solving systems of linear Inequalities
- Thread starter marko
- Start date
I don't know what your book means by "which quadrants is this contained in?"
You want +y by itself on the left side.
2x-y<0
-y < -2x +0
Multiply by -1 and reverse the inequaliye when you multiply by a negative.
y>2x
Graph y=2x
The "true" area is above the line. You can check that by putting it into the first equation. Say (x,y)=(0,5)
-5<0
True
It is true in 1/2 of I and III and all of II
x+y > 3
Subtract x
y>3-x
Graph y=3-x
Shade it
Check it
You want +y by itself on the left side.
2x-y<0
-y < -2x +0
Multiply by -1 and reverse the inequaliye when you multiply by a negative.
y>2x
Graph y=2x
The "true" area is above the line. You can check that by putting it into the first equation. Say (x,y)=(0,5)
-5<0
True
It is true in 1/2 of I and III and all of II
x+y > 3
Subtract x
y>3-x
Graph y=3-x
Shade it
Check it
