What am I doing wrong,inequalitie again

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January 26th 2010, 01:10 PM
Henryt999

What am I doing wrong,inequalitie again

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

$(-7)<x-4+2x<7$ $/+4$
$(-3)<3x<11$ $/ /3$
$(-1)<x<11/3$ but that is apparently very wrong...correct answer is x<3
Any idea where I wen´t wrong?
January 26th 2010, 01:16 PM
Krizalid

Quote:

Originally Posted by Henryt999

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

$(-7)<x-4+2x<7$ $/+4$

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

By the way, can't you generate the | ?
January 26th 2010, 01:18 PM
e^(i*pi)

Quote:

Originally Posted by Henryt999

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

$(-7)<x-4+2x<7$ $/+4$
$(-3)<3x<11$ $/ /3$
$(-1)<x<11/3$ 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.

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

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

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

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

$3x < 11$

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

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

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

Collect like terms: $-x > -3$

Multiply by -1: $x < 3$

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

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