Author Topic: Addition of absolute functions (Read 1116 times) Tweet Share

anonuser0511 · « **on:** March 27, 2010, 02:32:47 am »

$|x+3|+ |2-5x| = 7$

Been trying this for a while...
Wolfram tells me x = -1/2 and x = 1 which I can work out but I'm also getting x=3.

I've tried this

|x+3| is
x+3 when $x\ge-3$ eq. 1
-x-3 when $x<-3$ eq. 2

|2-5x| is
2-5x when $x\ge \frac{2}{5}$ eq. 3

5x-2 when $x< \frac{2}{5}$ eq. 4

then I find the common domains such as $x\ge \frac{2}{5}$ and i'll add eq. 1 and eq. 3 then solve for x.

If anyone could tell me the actual method + why x=3 is rejected that'd be awesome.

Mao · « **Reply #1 on:** March 27, 2010, 03:07:21 am »

Break it into three segments

$ \begin{align*} (-\infty,-3] & \implies -(x+3) + (2-5x) = 7 \implies -6x = 8 \implies x = -\frac{4}{3} \notin (-\infty,-3]\\ \left(-3,\frac{2}{5}\right] & \implies (x+3)+(2-5x) = 7 \implies -4x = 2 \implies x = -\frac{1}{2} \in \le ft(-3,\frac{2}{5}\right] \\ \left(\frac{2}{5},\infty\right) & \implies (x+3)-(2-5x) = 7 \implies 6x = 6 \implies x = 1 \in \left(\frac{2}{5},\infty\right) \end{align*} $

Hence, the solutions are -1/2 and 1.

[edit: fixed, fuck I'm an idiot, apologies for any confusion]

anonuser0511 · « **Reply #2 on:** March 28, 2010, 03:30:19 am »

thanks mao - you're my hero

anonuser0511 · « **Reply #3 on:** March 28, 2010, 09:22:20 pm »

wait t.t

Then how come when $x=-\frac{1}{2}$ is substituted into the formula the statement is still true?

Mao · « **Reply #4 on:** March 29, 2010, 01:58:25 am »

Because I'm an idiot, got the signs wrong completely. Let me fix that.

EDIT: done.

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Author Topic: Addition of absolute functions (Read 1116 times) Tweet Share

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