hidey there! okay, so...

the problem is -x - y < 3

How do you graph something like this?

thank you very much in advance. >.<

Printable View

- Aug 6th 2008, 01:21 PMedenpieGraphing linear inequalities
hidey there! okay, so...

the problem is -x - y < 3

How do you graph something like this?

thank you very much in advance. >.< - Aug 6th 2008, 03:48 PMticbol
I assume you know how to graph y = mx +b. So we convert the given inequality to resemble y = mx +b.

-x -y < 3

-y < x +3

Multiply both sides by (-1),

y > -x -3

The sense of inequality changed from "<" to ">" because we multiplied both sides of the previous inequality by a negative value.

To graph the y > -x -3:

a) First ignore the inequakity sign....think of it as an equal sign.

So y = -x -3.

b) Now graph that.

The line divides the x,y plane into two regions. The origin (0,0) is above the line. The (0,0) is in the region above the line.

c) Then consider now the inequality sign.

y > -x -3

Test which region satisfy the inequality. Choosing the (0,0) is popular, because at (0,0), x = 0 and y = 0.

So, when x = 0, and y = 0,

0 > -0 -3

0 > -3

Yes, that is true.

That means that the region above the line satisfies the inequality. That means the region above the line is the solution of the inequality.

So, mark the region above the line....by many lines or by any color.

That is the graph of -x -y < 3.

Note that since the inequality sign is only "<", .... {not "<=" , meaning less or equal to} ...., then the line itself is not part of the solution of the inequality.

------------------------

When you know already how to graph inequalities, you don't have to convert the original inequality into its y = mx =b form. You just play with the original inequality right away.

-x -y < 3

Think of it as -x -y = 3.

If x=0, then y = -3

If y=0, then x = -3.

Graph that.

Then consider the inequality as explained above.

Here, you'd see that for (0,0),

-0 -0 < 3

0 < 3

True. So the region containing the (0,0) is the solution of the inequality. The same as above.