# SOLUTION: Solve the open sentence and name and graph the solution set. Please and thank you, to I can have a clear understanding of what I am doing. |6r+8|<-4

Algebra ->  Algebra  -> Absolute-value -> SOLUTION: Solve the open sentence and name and graph the solution set. Please and thank you, to I can have a clear understanding of what I am doing. |6r+8|<-4      Log On

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 Algebra: Absolute value Solvers Lessons Answers archive Quiz In Depth

 Click here to see ALL problems on absolute-value Question 75867This question is from textbook Algebra I : Solve the open sentence and name and graph the solution set. Please and thank you, to I can have a clear understanding of what I am doing. |6r+8|<-4This question is from textbook Algebra I Answer by bucky(2189)   (Show Source): You can put this solution on YOUR website!|6r+8|<-4 . This is probably a bad example for showing you a way to work absolute value problems. Why? Because this problem can be done by inspection. Note that the left side of this inequality is entirely contained within the absolute value signs. That means that whatever value you select for r, the left side of the inequality is always going to be positive. Now look at the right side of the inequality. It is -4 and will always be negative. Now ask yourself, "When is the positive quantity on the left side less than a negative quantity on the right side?" The answer is "Cannot happen." Therefore, this inequality is never true ... regardless of the value of r. . There is, however, a point to this problem, and that point is to always check whatever answer you get to make sure that you don't make a mistake in interpreting the results that you work out. . For example, let's work this problem in a way that you can use to solve absolute value equations or inequalities. This way involves solving two separate inequality problems. The first inequality problem is to write the inequality with out the absolute value signs and precede the quantity that was in the absolute value signs by a + sign. For this problem that would result in: . +(6r + 8) < -4 . Note that because a plus sign precedes the parentheses, you can just remove them without making changes to the terms within to get: . 6r + 8 < -4 . Now eliminate the 8 on the left side by subtracting 8 from both sides to get: . 6r < -4 -8 which simplifies to: . 6r < -12 . Solve for +r by dividing both sides by 6 and you get: . r < -12/6 which simplifies to: . r < -2 . This solution tells us that r must lie to the left of -2 on the number line, if it is to be a solution. . Now recall that we were setting up two inequalities. The second one involves removing the absolute value signs and applying a negative sign to the quantity that was inside the absolute value signs. For this problem that would result in: . -(6r + 8) < -4 . Then we need to solve this inequality for +r. As a start, we can multiply both sides of this inequality by -1 to eliminate the negative sign. However, recall that whenever you multiply or divide both sides of an inequality by a negative number you must reverse the direction of the inequality sign. So when you multiply both sides of this inequality by -1 you end up with: . 6r + 8 > 4 . Subtract 8 from both sides to eliminate the 8 on the left side. You get: . 6r > -4 . Then to solve for +r, divide both sides by 6 and you get: . r > -4/6 which simplifies to . r > -2/3 . This tells you that another set of solutions involves all the values of r that lie to the right of -2/3 on the number line. . So far this looks like we're fine. We have found two restrictions on r ... valid solutions for r must be less than -2 and greater than -2/3. . This method of solving two inequalities (one with a + sign on the quantity inside the absolute value signs and another with a - sign on that same quantity) will work in general unless there is an unforeseen difficulty with the problem such as there was with this one. . So what we need to do is always check our answers. In this problem we can do that by picking a value for r that is less than -2 and plugging it into the original problem to see if it works. Then pick a value of r that is greater than -2/3 and see if it works in the original problem. And finally pick a value of r in between -2 and -2/3 and see if it works. . Let's try r = -10. That surely meets the criterion that r be less than -2. If we substitute -10 for r in the original inequality we get: . |6*(-10) + 8 | < -4 and this simplifies to: . |-60 + 8| < -4 which further simplifies to: . | -52 | < -4 . but the absolute value of -52 is +52 and it is not true that: . +52 < -4 . This is a clue that something isn't correct here and we must find the reason. . You can try the same checking by letting r be 0 (which meets the requirement that r be greater than -2/3. If you let r be 0, the inequality becomes 8 < -4 and this is also not true. . What happens if you let r = -1, which is a value between r = -2 and r = -2/3. If you do this the original inequality becomes 2 < -4 and this also is not correct. Nothing works. So the original statement of the problem is based on an incorrect assumption which we have already noted ... that the absolute value of any quantity cannot be less than a negative number. . This is one of the things that you can learn from this problem ... the absolute value of a quantity cannot be less than zero or a negative number. Another thing you can learn is a process for solving absolute value problems (use both + and - the quantity inside the absolute value signs). A third thing is that when you multiply or divide both sides of an inequality by a negative quantity, you reverse the direction of the inequality sign. And finally, always spot check your answer just to make sure that you haven't overlooked something, and that your solution does really work. (In this case it did not and we found a reason for that.) . Hope this helps you understand inequalities and absolute values a little better. .