Find the graph of the absolute value: [(x-3y)/(x+y)] <= 1

[happybeans]

. . . . .\(\displaystyle \left|\dfrac{x\, -\, 3y}{x\, +\, y}\right|\, \leq\, 1\)

I helped myself with the definition of the absolute value, but now I don't know how draw the graph.

 

[JeffM]

. . . . .\(\displaystyle \left|\dfrac{x\, -\, 3y}{x\, +\, y}\right|\, \leq\, 1\)

I helped myself with the definition of the absolute value, but now I don't know how draw the graph.

This question as written makes no sense. Find the absolute value? Your expression is an absolute value involving variables: it can take on any non-negative real value. Your inequation limits those values to the interval [0, 1]. But there is no single value implied.

Why don't we start by your giving the complete question exactly as it is given in your text.

Then you can tell us what you have done or tried or thought about.

 

[happybeans]

This question as written makes no sense. Find the absolute value? Your expression is an absolute value involving variables: it can take on any non-negative real value. Your inequation limits those values to the interval [0, 1]. But there is no single value implied.

Why don't we start by your giving the complete question exactly as it is given in your text.

Then you can tell us what you have done or tried or thought about.

I have to draw this into a graph. I did this
|x-3y| = x-3y; x-3y>0
-x+3y; x-3y <0


|x+y|=x+y; x+y>0
-x-y; x+y <0
Now, I do not know how to combine this into a graph.

 

[Dr.Peterson]

I have to draw this into a graph. I did this
|x-3y| = x-3y; x-3y>0
-x+3y; x-3y <0


|x+y|=x+y; x+y>0
-x-y; x+y <0
Now, I do not know how to combine this into a graph.

You have found that the lines x - 3y = 0 and x + y = 0 divide the plane into four regions, within each of which the signs of the numerator and denominator are always the same. For example, above both lines, x-3y<0 and x+y>0, so in that region, |(x-3y)/(x+y)| = -(x-3y)/(x+y). Graph the resulting inequality within that region, then repeat for each of the other regions.