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Can someone see why there is only get one solution when solving following equation in this way:

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

$$|x+1|+|2x-3|=|x-5| $$ $$\pm (x+1) \pm(2x-3)=\pm(x-5)$$ $$\pm x \pm 1 \pm 2x \mp 3 = \pm x \mp 5$$ $$\pm x \pm 2x \mp x \pm 1 \pm 5 \mp 3=0 $$ $$\pm 2x \pm 6 \mp 3 = 0$$ $$\pm 2x \pm 3=0$$ $$\pm 2x=\mp 3$$ $$x=\frac{\mp 3}{\pm 2} = -\frac{3}{2}$$

There should be another solution as well, $\frac{7}{4}$ by constructing two graphs and finding the intercepts.

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  • $\begingroup$ Why does not this method work? I think I use the correct definition of absolute value, and still it does not turn out nicely. $\endgroup$
    – Artem
    Commented Jun 4, 2013 at 21:35
  • $\begingroup$ thank you for the +1, @Patrick Da Silva! $\endgroup$
    – Artem
    Commented Jun 4, 2013 at 21:35
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    $\begingroup$ It's because you didn't use the correct definition! The sign of the absolute value changes depending on which interval you are, and when you write $$ \pm x \pm 2x \mp x $$ you have the possibilities $0, x, 2x, 3x, 4x$ that are possible if you consider all the signs. By looking at what happens over each interval you will get the signs right and not worry about multiple signs occurences. $\endgroup$ Commented Jun 4, 2013 at 21:36
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    $\begingroup$ @Artem : It would be nicer to just do things right! Don't you think? Break into intervals where you know the sign of the absolute value and you will have easy linear equations. $\endgroup$ Commented Jun 4, 2013 at 21:44
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    $\begingroup$ @Artem : the problem is that saying that $|x+1| = \pm (x+1)$ is ambiguous : the notation $\pm$ suggets that both values are possible, but they are not. The right notation would be that $$ |x+1| = \begin{cases} x+1 & \text{ if } x \ge -1 \\ -(x+1) & \text{ if } x < -1 \end{cases} $$ because then there is no ambiguity in the notation, hence my idea of breaking into intervals to understand the latter equation better. $$ $\endgroup$ Commented Jun 4, 2013 at 21:59

1 Answer 1

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You should try to find solutions over the intervals $]-\infty,-1]$, $[-1,3/2]$, $[3/2,5]$ and $[5,\infty[$ instead, because over each of those intervals you know the sign of the absolute value and you will have a linear equation to solve.

For instance, over the interval $[-1,3/2]$, $|x+1| = x+1$, $|2x-3| = -(2x-3)$ and $|x-5| = -(x-5)$. This gives you the equation $$ x+1 -(2x-3) = -(x-5) \quad \Longrightarrow \quad x+1 = 2x-3 - (x-5) = x+2 $$ which has no solution, hence there is no solution over the interval $[-1,3/2]$. Work out all four cases similarly. You need to worry that the solution that you find is actually in the interval you are working on though.

Hope that helps,

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  • $\begingroup$ thank you, @Patrick Da Silva! $\endgroup$
    – Artem
    Commented Jun 4, 2013 at 21:37

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