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anonymous
 one year ago
Absolute Values
anonymous
 one year ago
Absolute Values

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Solve for:\[x+1\le 2x5\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0In 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0But don't two cases cancel each other out though..?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No apparently not though

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1433509203824:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Are these the posibilities? \[x+1\le 2x5\]\[(x+1)\le 2x5\]\[x+1\le (2x5)\]\[(x+1)\le (2x5)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1433509233981:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes exactly, so you have to work out four of those linear inequalities for x

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Exactly. At one point, both are negatives, so its the same as having positive on both, right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Or will that affect somehow?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Just resolve them, how you would normally approach them, you will just result in four individual 'x' values

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok. Let me solve for some \[x+1≤2x−5\]\[x\le2x6\]\[x\le 6\]\[x\ge 6\] \[−(x+1)≤2x−5\]\[x1\le2x5\]\[x\le 2x4\]\[3x\le 4\]\[x\ge \dfrac{4}{3}\] \[x+1≤−(2x−5)\]\[x+1\le 2x+5\]\[x\le2x+4\]\[x\le4\]\[x\ge 4\] \[−(x+1)≤−(2x−5)\]\[x1\le 2x+5\]\[x\le 2x+6\]\[x\le 6\] Do we just combine like terms? @DelTaVsPi?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I just briefly skimmed your working out, but other than that, the solutions can be presented as you have done them

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I made a mistake in calculations. let me fix

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0square both sides and work as a quadratic or draw both easier

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The last one, I've noticed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[x+1≤2x−5\]\[x≤2x−6\]\[−x≤−6\]\[x≥6\] \[−(x+1)≤2x−5\]\[−x−1≤2x−5\]\[−x≤2x−4\]\[−3x≤−4\]\[x≥\dfrac{4}{3}\] \[x+1≤−(2x−5)\]\[x+1≤−2x+5\]\[x≤2x+4\]\[3x≤4\]\[x\le\frac{4}{3}\] \[−(x+1)≤−(2x−5)\]\[−x−1≤−2x+5\]\[−x≤−2x+6\]\[x≤6\] 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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