I know the method of solving the equation like this $|2x+1|=|3x+9|$ but the problem is if the same equation would be like this $| 2x+1 | = x | 3x+9 |$, how can I solve this?
Solving equations with multiple absolute values
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algebra-precalculus
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1Again, the point that has been raised to you in your past few questions: treat the absolute value as a function with two cases. $|2x+1|=\mathrm{something}$ can mean either of $2x+1=\mathrm{something}$ or $-(2x+1)=\mathrm{something}$. You will have four "cases" since you have two absolute values; work from that. – 2010-08-23
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0actuall we can solve the first example as 2x+1=3x+9 or 2x+1=-(3x+1) we have only two cases to solve , now i want to discover that how we can solve the second example in this way – 2010-08-23
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1The same technique still applies, Zia. – 2010-08-23
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0Lastly, making a good plot: http://tinyurl.com/28lkzje helps greatly when trying to solve such things. – 2010-08-23
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1Eleven exclamation marks look like someone's shouting. What shall I picture for myself in the face of eleven interrogation marks? – 2010-08-23
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0Rasmus: extreme curiosity? – 2010-08-23
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0Solving an equation with n absolute values by using $2^n$ cases can be a lot harder than with $n+1$ cases (since $2^n-n-1$ of the $2^n$ cases for $n>1$ are impossible anyway); the solutions using only two cases are restricted to equations with an essentially-multiplicative relationship between the absolute values and won't generalize to equations like $|x+1|+|x-3|=4$. – 2010-08-23
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0I disagree that this is an *exact* duplicate of a previous question. Consider me to have voted to reopen. – 2010-08-24