# solve: lx-3l - lx-1l << 2

• March 11th 2009, 09:19 PM
differentiate
solve: lx-3l - lx-1l << 2
solve: lx-3l - lx-1l << 2

I got x<<1 but i think its wrong.

THank u
• March 12th 2009, 01:43 AM
earboth
Quote:

Originally Posted by differentiate
solve: lx-3l - lx-1l << 2

I got x<<1 but i think its wrong. .......True!

THank u

1. Use the definition of the absolute value:

$|x-3| = \left\{\begin{array}{lcr}x-3&if&x \geq 3\\-(x-3)&if&x < 3\end{array}\right.$

$|x-1| = \left\{\begin{array}{lcr}x-1&if&x \geq 1\\-(x-1)&if&x < 1\end{array}\right.$

2. Thus you get three different intervals. Now re-write the equation without absolute values:

x < 1~:~\begin{aligned}-(x-3)-(-(x-1)) \leq 2\\-x+3+x-1\leq2\\2\leq2\end{aligned}

1\leq x<3~:~\begin{aligned}-(x-3)-(x-1) \leq 2\\ -2x\leq 2\\x \geq 1\end{aligned}

x \geq 3~:~\begin{aligned}(x-3)-(x-1) \leq 2\\-3+1\leq2\\-2\leq2\end{aligned}

Therefore the solution is: $x \in \mathbb{R}$
• March 12th 2009, 02:10 AM
differentiate
Quote:

Originally Posted by earboth
1. Use the definition of the absolute value:

$|x-3| = \left\{\begin{array}{lcr}x-3&if&x \geq 3\\-(x-3)&if&x < 3\end{array}\right.$

$|x-1| = \left\{\begin{array}{lcr}x-1&if&x \geq 1\\-(x-1)&if&x < 1\end{array}\right.$

2. Thus you get three different intervals. Now re-write the equation without absolute values:

x < 1~:~\begin{aligned}-(x-3)-(-(x-1)) \leq 2\\-x+3+x-1\leq2\\2\leq2\end{aligned}

1\leq x<3~:~\begin{aligned}-(x-3)-(x-1) \leq 2\\ -2x\leq 2\\x \geq 1\end{aligned}

x \geq 3~:~\begin{aligned}(x-3)-(x-1) \leq 2\\-3+1\leq2\\-2\leq2\end{aligned}

Therefore the solution is: $x \in \mathbb{R}$

oh no, that was the solution I wrote on my exam paper but then I changed it into x<<1!!!! OMG!!!!!!!!! I still had all real x values in pencil, but I don't think they will give me the mark :( ARGH!! THank you. I can't believe I got the answer and then rubbed it out. I am so stupid!