Ask your own question, for FREE!
Mathematics 14 Online
OpenStudy (anonymous):

A & D???Which inequality will have a shaded area above its graph? 2x − y ≥ 4 4x + 3y < 6 x − 9y ≤ 1 x − 3y > 4

OpenStudy (texaschic101):

put in y = mx + b form A. 2x - y >= 4 -y >= -2x + 4 --- multiply by -1 to make y positive...and change inequality sign y <= 2x - 4 < less then means shading below the line....this is not the correct answer D. x - 3y > 4 -3y > -x + 4 -- divide by -3...change sign y < x - 4/3 < less then....not correct so it is not A or D try doing the same with the other answer choices and see what you get

OpenStudy (texaschic101):

just remember that when putting in y = mx + b form, when dividing/multiplying by a negative, it changes the inequality sign

OpenStudy (anonymous):

y>x/9-1/9

OpenStudy (anonymous):

@dan815

OpenStudy (anonymous):

I don't get why I had to put it into y=mx+b form

OpenStudy (anonymous):

@Hero

OpenStudy (loser66):

|dw:1430869493327:dw|

OpenStudy (loser66):

it means, if x =1, y =1 if x =2, y =2, ok?

OpenStudy (anonymous):

So I have to put it into y=mx+b?

OpenStudy (loser66):

|dw:1430869544638:dw|

OpenStudy (anonymous):

I get that

OpenStudy (loser66):

that means , y > x (GREATER THAN X)

OpenStudy (anonymous):

I also get that, But A & D are greater

OpenStudy (loser66):

hence, all you have to do is isolate y

OpenStudy (loser66):

and if they ask you ABOVE, that is your y must > the other

OpenStudy (anonymous):

Ohhhhhhhhhhhh.

OpenStudy (anonymous):

I didn't know it had to be above to y=x line, and y>____

OpenStudy (loser66):

for A) \(2x-4\geq y\)

OpenStudy (loser66):

your y is LESSER, it is not GREATER

OpenStudy (anonymous):

I get how to isolate

OpenStudy (anonymous):

I understand now. Thanks!

OpenStudy (loser66):

|dw:1430869746345:dw|

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!