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$$|2x − 3| − |x + 2| = 5$$

I have no idea. I didn't see anything like this in class. It is a practice question and something like it will come up on the exam can someone please show me the full solution and working (that is how I learn) Thanks in advance.

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Hint: using the definition of the absolute value we get $$|2x-3|=\begin{cases} 2x-3, & 2x-3\geq 0 \\ -(2x-3), & 2x-3<0 \end{cases}=\begin{cases} 2x-3, & x\geq \frac{3}{2} \\ -2x+3, & x<\frac {3}{2}\end{cases}$$ $$ |x+2|=\begin{cases} x+2, & x+2\geq 0 \\ -(x+2) & x+2<0 \end{cases} = \begin{cases} x+2, & x\geq -2 \\ -x-2, & x<-2\end{cases}$$

This is all we need to simplify the equation. "Something happens" at $x=-2$ and $x=\frac{3}{2}$, one of the terms $|2x-3|$ or $|x+2|$ will change there. Using this we get 3 cases we have to consider:

  1. $x<-2$
  2. $-2\leq x < \frac{3}{2}$
  3. $\frac{3}{2}\leq x$
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  • $\begingroup$ Where do i go from there? sorry I am really lost. $\endgroup$ – jb3navides Aug 13 '15 at 11:34
  • $\begingroup$ @jb3navides in each of these cases you can identify $|2x-3|$ and $|x+2|$ with an easier term, e.g. in the first case $x<-2$ you know, that $|x+2|=-x-2$ from the definition of the absolute value. As $x<-2$ also implies $x<\frac{3}{2}$ we get $|2x-3|=-2x+3$. Now you can rewrite your equation using these easier terms and solve the equation. The same needs to be done for the other two cases. $\endgroup$ – Hirshy Aug 13 '15 at 11:45
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Note that for all $x \geq 3/2$ we have $|2x-3| - |x+2| = 5$ if and only if $x-5 = 5$, i.e. $x = 10$; for all $-2 \leq x < 3/2$ we have $|2x-3| - |x+2| = 5$ if and only if $-3x = 4$, i.e. $x = -4/3$; for all $x < -2$ we have $|2x-3| - |x+2| = -x + 5 = 5$, i.e. $x = 0$.

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Hint

if $x \geq 0$ then you would have $$2x-3 -(x+2) = 5$$

if $x < 0$ then you would have $$-(2x-3) +(x+2) =5$$

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    $\begingroup$ No, just looking at $x\geq 0$ and $x<0$ is not enough for this question. $\endgroup$ – Hirshy Aug 13 '15 at 10:09

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