5

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.

Artem
  • 1,208
  • Why does not this method work? I think I use the correct definition of absolute value, and still it does not turn out nicely. – Artem Jun 04 '13 at 21:35
  • thank you for the +1, @Patrick Da Silva! – Artem Jun 04 '13 at 21:35
  • 1
    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. – Patrick Da Silva Jun 04 '13 at 21:36
  • Two $\pm$ occurring in different places do not necessarily represent the same choice. It is partially by accident that one solution was identified. – André Nicolas Jun 04 '13 at 21:37
  • It would be nice if it would work to use multiple signs and in that way find the solutions. It should work as I have solved other equations in that way where I got two answers by using multiple signs. – Artem Jun 04 '13 at 21:42
  • 1
    @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. – Patrick Da Silva Jun 04 '13 at 21:44
  • I suppose I just need to get used to the fact that this is the case! Although, don't you agree that it is easy to think that because $|x+1|=\sqrt{(x+1)^2}=\pm (x+1) $? However, I agree why it should be broken up into different intervals, as different values will cause each interval to either be positive or negative. Thank you again, @Patrick Da Silva! – Artem Jun 04 '13 at 21:54
  • 1
    @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. $$ – Patrick Da Silva Jun 04 '13 at 21:59

1 Answers1

4

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,

Patrick Da Silva
  • 41,413