Solving Linear Inequalities: Introduction and Formatting (page 1 of 7)

Similar documents
Math 1 Variable Manipulation Part 5 Absolute Value & Inequalities

September 12, Math Analysis Ch 1 Review Solutions. #1. 8x + 10 = 4x 30 4x 4x 4x + 10 = x = x = 10.

3.1 Inequalities - Graphing and Solving

Solving Multi-Step Linear Equations (page 3 and 4)

MITOCW watch?v=ztnnigvy5iq

ABSOLUTE VALUE EQUATIONS AND INEQUALITIES

Name: Block: Unit 2 Inequalities

1) 2) Algebra (3-2) Solving Inequalities with Additon and Subtraction

MA 1128: Lecture 08 03/02/2018. Linear Equations from Graphs And Linear Inequalities

Inequalities - Solve and Graph Inequalities

Practical Algebra. A Step-by-step Approach. Brought to you by Softmath, producers of Algebrator Software

Solving Equations. A: Solving One-Variable Equations. One Step x + 6 = 9-3y = 15. Two Step 2a 3 6. Algebra 2 Chapter 1 Notes 1.4 Solving Equations

Since the logs have the same base, I can set the arguments equal and solve: x 2 30 = x x 2 x 30 = 0

June If you want, you may scan your assignment and convert it to a.pdf file and it to me.

may be sent to:

Self-Directed Course: Transitional Math Module 4: Algebra

Chapter 5 Simplifying Formulas and Solving Equations

MITOCW MITRES_18-007_Part5_lec3_300k.mp4

P.1 Prerequisite skills Basic Algebra Skills

Math 5a Reading Assignments for Sections

Multi-Step Equations and Inequalities

Section 2.6 Solving Linear Inequalities

LESSON EII.C EQUATIONS AND INEQUALITIES

Chapter 1 Review of Equations and Inequalities

HW#2: Quads 7 #1 6. How do you find the answer to a Quadratic Inequality? 02Quad7 SolvingQuadraticInequalities Notes.notebook.

Inequalities - Absolute Value

MITOCW MIT18_02SCF10Rec_50_300k

A constant is a value that is always the same. (This means that the value is constant / unchanging). o

MITOCW ocw f99-lec17_300k

Lesson ACTIVITY: Tree Growth

Lesson #9 Simplifying Rational Expressions

Section 1: Sets and Interval Notation

PROFESSOR: WELCOME BACK TO THE LAST LECTURE OF THE SEMESTER. PLANNING TO DO TODAY WAS FINISH THE BOOK. FINISH SECTION 6.5

9. TRANSFORMING TOOL #2 (the Multiplication Property of Equality)

2-7 Solving Absolute-Value Inequalities

A. Incorrect! This inequality is a disjunction and has a solution set shaded outside the boundary points.

6.4 Division of Polynomials. (Long Division and Synthetic Division)

Finding Limits Graphically and Numerically

MITOCW watch?v=pqkyqu11eta

STANDARDS OF LEARNING CONTENT REVIEW NOTES HONORS ALGEBRA II. 1 st Nine Weeks,

Student Instruction Sheet: Unit 2, Lesson 2. Equations of Lines, Part 2

Chapter 5 Inequalities

Math 119 Main Points of Discussion

Ch. 3 Equations and Inequalities

Equations and inequalities

College Algebra Through Problem Solving (2018 Edition)

MITOCW MITRES_6-007S11lec09_300k.mp4

1.1 Linear Equations and Inequalities

17. 8x and 4x 2 > x 1 < 7 and 6x x or 2x x 7 < 3 and 8x x 9 9 and 5x > x + 3 < 3 or 8x 2

Instructor (Brad Osgood)

31. TRANSFORMING TOOL #2 (the Multiplication Property of Equality)

Investigating Inequalities:

Module 3 Study Guide. GCF Method: Notice that a polynomial like 2x 2 8 xy+9 y 2 can't be factored by this method.

MITOCW ocw-18_02-f07-lec02_220k

MITOCW ocw f99-lec01_300k

Algebra Revision Guide

WSMA Algebra - Expressions Lesson 14

Chapter 13 - Inverse Functions

The First Derivative Test

2.2 Graphs of Functions

MITOCW watch?v=fkfsmwatddy

32. SOLVING LINEAR EQUATIONS IN ONE VARIABLE

Algebra I Chapter 6: Solving and Graphing Linear Inequalities

University of Colorado at Colorado Springs Math 090 Fundamentals of College Algebra

C if U can. Algebra. Name

A-Level Notes CORE 1

STANDARDS OF LEARNING CONTENT REVIEW NOTES. ALGEBRA I Part II. 2 nd Nine Weeks,

Lesson 21 Not So Dramatic Quadratics

Definition: Absolute Value The absolute value of a number is the distance that the number is from zero. The absolute value of x is written x.

8/6/2010 Assignment Previewer

Pre-Algebra Notes Unit Three: Multi-Step Equations and Inequalities (optional)

MA 180 Lecture Chapter 1 College Algebra and Calculus by Larson/Hodgkins Equations and Inequalities

Matrix Dimensions(orders)

Definition: Quadratic equation: A quadratic equation is an equation that could be written in the form ax 2 + bx + c = 0 where a is not zero.

