# Thread: What am I doing wrong,inequalitie again

1. ## What am I doing wrong,inequalitie again

Solve following:
$\displaystyle Ix-4I+2x<7$
Those I:s mean absolute value...
Well I did this..

$\displaystyle (-7)<x-4+2x<7$ $\displaystyle /+4$
$\displaystyle (-3)<3x<11$ $\displaystyle / /3$
$\displaystyle (-1)<x<11/3$ but that is apparently very wrong...correct answer is x<3
Any idea where I wen´t wrong?

2. Originally Posted by Henryt999
Solve following:
$\displaystyle Ix-4I+2x<7$
Those I:s mean absolute value...
Well I did this..

$\displaystyle (-7)<x-4+2x<7$ $\displaystyle /+4$
This is wrong, you should've done this: $\displaystyle |x-4|<7-2x\implies 2x-7<x-4<7-2x,$ and solve each one separately.

By the way, can't you generate the | ?

3. Originally Posted by Henryt999
Solve following:
$\displaystyle Ix-4I+2x<7$
Those I:s mean absolute value...
Well I did this..

$\displaystyle (-7)<x-4+2x<7$ $\displaystyle /+4$
$\displaystyle (-3)<3x<11$ $\displaystyle / /3$
$\displaystyle (-1)<x<11/3$ but that is apparently very wrong...correct answer is x<3
Any idea where I wen´t wrong?
You can use the | key to mark the beginning and end of the modulus. On an English (UK) keyboard it's assigned to Shift+\ which is in the bottom left, next to shift.

$\displaystyle |x-4| + 2x < 7$

$\displaystyle |x-4| < 7 - 2x$

$\displaystyle \pm (x-4) < 7-2x$

A)
$\displaystyle (x-4) < 7-2x$

$\displaystyle 3x < 11$

$\displaystyle x < \frac{11}{3}$

B)
$\displaystyle -(x-4) < 7-2x$

Multiply by $\displaystyle -1$: $\displaystyle (x-4) > 2x-7$

Collect like terms: $\displaystyle -x > -3$

Multiply by -1: $\displaystyle x < 3$

Since $\displaystyle 3 < \frac{11}{3}$ discard solution A.