Download Linear Inequalities: Solving and Representing with Interval Notation and Graphs and more Study notes Algebra in PDF only on Docsity! 1310, Section 2.6 Linear Inequalities Linear – means that the variable is to the first power Inequality – means that the comparative symbol is one of the following: So we’re comparing and there’s a boundary involved. For example if I say x > 5, I mean for you to think of numbers strictly bigger than 5.0….5.55 is an example of this type of x. I can use interval notation to summarize all these x’s: ),5( Or I can use a graph with a numberline featuring a hollow dot at 5: (Notice, this can become an x axis if I put in the y axis) Inequalities have the feature that you may add and subtract a number to both sides and you will get an equivalent inequality. You may multiply and divide by a positive number and you will get an equivalent inequality. If you multiply or divide by a negative number you induce a flip about the centerpoint zero and we symbolize this by flipping the inequality symbol. So: Solve 83x5 Subtract 3 from both sides… Now the claim is that negative 5 times x is greater than positive 5. What kind of numbers are our x’s if the product of x and 5 is a positive? 1 5 Indeed, the x’s are negatives. Divide both sides by 5 What happens if we forget to flip the inequality? Put your answer in interval notation and sketch a graph of your answer: How about: 6x + 3 < 9x + 15 Now we can have compound inequalities, too. You must read these carefully. 2 But – I want you to start thinking of absolute value as a distance. Indeed as a distance from zero. If I tell you to read out loud and draw a picture of the following equation: 3x I’ll be expecting to hear: “The distance from x to zero is 3” and I’ll expect a picture that looks like this: There are two places on the number line that are 3 away from zero. The answer is x = 3 or x = 3. When you get absolute value equations, then the “problem” is then two problems: the negative problem and the positive problem. Look at 115x3 . From order of operations, you know that you need to remove the absolute value problem delimiters before continuing. We’ll do that by reading the problem and translating our words to set of problems. This says: The distance from 3x 5 to zero is 11. Well, now there are two places that are 11 away from zero on the number line. So we’ll solve both problems: 3x 5 = 11 3x 5 = 11 5 -3 3 0 Let’s do another: |3- x| =15
And another: 2|2x - 3 =12
