ABS(2x + y) - ABS(x + 3y) = 8

ABS = absolute value

I need to know where to start

Printable View

- May 4th 2009, 02:35 PMmathprobAbsolute value graph
ABS(2x + y) - ABS(x + 3y) = 8

ABS = absolute value

I need to know where to start - May 4th 2009, 03:30 PMVonNemo19
$\displaystyle \mid2x+y\mid-\midx+3y\mid=0$

This is what we wish to solve. $\displaystyle \mid2x+y\mid-\midx+3y\mid=0$

We first must recognize that $\displaystyle \mid2x+y\mid=2x+y$$\displaystyle -(2x+y)$ . So, knowing this, can you finish up?*or* - May 4th 2009, 03:43 PMmathprob
- May 4th 2009, 04:31 PMVonNemo19
When dealing with absolute value, you have to be aware of the folling fact-that $\displaystyle \mid{x}\mid=\mid{-x}\mid=x$ So, therefore $\displaystyle x$ can have 2 different values, but the corresponding value of $\displaystyle y$, or $\displaystyle f(x)$ will have only one. So knowing that....

We've got to set up different possibilities for the given expressions.

$\displaystyle \mid2x+y\mid=\pm(2x+y)$ and $\displaystyle \mid{x+3y}\mid=\pm(x+3y)$.

So, with this, can you take it from here? - May 4th 2009, 04:39 PMVonNemo19
$\displaystyle \sqrt{a^2+b^2}$

- May 4th 2009, 04:45 PMmathprob
i dont really know what you want me to do? what do i do after i get +- ?

what does the square root formula have to do with this - May 4th 2009, 04:51 PMVonNemo19
Solve for y for the different possibilities and then plot your graph.

Pythagoras has nothing to do with this. I was showing my brother how to type in math code.(Wink) - May 4th 2009, 04:53 PMmathprob
- May 4th 2009, 05:11 PMVonNemo19
It stays right were it is. you leave everything the same but the two expressions in the inequality signs

- May 4th 2009, 05:16 PMmathprob