P arenthesis E xponents M ultiplication D ivision A ddition S ubtraction

ABE Math Review Package

Manipulating Equations

Chapter 1: January 26 January 30

Section 1.1. Chapter 1. Quadratics. Parabolas. Example. Example. ( ) = ax 2 + bx + c -2-1

MITOCW ocw f99-lec05_300k

AP Physics C Mechanics Calculus Basics

Pre-Algebra Notes Unit Three: Multi-Step Equations and Inequalities

Basic Equations and Inequalities. An equation is a statement that the values of two expressions are equal.

D. Correct! You translated the phrase exactly using x to represent the given real number.

Rational Functions. Elementary Functions. Algebra with mixed fractions. Algebra with mixed fractions

Math 2 Variable Manipulation Part 7 Absolute Value & Inequalities

Expressions, Equations and Inequalities Guided Notes

North Carolina State University

MITOCW MITRES_18-007_Part1_lec5_300k.mp4

Descartes' Rule of Signs

3: Linear Systems. Examples. [1.] Solve. The first equation is in blue; the second is in red. Here's the graph: The solution is ( 0.8,3.4 ).

Algebra 1 S1 Lesson Summaries. Lesson Goal: Mastery 70% or higher

Chapter 3. September 11, ax + b = 0.

MITOCW ocw-18_02-f07-lec25_220k

MITOCW MITRES18_006F10_26_0101_300k-mp4

Order of Operations. Real numbers

6.041SC Probabilistic Systems Analysis and Applied Probability, Fall 2013 Transcript Tutorial:A Random Number of Coin Flips

MITOCW watch?v=poho4pztw78

33. SOLVING LINEAR INEQUALITIES IN ONE VARIABLE

To Create a Simple Formula using the Point and Click Method:

Transcription:

Solving Linear Inequalities: Introduction and Formatting (page 1 of 7) Sections: Introduction and formatting, Elementary examples, Advanced examples Solving linear inequalities is almost exactly like solving linear equations. Solve x + 3 < 0. If they'd given me "x + 3 = 0", I'd have known how to solve: I would have subtracted 3 from both sides. I can do the same thing here: Then the solution is: x < 3 The formatting of the above answer is called "inequality notation", because the solution is written as an inequality. This is probably the simplest of the solution notations, but there are three others with which you might need to be familiar. Copyright Elizabeth Stapel 2002-2011 All Rights Reserved "Set notation" writes the solution as a set of points. The above solution would be written in set notation as "{x x is a real number, x < 3}", which is pronounced as "the set of all x-values, such that x is a real number, and x is less than minus three". The simpler form of this notation would be something like "{x x < 3}", which is pronounced as "all x such that x is less than minus three". "Interval notation" writes the solution as an interval (that is, as a section or length along the number line). The above solution, "x < 3", would be written as " ", which is pronounced as "the interval from negative infinity to minus three", or just "minus infinity to minus three". Interval notation is easier to write than to pronounce, because of the ambiguity regarding whether or not the endpoints are included in the interval. (To denote, for instance, "x < 3", the interval would be written " ", which would be pronounced as "minus infinity through (not just "to") minus three" or "minus infinity to minus three, inclusive", meaning that 3 would be included. The right-parenthesis in the "x < 3" case indicated that the 3 was not included; the right-bracket in the "x < 3" case indicates that it is.) The last "notation" is more of an illustration. You may be directed to "graph" the solution. This means that you would draw the number line, and then highlight the portion that is included in the solution. First, you would mark off the edge of the solution interval, in this case being 3. Since 3 is not included in the solution (this is a "less than", remember, not a "less than or equal to"), you would mark this point with an open dot or with an open parenthesis pointing in the direction of the rest of the solution interval:

...or: Then you would shade in the appropriate side:...or: Why shade to the left? Because they want all the values that are less than 3, and those values are to the left of the boundary point. If they had wanted the "greater than" points, you would have shaded to the right. In all, we have seen four ways, with a couple variants, to denote the solution to the above inequality: notation format pronunciation inequality x < 3 x is less than minus three set interval i) {x x is a real number, x < 3}...or: ii) {x x < 3} i) the set of all x, such that x is a real number and x is less than minus three ii) all x such that x is less than minus three the interval from minus infinity to minus three either of the following graphs: graph Here is another example, along with the different answer formats: Solve x 4 > 0. If they'd given me "x 4 = 0", then I would have solved by adding four to each side. I can do the same here:

Then the solution is: x > 4 Just as before, this solution can be presented in any of the four following ways: notation format pronunciation inequality x > 4 x is greater than or equal to four set interval i) {x x is a real number, x > 4}...or: ii) {x x > 4} i) the set of all x, such that x is a real number, and x is greater than or equal to four ii) all x such that x is greater than or equal to four the interval from four to infinity, inclusive of four either of the following graphs: graph Regarding the graphs of the solution, the square bracket notation goes with the parenthesis notation, and the closed (filled in) dot notation goes with the open dot notation. While your present textbook may require that you know only one or two of the above formats for your answers, this topic of inequalities tends to arise in other contexts in other books for other courses. Since you may need later to be able to understand the other formats, make sure now that you know them all. However, for the rest of this lesson, I'll use only the "inequality" notation; I like it best. Solve x / 4 > 1 / 2. If they had given me " x / 4 = 1 / 2 ", I would have multiplied both sides by 4. I can do the same thing here: Then the solution is: x > 2 Solve 2x < 5.

