So I am having a problem in solving this type of equation. The problem I am dealing with is...
$$\left|(2x-1) + \frac{3x-1}x\right| = \left|2x-1\right| + \left|\frac{3x-1}x\right|$$
Please help me with this problem. Thank you so much.
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1Real numbers? Or complex? Probably reals, I'm guessing, but it does make a difference... – MPW Apr 20 '16 at 03:45
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See Equality holds in triangle inequality iff both numbers are positive, both are negative or one is zero (and maybe some other posts linked there). – Martin Sleziak Apr 20 '16 at 05:33
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Hint: If considered over the reals, $|a+b| = |a| + |b|$ holds if and only if $a$ and $b$ are of same sign, or at least one of $a, b$ is $0$.
For you question, determine the intervals over which $2x-1$ is positive/zero/negative, and the intervals over which $(3x-1)/x$ is positive/zero/negative, and then take suitable unions/intersections. (Care must be taken at then endpoint $x = 0$, since $x$ appears in denominators.)
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Another approach, assuming $x$ and $y$ are real: $$\begin{aligned} |x+y| = |x| + |y| &\iff |x+y|^2 = (|x|+|y|)^2 \\ &\iff (x+y)^2 = |x|^2 + 2|x||y| + |y|^2 \\ &\iff x^2 + 2xy + y^2 = x^2 + 2|x||y| + y^2 \\ &\iff 2xy = 2|x||y| \\ &\iff xy = |x||y| \\ &\iff xy = |xy| \\ &\iff xy \geq 0 \\ \end{aligned}$$ In your case, the original problem is therefore equivalent to $$(2x-1)\left(\frac{3x-1}{x}\right) \geq 0$$ which should be straightforward to solve.
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Hint: A brute force approach is to drop one set of abs bars and replace the equation by two alternatives, one with "$+$" and one with "$-$" before the group that carried the bars.
The alternatives are logically "ORed", so each solution to each alternative is a solution to the original.
This can then be continued to double the number of equations again and again.
MPW
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