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Find what value of $x$ satisfy:

$(1/3)(1-x) \geq 2(x-3)$

First I multiplied both sides by $3$ so that $1/3$ became $3/3=1$. So I tried to find $x$ this way: $(1-x) \geq 6(x-3)$. I tried solving it with sign diagrams. Both were positive/$0$ when $x\leq1$, $x\geq3$

The answer is incorrect as the sign variation are heading both a different direction and don't get touch each other anywhere.

When I couldn't get out I also tried $(1/3)(1-x) \geq 2(x-3)$ [without multiplying]. But ended up with the sign diagrams. Which makes sense.

I decided to look at the answer but don't understand a thing. Can anyone please explain me in steps what I did wrong or how you would solve this differently? I can't draw my sign diagrams here so I tried describing as detailed as possible. I hope it's understandable.

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1 Answer 1

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$$\frac{1-x}3\ge2(x-3)\iff1-x\ge6x-18\iff7x\le19\iff x\le?$$

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  • $\begingroup$ (/7)--> x≥19/7. Thanks you! I think my mistake was not multiplying 6(x-3).Cause I did 6(3-3=0) 6(0)=0 /positive. I'm gonna start at 6x-18 :) $\endgroup$
    – user160137
    Commented Jul 21, 2014 at 15:31
  • $\begingroup$ I still don't understand how they meet in the sign diagram. I again get: x≤1, x≥3 using a sign diagram. I have to use a sign diagram at my exam. Could you help me explain what I do wrong at it? $\endgroup$
    – user160137
    Commented Jul 21, 2014 at 15:34
  • $\begingroup$ @user160137, For $$A(x-a)=B(x-b)$$ we can not just check for $x=a,b$ $\endgroup$
    – lab bhattacharjee
    Commented Jul 21, 2014 at 15:49
  • $\begingroup$ a and b tell nothing about each other I get it. And this question is just like solving a equation with instead of = it's replaced by ≥. $\endgroup$
    – user160137
    Commented Jul 21, 2014 at 16:07
  • $\begingroup$ It should be $19\ge7x$, not $7x\ge19$. $\endgroup$
    – Barry Cipra
    Commented Jul 22, 2014 at 19:17

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