Absolute value for most of us is a brief unit in math, and we hardly touch it again. However, if you need a refresher course or are diving into more complicated absolute value equations, allow us to help you!<\/p>\n
To find absolute zero, count how far away a number is from zero on a number line. Both 6 and -6 are six units away from zero. You can represent this distance using the symbol \u2502x<\/em>\u2502.\u00a0 So, when you're writing absolute value for 6 and -6, it would look like this:\u00a0\u25026\u2502= 6 and \u2502-6\u2502 = 6.<\/p>\n Absolute value is never a negative number. It is always positive. The absolute value bars tell us that we do not want a negative number as a solution.<\/p>\n Let's take\u00a0\u25026\u2502= 6 again, but turn it to\u00a0\u2502-6\u2502= x. What is the absolute value of -6? Already we know that it's six, but let's explore why.<\/p>\n If the number\u00a0in the absolute value bars is greater than 0 or is 0, then you do not have to do anything to the solution. If the number in the absolute value bar is less than zero, then you use \"-x\" to flip the solution into a positive.<\/p>\n \u2502-6\u2502=\u00a0 -x<\/p>\n But wait, we can't have a negative number as a solution. Multiply both sides by -x.<\/p>\n -(-6) = x<\/p>\n \u25026\u2502= 6<\/p>\n Perhaps the most common need for using absolute value is considering distance differences.\u00a0 If two cars driving on an interstate highway are 30 miles apart, and one just passed mile marker 70, where is the other?\u00a0 Since we don't know if the one at 70 is leading or not, the other is either at 40 or 100.<\/p>\n In reverse, if you know Car\u00a0A<\/em>\u00a0is at mile marker 70 and Car B is at 40, easy math of\u00a0A<\/em>\u00a0\u2013\u00a0B<\/em>\u00a0tells us they are 30 miles apart.\u00a0 Unfortunately,\u00a0subtracting\u00a0A<\/em>\u00a0\u2013\u00a0B<\/em>\u00a0does not always work. \u00a0If someone tells you that\u00a0B<\/em>\u00a0is at 70 and\u00a0A<\/em>\u00a0is at 40, then\u00a0A\u00a0<\/em>\u2013\u00a0B<\/em>\u00a0= -30.\u00a0 Of course, the distance between them cannot be -30.<\/p>\n The absolute value function eliminates the need to know which is larger.<\/p>\n The symbolic relation should be \u2502A \u2013 B<\/em>\u2502, or you can use \u2502B - A<\/em>\u2502.\u00a0 Both will give you a positive 30.<\/p>\n This is necessary for situations where you cannot use negative values, like square roots.\u00a0 For example, \u221a(B \u2013 A) in the first illustration will give you \u221a(-30), which is not a real number.\u00a0 To write it properly, you would use the absolute value:<\/p>\n \u221a(\u2502B \u2013 A\u2502)<\/p>\n The most simple absolute value equation is one that asks you to solve for one number.<\/p>\n \u2502x\u2502 = 6.<\/p>\n What value can you substitute for x to get 6? \u00a0There are two solutions: x = -6 and x = 6.\u00a0 You express this solution in brackets: {-6, 6}.<\/p>\n Now consider \u2502x + 1\u2502 = 6.<\/p>\n Again the operation inside the absolute value bars must equal -6 or 6.\u00a0 To solve two separate equations.<\/p>\n x + 1 = 6\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 and\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 x + 1 = -6<\/p>\n x = 5\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0x = -7\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 The solution set is {-7, 5}<\/p>\n \u25023x + 5\u2502 = 6.\u00a0 \u00a0What\u2019s inside the absolute value bars must equal -6 or 6. Break this up into two equations and solve.<\/p>\n 3x + 5 = 6\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 and\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 3x + 5 = -6<\/p>\n x = 1\/3\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0\u00a0x = -11\/3<\/p>\n The solution set is {-11\/3, 1\/3}<\/p>\n Check<\/strong><\/p>\n \u25023(1\/3) + 5\u2502 = \u25021 + 5\u2502 = 6\u00a0checks<\/em><\/p>\n and<\/p>\n \u25023(-11\/3) + 5\u2502 = \u2502-11 + 5\u2502 = 6\u00a0\u00a0checks<\/em><\/p>\n 4\u25022x - 3\u2502 = 12<\/p>\n Here you must first isolate the absolute value by dividing both sides by 4.<\/p>\n \u25022x - 3\u2502 = 3<\/p>\n Then set up two separate equations.<\/p>\n (2x \u2013 3) = 3\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 and\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (2x \u2013 3) = -3<\/p>\n x = 3\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 x = 0<\/p>\n Check\u00a0<\/strong><\/p>\n 4\u25022(3) - 3\u2502 = 4\u25023\u2502 = 12 \u00a0\u00a0checks<\/em><\/p>\n and<\/p>\n 4\u25022(0) - 3\u2502 = 4\u2502- 3\u2502 = 12\u00a0\u00a0checks<\/em><\/p>\n For these examples the solution set had two values. However, consider these next two examples.<\/p>\n \u2502x + 1\u2502 = 0<\/p>\n Zero is neither positive nor negative. The only solution is x + 1 = 0 and x = -1. The solution set is {-1}.<\/p>\n \u2502x + 1\u2502 = -3<\/p>\n The left side has to be positive but the right side is negative. There is no number whose absolute value is -3. There is no solution. The solution set is { } or \uab3e (empty set).<\/p>\n \u25022x + 5\u2502 = 11 \u2013 x<\/p>\n Set up two separate equations.\u00a0 For the second equation, you must negate the entire left side.<\/p>\n 2x + 5 = 11 \u2013 x\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 and\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 2x + 5 = -(11 \u2013 x)<\/p>\n x = 2\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 x = -16<\/p>\n <\/p>\n Check\u00a0<\/strong><\/p>\n \u25022(2) + 5\u2502 =? \u00a011 \u2013 (2)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0and\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0 \u25022(-16) + 5\u2502 =? \u00a0-(11 \u2013 x)<\/p>\n \u25024 + 5\u2502 =? 11 - 2\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u2502-32 + 5\u2502 =? \u00a0-(11 \u2013 (-16))<\/p>\n \u25029\u2502 = 9\u00a0\u00a0checks<\/em><\/p>\n \u2502-27\u2502 = -(27) \u00a0Left side is positive, but the right side is negative. \u00a0It does not check.<\/em><\/p>\n Even though the algebra is right, one of the proposed solutions is not correct.\u00a0 The solution set is {2}.<\/p>\n Absolute values force results of operations to be positive that otherwise may be negative.\u00a0This means you must absolutely check your work. Sometimes your solution works; sometimes, one does but not the other; sometimes, neither works.<\/p>\n Solving absolute value equations is easy, but it can get tricky if you're faced with an equation that uses more than one value. Keep the following tips in mind:<\/p>\nMake it Opposite<\/h2>\n
Why Use Absolute Values<\/h2>\n
Absolute Values in Equations<\/h2>\n
Basic Equations<\/h3>\n
A Little More Complication<\/h3>\n
Isolate the Absolute Value<\/h3>\n
How Many Solutions?<\/h3>\n
Variable on Both Sides<\/h3>\n
Extraneous Solutions<\/h3>\n
Summary<\/h2>\n
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