Remember how I said that solving linear inequalities is "almost" exactly like solving linear equations? Well, this is the one place where it's different. To explain what I'm about to do, consider the following: Copyright Elizabeth Stapel 2002-2011 All Rights Reserved 3 > 2 What happens to the above inequality when I multiply through by 1? The temptation is to say that the answer will be " 3 > 2". But 3 is not greater than 2; it is in actuality smaller. That is, the correct inequality is actually the following: 3 < 2 As you can see, multiplying by a negative (" 1", in this case) flipped the inequality sign from "greater than" to "less than". This is the new wrinkle for solving inequalities: Solve 2x < 9. When solving inequalities, if you multiply or divide through by a negative, you must also flip the inequality sign. If they had given me "2x = 9", I would have divided the 2 from each side. I can do the same thing here: Then the solution is: x < 9 / 2...or, if you prefer decimals (and if your instructor will accept decimal equivalents instead of fractions): x < 4.5 Solve x / 4 > 1 / 2. If they had given me " x / 4 = 1 / 2 ", I would have multiplied both sides by 4. I can do the same thing here:

Then the solution is: x > 2 Solve 2x < 5. Remember how I said that solving linear inequalities is "almost" exactly like solving linear equations? Well, this is the one place where it's different. To explain what I'm about to do, consider the following: Copyright Elizabeth Stapel 2002-2011 All Rights Reserved 3 > 2 What happens to the above inequality when I multiply through by 1? The temptation is to say that the answer will be " 3 > 2". But 3 is not greater than 2; it is in actuality smaller. That is, the correct inequality is actually the following: 3 < 2 As you can see, multiplying by a negative (" 1", in this case) flipped the inequality sign from "greater than" to "less than". This is the new wrinkle for solving inequalities: When solving inequalities, if you multiply or divide through by a negative, you must also flip the inequality sign. To solve " 2x < 5", I need to divide through by a negative (" 2"), so I will need to flip the inequality: Then the solution is: x > 5 / 2 Solve (2x 3) / 4 < 2. First, I'll multiply through by 4. Since the "4" is positive, I don't have to flip the inequality sign: (2x 3) / 4 < 2 (4) (2x 3) / 4 < (4)(2) 2x 3 < 8 2x < 11 x < 11 / 2 = 5.5 The velocity of an object fired directly upward is given by V = 80 32t, where t is in seconds. When will the velocity be between 32 and 64 feet per second? I will set up the compound inequality, and then solve for t:

32 < 80 32t < 64 32 80 < 80 80 32t < 64 80 48 < 32t < 16 48 / 32 > 32t / 32 > 16 / 32 1.5 > t > 0.5 Note that, since I had to divide through by a negative, I had to flip the inequality signs. Note also that you might (as I do) find the above answer to be more easily understood if written the other way around: 0.5 < t < 1.5 Looking back at the original question, it did not ask for the value of the variable "t", but asked for the times when the velocity was between certain values. So the actual answer is: The velocity will be between 32 and 64 feet per second between 0.5 seconds after launch and 1.5 seconds after launch. Always remember when doing word problems, that, once you've found the value for the variable, you need to go back and re-read the problem to make sure that you're answering the actual question. The inequality "0.5 < t < 1.5" did not answer the actual question regarding time. I had to interpret the inequality and express the values in terms of the original question. Solve 5x + 7 < 3(x + 1). First I'll multiply through on the right-hand side, and then solve as usual: 5x + 7 < 3(x + 1) 5x + 7 < 3x + 3 2x + 7 < 3 2x < 4 x < 2 Since I divided through by a positive "2" to get the final answer, I didn't have to flip the inequality sign. Solve 3(x 2) + 4 > 2(2x 3). First I'll multiply through and simplify; then I'll solve: 3(x 2) + 4 > 2(2x 3) 3x 6 + 4 > 4x 6 3x 2 > 4x 6 2 > x 6 (*) 4 > x x < 4

Why did I move the "3x" over to the right-hand side (to get to the line marked with a star), instead of moving the "4x" to the left-hand side? Because by moving the smaller term, I was able to avoid having a negative coefficient on the variable, and therefore I was able to avoid having to remember to flip the inequality when I divided off that coefficient. I find it simpler to work this way; I make fewer errors. But it's just a matter of taste. Why did I switch the inequality in the last line and put the variable on the left? Because I'm more comfortable with inequalities when the answers are formatted this way. Again, it's only a matter of taste. The form of the answer in the previous line, "4 > x", is perfectly acceptable. As long as you remember to flip the inequality sign when you multiply or divide through by a negative, you shouldn't have any trouble with solving linear inequalities.