point seems to be on the narrower line, so this must be the equa>on for the narrower parabola.

.

*Remember, plugging in 0 for x gets you the y-‐intercept, since all y-‐intercepts have x values of 0















–3 0 3

0 3 6 - 3 - 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–3 0 3

Which (x,y) values make the first func>on work?















–3 0 3

0 3 6 - 3 - 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–3 0 3

y















–3 0 3

0 3 6 - 3 - 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–3 0 3
















–3 0 3

0 3 6 - 3 - 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–3 0 3

y

Which (x,y) values make the second func>on work?















–3 0 3

0 3 6 - 3 - 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–3 0 3

y















–3 0 3

0 3 6 - 3 - 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–3 0 3
















–3 0 3

0 3 6 - 3 - 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–3 0 3

y
















–3 0 3

0 3 6 - 3 - 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–3 0 3

y

If the red part is all the solu>ons for the first func>on, and the blue is the solu>on for all of the second func>on, where is the solu>on for both?















–3 0 3

0 3 6 - 3 - 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–3 0 3
















–3 0 3

0 3 6 - 3 - 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–3 0 3

y
















–3 0 3

0 3 6 - 3 - 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–3 0 3

y

If the red part is all the solu>ons for the first func>on, and the blue is the solu>on for all of the second func>on, where is the solu>on for both?

Shouldn’t it be where they both overlap?!?

How can these solu>ons be represented on a number line? Locate the number line labeled with _________________ below the graph on your resource page. Use a colored marker to highlight the solu>ons to the inequality on the number line.















–3 0 3

0 3 6 - 3 - 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–3 0 3
















–3 0 3

0 3 6 - 3 - 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–3 0 3

Just consider what x values are used for all of the solu>ons for the swoosh part
















–3 0 3

0 3 6 - 3 - 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–3 0 3


Where does it start, where does it stop? Try to think of it as a domain.
















–3 0 3

0 3 6 - 3 - 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–3 0 3

















–3 0 3

0 3 6 - 3 - 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–3 0 3
















–3 0 3

0 3 6 - 3 - 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–3 0 3

















–3 0 3

0 3 6 - 3 - 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–3 0 3

What about the inequality ____________________ What are the solu>ons to this inequality? Represent your solu>ons algebraically and on a number line.















–3 0 3

0 3 6 - 3 - 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–3 0 3
















–3 0 3

0 3 6 - 3 - 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–3 0 3

















–3 0 3

0 3 6 - 3 - 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–3 0 3
















–3 0 3

0 3 6 - 3 - 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–3 0 3

Think about what x values are not included, while also considering which ones are
















–3 0 3

0 3 6 - 3 - 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–3 0 3

















–3 0 3

0 3 6 - 3 - 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–3 0 3
















–3 0 3

0 3 6 - 3 - 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–3 0 3

Think about what x values are not included, while also considering which ones are















–3 0 3

0 3 6 - 3 - 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–3 0 3















–3 0 3

0 3 6 - 3 - 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–3 0 3















–3 0 3

0 3 6 - 3 - 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–3 0 3















–3 0 3

0 3 6 - 3 - 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–3 0 3















–3 0 3

0 3 6 - 3 - 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–3 0 3

There are many answers, and all of them are bigger than 3, so you can write it as x > 3.















–3 0 3

0 3 6 - 3 - 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–3 0 3















–3 0 3

0 3 6 - 3 - 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–3 0 3
















–3 0 3

0 3 6 - 3 - 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–3 0 3















–3 0 3

0 3 6 - 3 - 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–3 0 3















–3 0 3

0 3 6 - 3 - 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–3 0 3
















–3 0 3

0 3 6 - 3 - 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–3 0 3

The answer is 3, because when you plug it into the inequality, it creates this ridiculous statement: 3 > 3… and that is completely absurd















–3 0 3

0 3 6 - 3 - 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–3 0 3















–3 0 3

0 3 6 - 3 - 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–3 0 3
















–3 0 3

0 3 6 - 3 - 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–3 0 3
















–3 0 3

0 3 6 - 3 - 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–3 0 3















–3 0 3

0 3 6 - 3 - 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–3 0 3















–3 0 3

0 3 6 - 3 - 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–3 0 3















–3 0 3

0 3 6 - 3 - 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–3 0 3
















–3 0 3

0 3 6 - 3 - 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–3 0 3
















–3 0 3

0 3 6 - 3 - 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–3 0 3















–3 0 3

0 3 6 - 3 - 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–3 0 3















–3 0 3

0 3 6 - 3 - 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–3 0 3















–3 0 3

0 3 6 - 3 - 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–3 0 3















–3 0 3

0 3 6 - 3 - 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–3 0 3















–3 0 3

0 3 6 - 3 - 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–3 0 3















–3 0 3

0 3 6 - 3 - 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–3 0 3

Something to think about: |x| = x. That means |-‐8| = 8 and |8| =8 also















–3 0 3

0 3 6 - 3 - 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–3 0 3















–3 0 3

0 3 6 - 3 - 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–3 0 3


x= 1 and x= -‐3 4|x+1| -‐2 > 6 4|x+1| > 8 |x+1| >2

X+1 > 2 X> 1

X+1> -‐2 X> -‐3

Once you get to this point, make two separate inequali>es

So 1 and -‐3 are the boundaries.















–3 0 3

0 3 6 - 3 - 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–3 0 3















–3 0 3

0 3 6 - 3 - 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–3 0 3


x= 1 and x= -‐3 4|x+1| -‐2 > 6 4|x+1| > 8 |x+1| >2

X+1 > 2 X> 1

X+1> -‐2 X> -‐3

















–3 0 3

0 3 6 - 3 - 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–3 0 3















–3 0 3

0 3 6 - 3 - 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–3 0 3















–3 0 3

0 3 6 - 3 - 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–3 0 3


x= 1 and x= -‐3 4|x+1| -‐2 > 6 4|x+1| > 8 |x+1| >2

X+1 > 2 X> 1

X+1> -‐2 X> -‐3

















–3 0 3

0 3 6 - 3 - 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–3 0 3















–3 0 3

0 3 6 - 3 - 6

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–5 –4 –3 –2 –1 0 1 2 3 4 5 x

–3 0 3

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