Absolute Values

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Absolute Values

Mathematics
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Solve for:\[|x+1|\le |2x-5|\]
In this particular question, you will have to work out for x four times. The reason is that, once you move the 'absolute signs' over the equal sign, you already have two equations to work with.
But don't two cases cancel each other out though..?

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Other answers:

No apparently not though
|dw:1433509203824:dw|
Are these the posibilities? \[x+1\le 2x-5\]\[-(x+1)\le 2x-5\]\[x+1\le -(2x-5)\]\[-(x+1)\le -(2x-5)\]
|dw:1433509233981:dw|
Yes exactly, so you have to work out four of those linear inequalities for x
Exactly. At one point, both are negatives, so its the same as having positive on both, right?
Or will that affect somehow?
Just resolve them, how you would normally approach them, you will just result in four individual 'x' values
Ok. Let me solve for some \[x+1≤2x−5\]\[x\le2x-6\]\[-x\le -6\]\[x\ge 6\] \[−(x+1)≤2x−5\]\[-x-1\le2x-5\]\[-x\le 2x-4\]\[-3x\le -4\]\[x\ge \dfrac{4}{3}\] \[x+1≤−(2x−5)\]\[x+1\le -2x+5\]\[x\le2x+4\]\[-x\le4\]\[x\ge -4\] \[−(x+1)≤−(2x−5)\]\[-x-1\le -2x+5\]\[-x\le -2x+6\]\[x\le 6\] Do we just combine like terms? @DelTaVsPi?
I just briefly skimmed your working out, but other than that, the solutions can be presented as you have done them
I made a mistake in calculations. let me fix
square both sides and work as a quadratic or draw both easier
The last one, I've noticed
\[x+1≤2x−5\]\[x≤2x−6\]\[−x≤−6\]\[x≥6\] \[−(x+1)≤2x−5\]\[−x−1≤2x−5\]\[−x≤2x−4\]\[−3x≤−4\]\[x≥\dfrac{4}{3}\] \[x+1≤−(2x−5)\]\[x+1≤−2x+5\]\[x≤-2x+4\]\[3x≤4\]\[x\le\frac{4}{3}\] \[−(x+1)≤−(2x−5)\]\[−x−1≤−2x+5\]\[−x≤−2x+6\]\[x≤6\] Combine the like terms, the ones that have the smaller set of possible values \[x\ge6, x\le\dfrac{4}{3}\]

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