Question 49225
For absolute value problems, you will usually have two answers.  Since we don't know if the value of "x" is negative or positive, both cases must be considered.
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2=|4+3x| 
Case # 1. [Just in case x = a positive number (x>0)]
2=4+3x [For case #1, just drop the absolute value bars and solve for x]
2-4=4-4+3x
-2=+3x
-2/3=+3x/3
(-2/3) = x
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Case #2 [Just in case x = a negative number (x<0)]
2=|4+3x|
(-1)(2) = 4+3x [Drop the absolute bars, mulitiply one of the sides by (-1)] 
-2 = 4+3x [Solve for x]
-2-4=4-4+3x
-6=3x
-2=x
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x = (-2/3) or x = -2
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Check by pluggin the values of "x" back into the original equation.
2=|4+3x|
2=4+3(-2/3)
2=4-2
2=2 [Checks out]
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2=|4+3x|
-2=4+3(-2)
-2=4-6
-2=-2 [Also checks out, so both answers are correct]

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|5x+1|<5 How do you isolate x when it is not an equation (<5 not =5)?
Solve this the same as above, except when multiplying or dividing by a negative number, switch the inequality sign.
|5x+1|<5 [Needs two cases in case x>0 or in case x<0]
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Case #1. [Just in case x = a positive number (x>0)]
|5x+1|<5
5x+1<5 [Drop the absolute bars]
5x+1-1<5-1  [Solve for x]
5x<4
5x/5<4/5
x<4/5
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Case #2.[Just in case x = a negative number (x<0)]

|5x+1|<5 
5x+1<5  [Drop the absolute bars[
5x+1>(-1)5 [Multiply one side by (-1)and immediately switch the inequality sign]
5x+1>-5  [Solve for x]
5x+1-1>-5-1
5x>-6
5x/5>-6/5
x>-6/5
-6/5 or -1.2
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x<4/5 AND x>-6/5 [Final answer]
[If the inequality is "less thAND" the graph will be "and".  If the inequality is greatOR, the graph will be an "or".]
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Also, what needs to be manipulated to graph this on a horizontal graph? 
To graph you know that the two points will overlap because it is an AND graph (less thAND).  Every point from -6/5 through 4/5 is included in the graph.

<______-2______(-1.2=======-1======0=====4/5) ___1______2_____>
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The interval notation includes every point between (-1.2, 4.5)

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3+|x-3|=7 How do I solve this algebraically? 
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Case #1.[Just in case x = a positive number (x>0)]

3+|x-3|=7 
3+x-3=7 [Drop the absolute bars]
x=7 [Solve for x]

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Case #2 [Just in case x = a negative number (x<0)]
3+|x-3|=7
3+x-3=7 [Drop the absolute bars]
3+x-3=(-1)(7) [Multiply one side by (-1)]
3+x-3=-7 [Solve for x]
x=-7
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x = 7 or x = -7