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Mathematics 15 Online
OpenStudy (dancergirl45):

The graph below represents which system of inequalities? graph of two infinite lines that intersect at a point. One line is solid and goes through the points negative 3, 0, negative 4, negative 1 and is shaded in below the line. The other line is solid, and goes through the points 1, 1, 2, negative 1 and is shaded in below the line. A y ≤ −2x + 3 y ≤ x + 3 B y ≥ −2x + 3 y ≥ x + 3 C y ≤ −3x + 2 y ≤ −x + 2 D y > −2x + 3 y > x + 3

OpenStudy (dancergirl45):

OpenStudy (dancergirl45):

@xMissAlyCatx

OpenStudy (dancergirl45):

@AloneS

OpenStudy (dancergirl45):

@welshfella

OpenStudy (xmissalycatx):

Hello there! What do you think it is? :)

OpenStudy (dancergirl45):

Well I know it is not D

OpenStudy (dancergirl45):

So A or c

OpenStudy (xmissalycatx):

That's true. So D is eliminated. Now if you look at your lines they are both solid, not dotted so that means that which ever equations you choose would NOT be > or < but \[\le or \ge \]

OpenStudy (xmissalycatx):

Now what you need to do is look at your slope of the line that was pointing downwards.. (the one with the green below it). If you use the rise over run method .. what could you determine the slope is of it?

OpenStudy (dancergirl45):

3/3=1

OpenStudy (dancergirl45):

@xMissAlyCatx

OpenStudy (xmissalycatx):

That is incorrect. The slope of that line is -2 over 1 (or -2)

OpenStudy (dancergirl45):

ok so that leaves a or b

OpenStudy (dancergirl45):

@xMissAlyCatx

OpenStudy (dancergirl45):

@AloneS PLEASE HELP

OpenStudy (xmissalycatx):

Okay so now, if you look at both of your lines, are the colors shaded UNDER or OVER both of the lines?

OpenStudy (xmissalycatx):

If they are over, it would be greater than or equal to.. If they are under, it is less than or equal to..

OpenStudy (dancergirl45):

under

OpenStudy (dancergirl45):

so A?????

OpenStudy (xmissalycatx):

Yes! :D

OpenStudy (dancergirl45):

Thank you so much can u check my ansewr on another one please

OpenStudy (xmissalycatx):

Yes. Tag me in it.

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