boogie here with an inequality

[boogie]

cant post graph, so,,

2x-3y<=6

2x-3*0=6
2x=6x
x=3 (3,0)

2*0-3y=6
-3y=6
y=-2 (0,-2)

2*1-3y=6
2-3y=6
2-3y-2=6-2
-3y=4
-3y/-3=4/-3
y=-4/3 (1,-4/3)
this gives me my boundary line
and i just need to put in a couple test points in my equation
(0,0) or(5,-10) and this will tell me the side of the graph that
ineed to shade, am i right?

also, if i have 2x-3y,=12
x+5y<=20
x=0 meaning two boundary lines and two shaded areas,
do i just attack them the same way, just one at a time, more test points and the same conclusion?

thank you DAN

 

[Deleted member 4993]

cant post graph, so,,

2x-3y<=6

2x-3*0=6
2x=6x
x=3 (3,0)

2*0-3y=6
-3y=6
y=-2 (0,-2)

2*1-3y=6
2-3y=6
2-3y-2=6-2
-3y=4
-3y/-3=4/-3
y=-4/3 (1,-4/3)
this gives me my boundary line
and i just need to put in a couple test points in my equation
(0,0) or(5,-10) and this will tell me the side of the graph that
ineed to shade, am i right?

also, if i have 2x-3y,=12
x+5y<=20
x=0 meaning two boundary lines and two shaded areas,
do i just attack them the same way, just one at a time, more test points and the same conclusion?

thank you DAN

2x-3y<=6

3y >= 2x - 6

y >= (2/3) * x - 2

Plot y = (2/3)*x -2 on graph paper.

All the points above the line, including the points on the line, satisfies the given condition.

 

[Mrspi]

Once you have the graph of the "boundary line," and are trying to determine which side of that line to shade, just pick a point on one side of the line (the origin is always an easy point to test as long as the boundary line doesn't contain the origin) and test to see if that point satisfies the inequality. If it does, then shade the side which contains your test point. If it doesn't, then shade the OTHER side.

 

[boogie]

thanks

ill do it. more equations tomorrow after class.

2x-3y<=6

3y >= 2x - 6

y >= (2/3) * x - 2

Plot y = (2/3)*x -2 on graph paper.

All the points above the line, including the points on the line, satisfies the given condition.

 

[HallsofIvy]

cant post graph, so,,

2x-3y<=6

2x-3*0=6
2x=6x
x=3 (3,0)

2*0-3y=6
-3y=6
y=-2 (0,-2)

Yes, that's good. And since "two points determine a line", you don't really need to plot a third point. 2x- 3y= 6 is the line through (3, 0) and (0, -2).

2*1-3y=6
2-3y=6
2-3y-2=6-2
-3y=4
-3y/-3=4/-3
y=-4/3 (1,-4/3)
this gives me my boundary line
and i just need to put in a couple test points in my equation
(0,0) or(5,-10) and this will tell me the side of the graph that
ineed to shade, am i right?

Yes, although because this is a linear inequality you don't really need two points. Since 2(0)- 3(0)= 0< 6, you shade the side of the line that contains (0, 0).

also, if i have 2x-3y,=12

I take it the ",=" was a "fingerfehler" and you mean "<=".

x+5y<=20
x=0 meaning two boundary lines and two shaded areas,

What does "x= 0" have to do with this? Each inequality gives a shaded area but the x and y that satisfy both is only one area.

do i just attack them the same way, just one at a time, more test points and the same conclusion?

Pretty much. If x= 0, -3y= 12 so (0, -4) is on 2x- 3y= 12. If y= 0 2x= 12 so (12,0) is on 2x- 3y= 12. The graph of 2x- 3y= 12 is the line through both points. 2(0)- 3(0)= 0< 12 so you should shade the area below that line, blue, say. If x= 0, 5y= 20 so (0, 4) is on the second line. If y= 0, x= 20 so (20, 0) is on the second line. Draw the line through those two points. Again 0+ 5(0)= 0< 20 so shade below the line, green, say.

(x, y) that satisfies both inequalities is the intersection of those two areas- the are shaded both green and blue, so cyan.

thank you DAN