# Find value of $x$ for: $(1/3)(1-x) \geq 2(x-3)$

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.

$$\frac{1-x}3\ge2(x-3)\iff1-x\ge6x-18\iff7x\le19\iff x\le?$$
• @user160137, For $$A(x-a)=B(x-b)$$ we can not just check for $x=a,b$ Commented Jul 21, 2014 at 15:49
• It should be $19\ge7x$, not $7x\ge19$. Commented Jul 22, 2014 at 19:17