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Mathematics 24 Online
OpenStudy (anonymous):

graph the solution set on the axes below. yge-x y<2x-3

myininaya (myininaya):

Do you know how to graph the linear equations y=-x and y=2x-3?

myininaya (myininaya):

Recall slope-intercept form is y=mx+b Both of these are in slope-intercept form.

OpenStudy (anonymous):

yes

myininaya (myininaya):

Put both of those lines on graphing paper together.

myininaya (myininaya):

Let's go back and look at that inequality part. y>=-x y<2x-3

myininaya (myininaya):

Notice one of those inequalities have an equal sign. For that line, you are going to make it a solid line. For the one without the equal sign, you will make it a broken line.

myininaya (myininaya):

So that will take care of the equals part. We still need to look at the greater than or less than signs. Let me know when you have got to this point then I will talk about this part.

OpenStudy (anonymous):

I already made the lines

myininaya (myininaya):

is one of them a broken/dashed line? one of them should be broken and one of them should be solid.

OpenStudy (anonymous):

yes ...the one with the equal sign with the solid line and the other one with the dashed line

myininaya (myininaya):

Ok. y> means you should shade above the line y< means you should shade below the line I would use different colors of shading for both inequalities. Then I would look to see where I have both colors of shading and that area will be the correct area of shading.

myininaya (myininaya):

for example |dw:1384636100143:dw|

OpenStudy (anonymous):

oh ok...so where both inequalities are shaded in the same region that would be the answer?

myininaya (myininaya):

that would be the correct area of shading

myininaya (myininaya):

you can look at the inequalities separately but you want the same region where they both have the shading

OpenStudy (anonymous):

oh ok thank you very much

myininaya (myininaya):

do you want to give a rough picture of what you have there is a draw tool

OpenStudy (anonymous):

|dw:1384636405672:dw|

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