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kwabon · Oct 27, 2009

hold up a sec, trying to use latex.

kwabon · Oct 27, 2009

\x +2\ + \x - 6\ < 10

\ \ those things are absolute values btw and epic fail at using latex.
can please someone explain to me, step by step. thanks

kwabon · Oct 27, 2009

Xcelz said:
not sure but square it all? this will get rid of the possible negative value in the brackets.

thats crazy method and not sure whether that will work, but thanks for your help anywho.

jet · Oct 27, 2009

$\text{You take positive and negative values under each absolute value sign, } \\ \text{giving 4 separate equations:} \\ -x - 2 - x + 6 < 10 \Rightarrow x > -3 \\ -x - 2 + x - 6 < 10 \Rightarrow - 8 < 10 \text{ this is meaningless} \\ x + 2 - x + 6 < 10 \Rightarrow 8 <10 \text{ also meaningless} \\ x + 2 + x - 6 < 10 \Rightarrow x < 7 \\ \therefore -3 < x < 7$

It's just like how you graph absolute value graphs in 4 unit.

kwabon · Oct 27, 2009

jetblack2007 said:
$\text{You take positive and negative values under each absolute value sign, } \\ \text{giving 4 separate equations:} \\ -x - 2 - x + 6 < 10 \Rightarrow x > -3 \\ -x - 2 + x - 6 < 10 \Rightarrow - 8 < 10 \text{ this is meaningless} \\ x + 2 - x + 6 < 10 \Rightarrow 8 <10 \text{ also meaningless} \\ x + 2 + x - 6 < 10 \Rightarrow x < 7 \\ \therefore -3 < x < 7$

It's just like how you graph absolute value graphs in 4 unit.

oh fuck simple calculation mistake in working
gahhh :-(

untouchablecuz · Oct 27, 2009

alternatively solve with y=10

$y=|x+2|+|x-6|\begin{cases}y=2x-4 \text{ if } x\geq 6 \\y=8 \text{ if }-2\leq x<6 \\y=-2x+4 \text{ if } x<-2 \end{cases}$

then use the graph to determine the range of x values

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