
Solving AbsoluteValue Equations (page 1 of 2) When you take the absolute value of a number, you always end up with a positive number (or zero). Whether the input was positive or negative (or zero), the output is always positive (or zero). For instance,  3  = 3, and  –3  = 3 also. This property — that both the positive and the negative become positive — makes solving absolutevalue equations a little tricky. But once you learn the "trick", they're not so bad. Let's start with something simple:
I've pretty much already solved this:  3  = 3 and  –3  = 3, so x must be 3 or –3. But how are you supposed to solve this if you don't already know the answer? You use the positive / negative property of the absolute value to split the equation into two cases, and you use the fact that the minus sign " – " indicates "the opposite sign", not necessarily a negative number. For example, if you have x = –6,
then Whatever the value of x might be, taking the absolute value of x makes it positive. Since x might have been positive and might have been negative, you have to acknowledge this fact when you take the absolutevalue bars off, and you do this by splitting the equation into two cases. If the value of x was positive to start with, then you can bring that value out of the absoluevalue bars without changing its sign, giving you x = 3. But x might also have been negative, in which case you would have to change the sign on x for the absolute value to come out positive, so you also have –x = 3, which solves as x = –3. Then the solution is x = –3, 3. Absolutevalue equations always work this way: to be able to remove the absolutevalue bars, you have to isolate the absolute value onto one side, and then split the equation into the two possible cases. You can, by the way, verify the above solution graphically. When you attempt to solve the absoluvevalue equation  x  = 3, you are, in effect, setting two line equations equal to each other and finding where they cross. In this case, as you can see below: ...the two lines, y_{1} =  x  and y_{2} = 3, cross at two xvalues: x = –3 and x = 3. Of course, any solution can be verified by plugging it back into the original exercise: If x = –3, then  x  =  –3  = 3, and if x = 3, then  x  =  3  = 3. Let's look at some more exercises: Copyright © Elizabeth Stapel 20022011 All Rights Reserved
To clear the absolutevalue bars, I must split the equation into its two possible two cases, one case for each sign: (x + 2) = 7 or –(x + 2) =
7 Then the solution is x = –9, 5. To confirm this graphically, you would look for the intersections of y_{1} =  x + 2  and y_{2} = 7:
First, I'll isolate the absolutevalue part; that is, I'll get the absolutevalue expression by itself on one side of the "equals" sign, with everything else on the other side: 
2x – 3  – 4 = 3 Now I'll clear the absolutevalue bars by splitting the equation into its two cases, one for each sign: (2x – 3) = 7 or –(2x – 3) = 7 So the solution is x = –2, 5. To confirm this graphically, look for the intersections of y_{1} =  2x – 3  – 4 and y_{2} = 3:
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