5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 Jun 4 '13 at 21:35
  • $\begingroup$ thank you for the +1, @Patrick Da Silva! $\endgroup$ – Artem Jun 4 '13 at 21:35
  • 1
    $\begingroup$ 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. $\endgroup$ – Patrick Da Silva Jun 4 '13 at 21:36
  • 1
    $\begingroup$ @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. $\endgroup$ – Patrick Da Silva Jun 4 '13 at 21:44
  • 1
    $\begingroup$ @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. $$ $\endgroup$ – Patrick Da Silva Jun 4 '13 at 21:59
4
$\begingroup$

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,

$\endgroup$
  • $\begingroup$ thank you, @Patrick Da Silva! $\endgroup$ – Artem Jun 4 '13 at 21:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.