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I have an equation:

$ |2x - 1| - |x + 5| = 3 $

To solve it, I divided the Reals range into three intervals, at which the absolute value expressions change their signs: $x < -5; -5 \le x < \frac{1}{2}; x \ge \frac{1}{2}$, derived the three equations and solved them. Two of the solutions are correct (9 and $-\frac{7}{3}$), but the problem is with the first interval $x < -5$, where both expressions are negative:

$ -(2x - 1) + (x + 5) = 3 \\ -2x + 1 + x + 5 = 3 \\ -x + 6 = 3 \\ -x = -3 \\ x = 3 $

The answer is incorrect, with $x = 3$ the whole expression is equal to -3.

So could anyone, please, tell me, where is the error in my logic?

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    $\begingroup$ There is no mistake. If there is a solution with $x<-5$ you will get it proceeding in the way you have done. Since you got $x=3>-5$ you can conclude that the is no solution with $x<-5.$ $\endgroup$ – mfl Nov 21 '17 at 19:44
  • $\begingroup$ @mfl thank you, that was it $\endgroup$ – user907860 Nov 21 '17 at 19:48
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There is no mistake in your math. If a solution existed in the interval $x< -5$ then you would have found it using the exact same steps that you already showed.

However, you now know that no such solution can exist in that interval. (since if it did then $x$ would have to equal 3 like you showed, however, 3 isn't in the interval $x<-5$)

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You might simply determine the variations on each interval, after you've simplified the expression of the function: $$\begin{array}{r|rccccc} x &&-5&&\frac12&&\\ \hline f(x)& -x+6&-3x+4&&x-6 \\ &\searrow & \\[-1ex] &&11\\[-1ex] &&&\mkern-30mu\searrow& &\nearrow \\ &&&&-\frac{11}2& \end{array} $$ By the Intermediate Value Theorem, we see the equation $f(x)=3$ has one solution on the interval $(-5,\frac12)$ and one solution on $(\frac12,+\infty)$.

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