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There was an equation I was trying to solve. I have converted the equation to basic terms for now.

\begin{align} & -2x + 6 > 0 \tag{1} \\ \implies & -2x > -6 \tag{2} \\ \implies & x > 3 \tag{3} \end{align}

Now, solving it another way, \begin{align} & -2x + 6 > 0 \tag{4}\\ \implies & 2x - 6 < 0 \tag{5} \\ \implies & 2x < 6 \tag{6} \\ \implies & x < 3 \tag{7} \end{align}

For the same equation, I am getting two different ranges of $x$. Am I missing some rule/property?

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  • $\begingroup$ $x < 3$ instead: the inequality reverses when you multiply through by a negative. $\endgroup$ – Randall Dec 26 '17 at 3:07
  • $\begingroup$ Actually I get the step you are referring to,but then too if I divide the equation with a (-2) in the equation -2x>-6 do have to flip the inequality sign? And are there some rules I can go through for inequalities. $\endgroup$ – Mary Lopez Dec 26 '17 at 3:12
  • $\begingroup$ I get going from $-2x+6>0$ to $2x-6 < 0$, that's fine. Then you get $2x < 6$ and so $x < 3$, done. $\endgroup$ – Randall Dec 26 '17 at 3:13
  • $\begingroup$ The point you are saying is correct. But I need to know what's wrong in the first flow. I know that x<3 is correct. And my question here is -2x+6>0 ->. -2x>-6 now I divide the equation with a (-2). So whenever I am dealing with a negative do I flip the sign is that the rule? $\endgroup$ – Mary Lopez Dec 26 '17 at 3:22
  • $\begingroup$ The thing wrong in the first flow is in my first comment. $\endgroup$ – Randall Dec 26 '17 at 3:24
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If you have $x<y$ then $ax < ay$ if and only if $a > 0$. If you multiply (or divide) by a negative number then the inequalities change direction. For example if I take $-3<-2$ and multiply by $-1$ then I would get $3 < 2$ which is evidently false. If I change the direction of the inequality though I would get $2<3$ which is correct.

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